# Equilibrium

## Chemistry Class 11

### NCERT

1   A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.$\\$ (a) What is the initial effect of the change on vapour pressure?$\\$ (b) How do rates of evaporation and condensation change initially?$\\$ (c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

##### Solution :

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.$\\$ (b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.$\\$ (c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

2   What is $K c$ for the following equilibrium when the equilibrium concentration of each substance is:$[SO_2]=0.60 M,[O_2]=0.82M$ and $[SO_3]=1.90 M?$$\\ 2SO_2(g)+O_2(g)\leftrightarrow 2SO_3(g)$$\\$

The relation between $K _p$ and $K _c$ is given as:$\\$ $K_ p = K _c ( RT ) \Delta n$$\\ (a) Here, \Delta n = 3 - 2 = 1$$\\$ $R = 0.0831$ barLmol$^{-1} K ^{-1}$$\\ T = 500 K$$\\$ $K_ p = 1.8 × 10 -2$$\\ Now,\\ K _p = K _c ( RT ) ^{\Delta n}$$\\$ $\Rightarrow 1.8*10^{-2}=K_c(0.0831*500)^{1}$$\\ \Rightarrow K_c=\dfrac{1.8*10^{-2}}{0.0831 *500}$$\\$ $=4.33*10^{-4}$(approximately)$\\$ (b)Here,$\\$ $\Delta n=2-1=1$$\\ R=0.0831 barLmol^{-1}K^{-1}T$$\\$ $=1073 K$$\\ K_p=167$$\\$ Now,$\\$ $K_p=K_c(RT)^{\Delta n}\\ \Rightarrow 167=K_c(0.0831*1073)^{\Delta n}\\ \Rightarrow K_c=\dfrac{167}{0.0831*1073}$$\\ =1.87(approximately) 6 For the following equilibrium, K_ c = 6.3 * 10 ^{14} at 1000K \\ NO ( g ) + O _3 ( g ) \leftrightarrow NO_ 2 ( g ) + O _2 ( g ) \\ Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K _c , for the reverse reaction? ##### Solution : It is given that K _c for the forward reaction is 6.3 *10 ^{14} \\ Then, K_c for the reverse reaction will be, K '_ C =\dfrac{1}{K_C} \\ =\dfrac{1}{6.3*10^{14}} \\ =1.59*10^{-15} 7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? ##### Solution : For a pure substance (both solids and liquids),\\ [Pure substance]=\dfrac{\text{Number of moles}}{\text{Volume}} \\ =\dfrac{\text{Mass/molecular mass}}{\text{Volume}} \\ =\dfrac{\text{Mass}}{\text{Volume * Molecular mass}} \\ =\dfrac{\text{Density}}{\text{Molecular mass }} \\ Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression. 8 Reaction between N _2 and O _2 takes place as follows:\\ 2 N _2 ( g ) + O_ 2 ( g ) \leftrightarrow 2 N _2 O ( g ) \\ If a mixture of 0.482 mol of N _2 and 0.933 mol of O _2 is placed in a 10 L reaction vessel and allowed to form N _2 O at a temperature for which K _c = 2.0 * 10 ^{-37} , determine the composition of equilibrium mixture. ##### Solution : Let the concentration of N _2 O at equilibrium be x. The given reaction is:\\ 2 N _2 ( g ) + O_ 2 ( g ) \leftrightarrow 2 N _2 O ( g ) \\ Initial conc. \quad 0.482 mol \quad 0.933 mol \qquad 0$$\\$ At equilibrium $(0.482-x)$ mol $\quad (1.933-x)$mol $\quad x$ mol $\\$ Therefore ,at equlibrium,in the $10L$ vessel: $\\$ $[N_2]=\dfrac{0.482-x}{10},[O_2]=\dfrac{0.933-x/2}{10},[N_2O]=\dfrac{x}{10}$ $\\$ The value of equilibrium constant i.e. $K _c = 2.0 *10^{-37}$ is very small. Therefore, the amount of $N _2$ and $O_ 2$ reacted is also very small. Thus, $x$ can be neglected from the expressions of molar concentrations of $N _2$ and $O _2$ . Then,$\\$ $[N_2]=\dfrac{0.482}{10}=0.0482$ mol L$6{-1}$ and $[O_2]=\dfrac{0.933}{10}=0.0933$ mol L$^{-1}$ $\\$ Now $\\$ $K_C=\dfrac{[N_2O(g)]^2}{[N_2(g)][O_2(g)]}$ $\\$ $\Rightarrow 2.0*10^{-37}=\dfrac{(\dfrac{x}{10})^2}{(0.0482)^2(0.0933)}$ $\\$ $\Rightarrow \dfrac{x^2}{100}=2.0*10^{-37}*(0.0482)^2(0.0933)$ $\\$ $\Rightarrow x^2=43.35*10^{-40}$ $\\$ $\Rightarrow x=6.6*10^{-20}$ $\\$ $[N_2O]=\dfrac{x}{10}=\dfrac{6.6*10^{-20}}{10}=6.6*10^{-21}$

9   Nitric oxide reacts with $Br_ 2$ and gives nitrosyl bromide as per reaction given below:$\\$ $2 NO ( g ) + Br_ 2 ( g ) \leftrightarrow 2 NOBr ( g )$ $\\$ When $0.087$ mol of $NO$ and $0.0437$ mol of $Br _2$ are mixed in a closed container at constant temperature, $0.0518$ mol of $NOBr$ is obtained at equilibrium. Calculate equilibrium amount of $NO$ and $Br_ 2 .$

##### Solution :

The given reaction is:$\\$ $2NO (g) +Br _2(g) \leftrightarrow 2NOBr (g)$ $\\$ $2$mol $\quad 1$mol$\quad 2$mol $\\$ Now, $2$ mol of $NOBr$ are formed from $2$ mol of $NO$. Therefore, $0.0518$ mol of $NOBr$are formed from $0.0518$ mol of $NO.$ $\\$ Again,$2$ mol of $NOBr$ are formed from $1$ mol of Br.$\\$ Therefore, $0.0518$ mol of $NOBr$ are formed from $\dfrac{0.0518}{2}$ mol of $Br$, or $0.0259$ mol of $NO.$ $\\$ The amount of $NO$ and $Br$ present initially is as follows:$\\$ $[NO] = 0.087$ mol $[Br_ 2 ] = 0.0437$ mol$\\$ Therefore, the amount of $NO$ present at equilibrium is:$\\$ $[NO] = 0.087 - 0.0518 = 0.0352$ mol$\\$ And, the amount of $Br$ present at equilibrium is:$\\$ $[Br_ 2 ] = 0.0437 - 0.0259 = 0.0178$ mol $\\$

10   At $450 K, K _p = 2.0 *10 ^{10} /$bar for the given reaction at equilibrium.$\\$ $2 SO _2( g ) + O_ 2( g ) \leftrightarrow 2 SO _3( g )$ $\\$ What is $K_ c$ at this temperature?

##### Solution :

For the given reaction,$\\$ $\Delta n = 2 - 3 = - 1$ $\\$ $T = 450 K$ $\\$ $R = 0.0831$ bar $L$ bar$K ^{-1 }$mol $^{-1}$ $\\$ $K _p =2.0 * 10 ^{10}$ bar$^{-1}$ $\\$ We know that,$\\$ $K_ P + K_ C ( RT ) \Delta n$ $\\$ $\Rightarrow 2.0 * 10^{ 10}$ bar $^{-1 }=K _C (0.0831 L$ bar $K ^{-1 }$mol $^{- 1 }* 450 K )^{- 1}$ $\\$ $\Rightarrow K_C=\dfrac{2.0*10^{10} \ bar ^{-1}}{(0.0831 L \ bar ^{-1} \ mol ^{-1}* 450 K)^{-1}}$ $\\$ $=(2.0*10^{10} \ bar^{-1})(0.0831 L \ bar \ K^{-1} \ mol ^ {-1} * 450 K)$ $\\$ $=74.79*10^{14}L$ mol$^{1}$

11   A sample of $HI (g)$ is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of $HI (g)$ is $0.04$atm. What is K p for the given equilibrium?$\\$ $2 HI ( g ) \leftrightarrow H _2 ( g ) + I _2 ( g )$

##### Solution :

The initial concentration of HI is$0.2$ atm. At equilibrium, it has a partial pressure of $0.04$ atm.$\\$ Therefore, a decrease in the pressure of HI is $0.2 – 0.04 = 0.16.$ The given reaction is:$\\$ $2HI(g)\leftrightarrow H_2(g)+I_2(g)$ $\\$ Initial conc. $\quad$ 0.2 atm$\quad$0 $\quad$0$\\$ At equilibrium$\quad$ 0.04atm$\quad$0.16/2 $\quad$ 2.15/2$\\$ = 0.08atm= 0.08atm$\\$ Therefore,$\\$ Hence, the value of $K _p$ for the given equilibrium is 4.0.

##### Solution :

The balanced chemical equation corresponding to the given expression can be written as:$\\$ $4 NO (g) + 6H_ 2 O (g) \leftrightarrow 4NH _3(g)+ 5O _2(g)$

##### Solution :

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,$\\$ $C _2 H _6( g )\leftrightarrow C _2 H_ 4( g )+H_ 2( g )$$\\ Initial conc. \quad 4.0 atm \quad 0 \quad 0$$\\$ At equilibrium $\quad 4.0 - p \quad p \quad p$ $\\$ We can write,$\\$ $\dfrac{P_{C_2H_4}*P_{H_2}}{P_{C_2H_6}}=K_p\\ = \dfrac{p*p}{40-p}=0.04\\ =p^2=0.16-0.04p\\ =p^2+0.04p-0.16=0$ $\\$ Now, $p=\dfrac{-0.04\pm \sqrt{( 0.04 ) ^2 - 4 * 1 *( -0.16)}}{2 * 1}\\ =\dfrac{-0.04\pm 0.08}{2}\\ =\dfrac{0.76}{2}$ (Taking positive value)$\\$ $=0.38$ $\\$ Hence, at equilibrium,$\\$ $[ C _2 H_ 6 ] - 4 - p = 4 - 0.38\\ = 3.62 atm$

18   Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:$\\$ $CH _3 COOH ( l ) + C _2 H_ 5 OH ( l ) \leftrightarrow CH _3 COOC _2 H _5 ( l ) + H _2 O ( l )$$\\ (i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note: water is not in excess and is not a solvent in this reaction)\\ (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.\\ (iii)Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?\\ ##### Solution : (i) Reaction quotient, Q _C =\dfrac{[CH _3 COOC_ 2 H _5][H _2 O]}{[CH _3 COOH] [C _2 H _5 OH]}$$\\$ (ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess. The given reaction is:$\\$ $CH 3 COOH ( l ) +C_ 2 H _5 OH ( l ) \leftrightarrow CH _3 COOC _2 H _5( l )+ H _2 O ( l )$ $\\$ Initial conc $\dfrac{1}{V}M \quad \dfrac{0.18}{V}M\quad 0 \quad 0$ $\\$ At equilibrium $\dfrac{1 - 0.17}{V}M \quad \dfrac{0.18 - 0.17}{V}M \quad \dfrac{ 0.17}{V}M \quad \dfrac{0.17}{V}M $$\\ \qquad \qquad \qquad =\dfrac{0.829}{V}M =\dfrac{0.009}{V}M \\ Therefore, equilibrium constant for the given reaction is: \\ K_C=\dfrac{[CH _3 COOC _2 H _5][H _2 O]}{[CH_ 3 COOH][C_ 2 H_ 5 OH]}\\ =\dfrac{\frac{0.171}{V}*\frac{0.171}{V}}{\frac{0.829}{V}*\frac{0.829}{V}}=3.919\\ =3.92(approximately)\\ (iii)Let the volume of the reaction mixture be V.\\ CH 3 COOH ( l ) +C_ 2 H _5 OH ( l ) \leftrightarrow CH _3 COOC _2 H _5( l )+ H _2 O ( l ) \\ Initial conc \dfrac{1.0}{V}M \quad \dfrac{0.5}{V}M\quad 0 \quad 0 \\ At equilibrium \dfrac{10 - 0.214}{V}M \quad \dfrac{0.5 - 0.214}{V}M \quad \dfrac{ 0.214}{V}M \quad \dfrac{0.214}{V}M$$\\$ $\qquad \qquad \qquad =\dfrac{0.786}{V}M =\dfrac{0.286}{V}M$ $\\$ Therefore, equilibrium constant for the given reaction is: $\\$ $K_C=\dfrac{[CH _3 COOC _2 H _5][H _2 O]}{[CH_ 3 COOH][C_ 2 H_ 5 OH]}\\ =\dfrac{\frac{0.214}{V}*\frac{0.214}{V}}{\frac{0.786}{V}*\frac{0.786}{V}}=3.919\\ =0.2037$ =0.2037(approximately)$\\$ Since $Q_ C < K _C ,$ equilibrium has not been reached.

19   A sample of pure $PCl _5$ was introduced into an evacuated vessel at $473 K$. After equilibrium was attained, concentration of $PCl _5$ was found to be $0.5 * 10 ^{–1} mol \ L ^{–1}$ . If value of $K _c$ is $8.3 * 10 ^{–3}$ , what are the concentrations of $PCl_ 3$ and $Cl_ 2$ at equilibrium?$\\$ $PCl_ 5(g) \leftrightarrow PCl _3(g) + Cl_ 2 (g)$

##### Solution :

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:$\\$ $2BrCl(g) \leftrightarrow Br _2 (g) + Cl _2 (g)$$\\ Initial conc. \qquad 3.3 * 10 ^{- 3} \qquad 0 \qquad 0 \\ At equilibrium \qquad 3.3 * 10 ^{- 3}-2x \qquad x \qquad x \\ Now, we can write,\\ \dfrac{[Br_2][Cl_2]}{[BrCl]^2}=K_C\\ =\dfrac{x*x}{(3.3*10^{-3}-2x)^2}=32\\ =\dfrac{x}{3.3*10^{-3}-2x}= 5.66\\ = x = 18.678 * 10 ^{- 3 }- 11.32 x\\ = 12.32 x = 18.678 * 10 ^{- 3}\\ = x = 1.5 * 10 ^{- 3}$$\\$ Therefore, at equilibrium$\\$ $[ BrCl ] = 3.3 * 10 ^{- 3 }- (2* 1.5 * 10 ^{-3} )\\ = 3.3 * 10 ^{- 3 }- 3.0 * 10 ^{- 3}\\ = 0.3 * 10 ^{- 3}\\ = 3.0 * 10 ^{- 4} mol\ L ^{- 1}$

##### Solution :

(a) For the given reaction,$\\$ $\Delta G^ o = \Delta G^ o$ ( Products) – $\Delta G^ o$(Reactants)$\\$ $\Delta G ^o = 52.0 – {87.0 + 0}$$\\ = – 35.0 kJ mol^{ –1}$$\\$ (b) We know that,$\\$ $\Delta G^ o = RT \log K _c\\ \Delta G^ o = 2.303 RT \log K_ c\\ \log K_c=\dfrac{-35.0*10^{-3}}{-2.303*8.314*298}\\ =6.134\\ \therefore K_c=antilog(6.134)\\ =1.36*10^6$ $\\$ Hence, the equilibrium constant for the given reaction $K _c$ is $1.36 * 10^ 6 .$

25   Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?$\\$ (a) $PCl _5 ( g ) \leftrightarrow PCl _3 ( g ) + Cl _2 ( g )$$\\ (b) CaO ( s ) + CO 2 ( g ) \leftrightarrow CaCO _3 ( s )$$\\$ (c)$3 Fe ( s ) + 4 H_ 2 O ( g ) \leftrightarrow Fe _3 O _4 ( s )+ 4 H_ 2 ( g )$$\\ ##### Solution : (a) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.\\ (b) The number of moles of reaction products will decrease.\\ (c) The number of moles of reaction products remains the same.\\ 26 Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.\\ (i) COCl _2 ( g ) \leftrightarrow CO ( g ) +Cl _2 ( g )$$\\$ (ii)$CH _4 ( g ) + 2 S _2 ( g ) \leftrightarrow CS _2 ( g ) + 2 H _2 S ( g )$$\\ (iii) CO _2 ( g ) \leftrightarrow C ( s ) + 2 CO ( g )$$\\$ (iv) $2 H_ 2 ( g ) +CO ( g ) \leftrightarrow CH_ 3 OH ( g )$$\\ (v) CaCO_ 3 ( s ) \leftrightarrow CaO ( s ) + CO _2 ( g )$$\\$ (vi) $4 NH _3 ( g ) + 5 O _2 ( g ) \leftrightarrow 4 NO ( g ) + 6 H _2 O ( g )$$\\ ##### Solution : The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure. The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.\\ The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products. 27 The equilibrium constant for the following reaction is 1.6 * 10 ^5\ \ at \ \ 1024 K.$$\\$ $H _2 ( g ) + Br _2 ( g ) \leftrightarrow 2 HBr ( g )$$\\ Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K. ##### Solution : Given,\\ K_ p for the reaction i.e., H _2( g ) + Br _2( g ) \leftrightarrow 2 HBr ( g ) is 1.6 * 10^ 5 . \\ Therefore, for the reaction 2 HBr ( g ) \leftrightarrow H _2( g ) + Br _2( g ) , the equilibrium constant will be,\\ K'_p=\dfrac{1}{K_p}\\ =\dfrac{1}{1.6*10^5}\\ =6.25*10^{-6} \\ Now, let p be the pressure of both H _2 and Br _2 at equilibrium.\\ \qquad \quad 2 HBr ( g ) \leftrightarrow H _2( g ) +Br _2( g )$$\\$ Initial conc.$\qquad 10 \qquad 0 \qquad 0$ $\\$ At equilibrium $\quad 10 - 2 p \qquad p \qquad p$ $\\$ Now, we can write,$\\$ $\dfrac{p_{HBr}*p_2}{p^2_{HBr}}=k'_p\\ \dfrac{p*p}{(10-2p)^2}=6.25 * 10 {- 6}\\ \dfrac{p}{10-2p}=2.5*10^{-3}\\ p=2.5*10^{-2}-(5.0*10^-)p\\ p+(5.0*10^{-3})p=2.5*10^{-2}\\ (1005*10^{-3})p=2.5*10^{-2}\\ p=2.49*10^{-2} \ bar =2.5*10^{-2}\ bar$(approximately)$\\$ Therefore, at equilibrium,$\\$ $[H _2 ] = [Br _2 ] = 2.49 * 10 ^{- 2 }bar$$\\ [HBr] =10 - 2 *(2.49 * 10 ^{- 2 }) bar$$\\$ $= 9.95\ bar = 10\ bar$(approximately)$\\$

28   The equilibrium constant for the following reaction is $1.6 * 10 ^5\ \ at \ \1024 K.$$\\ H _2 ( g ) + Br _2 ( g ) \leftrightarrow 2 HBr ( g )$$\\$ Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

##### Solution :

(a)$K_c=\dfrac{[PCl_3(g)][Cl_2(g)]}{[PCl_5(g)]}$$\\ (b) Value of K _c for the reverse reaction at the same temperature is:\\ K'_c=\dfrac{1}{K_c}\\ =\dfrac{1}{8.3*10^{-3}}=1.2048*10^2\\ =120-48 \\ (c) (i) K _c would remain the same because in this case, the temperature remains the same.\\ (ii) K _c is constant at constant temperature. Thus, in this case, K _c would not change.\\ (iii) In an endothermic reaction, the value of K _c increases with an increase in temperature.\\ Since the given reaction in an endothermic reaction, the value of K _c will increase if the temperature is increased.\\ 32 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H _2 . In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,\\ CO ( g ) + H_ 2 O ( g ) \leftrightarrow CO_ 2 ( g ) + H_ 2 ( g )$$\\$ If a reaction vessel at $400°C$ is charged with an equimolar mixture of $CO$ and steam such that $p_{ CO }= p_{ H _2 O} = 4.0$ bar, what will be the partial pressure of $H _2$ at equilibrium? $K_ p = 10.1\ \ at\ \ 400°C$

##### Solution :

Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:$\\$ $CO ( g )+H _2 O ( g ) \leftrightarrow CO _2( g ) +H _2( g )$$\\ Initial conc. \qquad 4.0 bar \quad 4.0 bar \quad 0 \quad 0 \\ At equilibrium \qquad 4.0 -p \quad 4.0 -p \quad p \quad p \\ It is given that K_ p = 10.1$$\\$ Now , $\\$ $\dfrac{p_{CO_2}*p_2}{p_{CO}*p_{H_2O}}=10.1\\ =\dfrac{p}{(4.0-p)}=3.178\\ =p=12.712-3.178p\\ =4.178p=12.712\\ =p=3.04$ $\\$ Hence, at equilibrium, the partial pressure of $H _2$ will be $3.04 bar.$

33   Predict which of the following reaction will have appreciable concentration of reactants and products:$\\$ a) $Cl _2 ( g ) \leftrightarrow 2 Cl ( g ); K _c = 5 * 10 ^{- 39}$$\\ b) Cl_ 2 ( g ) + 2 NO ( g ) \leftrightarrow 2 NOCl ( g ); K _c =3.7 * 10^ 8$$\\$ c)$Cl_ 2 ( g ) + 2 NO _2 ( g ) \leftrightarrow 2 NO_ 2 Cl ( g ); K_ c = 1.8$

##### Solution :

If the value of $K _c$ lies between $10 ^{–3}$ and $10 ^3$, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

34   The value of $K _c$ for the reaction $3 O_ 2 ( g ) \leftrightarrow 2 O_ 3 ( g )$is $2.0 * 10 ^{- 50}$ at $25°C$. If the equilibrium concentration of $O _2$ in air at $25°C$ is $1.6* 10^{ –2}$ , what is the concentration of $O _3 ?$

The given reaction is:$\\$ $3 O_ 2 ( g ) \leftrightarrow 2 O_ 3 ( g )$$\\ Then, K _C=\dfrac{[O_3(g)]^2}{[o_2(g)]^3} \\ It is given that K _C = 2.0 * 10 ^{- 50 }and [ O _2( g ) ] = 1.6 *10 ^{- 2} .\\ Then, we have,\\ 2.0 * 10^{-50} =\dfrac{[O_3(g)]^2}{[1.6*10^{-2}]^3}\\ =[ O _3( g ) ] ^2 = 2.0 * 10 ^{- 50 } *( 1.6 * 10 ^{- 2}) ^ 3\\ = [ O_ 3( g ) ] ^2 = 8.192 * 10 ^{- 50}\\ = [ O _3( g ) ] = 2.86 * 10 ^{- 28} M \\ Hence, the concentration of O_ 3 is 2.86 * 10 ^{- 28} M . 35 The reaction, CO ( g )+ 3 H_ 2 ( g ) \leftrightarrow CH _4 ( g ) +H_ 2 O ( g ) is at equilibrium at 1300 K in a 1L flask.\\ It also contain 0.30 mol of CO, 0.10 mol of H _2 and 0.02 mol of H _2 O and an unknown amount of CH _4 in the flask. Determine the concentration of CH _4 in the mixture. The equilibrium constant, K _c for the reaction at the given temperature is 3.90. ##### Solution : Let the concentration of methane at equilibrium be x.\\ CO ( g )+ 3 H_ 2 ( g ) \leftrightarrow CH _4 ( g ) +H_ 2 O ( g )$$\\$ At equilibrium $\quad \dfrac{0.3}{1}0.3M \quad \dfrac{0.1}{1}=0.1M \quad x \quad \dfrac{0.02}{1}=0.02M$ $\\$ It is given that $K_c=3.90$ $\\$ Therefore, $\\$ $\dfrac{[CH_4(g)][H_2O(g)]}{[CO(g)][H_2(g)]^3}=K_c\\ =\dfrac{x*0.02}{0.3*(0.1)^3}=3.90\\ =x=\dfrac{3.90*0.3*(0.1)^3}{0.02}\\ =\dfrac{0.00117}{0.02}\\ =0.0585M\\ =5.85*10^{-2}M$ $\\$ Hence, the concentration of $CH_ 4$ at equilibrium is $5.85 * 10 ^{–2} M.$

36   What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:$\\$ $HNO _2 , CN ^- , HClO _4 , F ^- , OH ^- , CO_ 3 ^{2-}$ and $S ^-$

##### Solution :

A conjugate acid-base pair is a pair that differs only by one proton. The conjugate acid-base for the given species is mentioned in the table below.$\\$ Species $\qquad$ Conjugate acid-base$\\$ $HNO _2 \qquad \qquad NO _3^- (base)$$\\ CN^- \qquad \qquad HCN (acid)$$\\$ $HClO _4 \qquad \qquad ClO ^-_ 4 (base)$$\\ F ^– \qquad \qquad HF (acid)$$\\$ $OH ^– \qquad \qquad H _2 O (acid) /O ^{2–} (base)$$\\ CO _3^{ 2 -} \qquad \qquad HCO _3 ^- (acid)$$\\$ $S_{ 2–} \qquad \qquad HS^ – (acid)$

37   Which of the followings are Lewis acids? $H _2 O, BF_ 3 , H ^+ ,$ and $NH ^+_ 4$

##### Solution :

Lewis acids are those acids which can accept a pair of electrons. For example, $BF _3 , H ^+$ , and $NH ^+_4$ are Lewis acids.

38   What will be the conjugate bases for the Brönsted acids:$HF, H _2 SO_ 4$ and $HCO _3 ?$

##### Solution :

The table below lists the conjugate bases for the given Bronsted acids.$\\$ Bronsted acid $\qquad \qquad$ Conjugate base$\\$ $HF \qquad \qquad \qquad \qquad F^-$ $\\$ $H_2SO_4 \qquad \qquad \qquad \qquad HSO_4^-$ $\\$ $HCO^-_3 \qquad \qquad \qquad \qquad CO_3^{2-}$ $\\$

39   Write the conjugate acids for the following Brönsted bases:$NH _2^ – , NH _3$ and $HCOO^ – .$

##### Solution :

The table below lists the conjugate acids for the given Bronsted bases.$\\$ Bronsted base $\qquad \qquad$Conjugate acid$\\$ $NH ^-_ 2\qquad \qquad \qquad \qquad NH _3$$\\ NH_3 \qquad \qquad \qquad \qquad NH^-_4 \\ HCOO ^– \qquad \qquad \qquad HCOOH$$\\$

40   The species:$H _2 O, HCO _3 ^- , HSO_ 4^-$ and $NH _3$ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

##### Solution :

The table below lists the conjugate acids and conjugate bases for the given species.$\\$ Species $\qquad$ Conjugate acid $\qquad$ Conjugate base$\\$ $H_2O \qquad \qquad H_3O^+ \qquad \qquad \qquad OH^-$ $\\$ $HCO^-_3 \qquad \qquad H_2CO_3 \qquad \qquad CO_3^{2-}$ $\\$ $HSO^-_4 \qquad \qquad H_2SO_4 \qquad \qquad SO_4^{2-}$ $\\$ $NH_3 \qquad \qquad NH_4^+ \qquad \qquad \qquad NH_2^-$

41   Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:$\\$ (a) $OH ^–$ $\\$ (b) $F ^–$ $\\$ (c) $H ^+$ $\\$ (d) $BCl _3 .$ $\\$

##### Solution :

(a) $OH^ –$ is a Lewis base since it can donate its lone pair of electrons.$\\$ (b) $F^ –$is a Lewis base since it can donate a pair of electrons.$\\$ (c) $H ^+$ is a Lewis acid since it can accept a pair of electrons.$\\$ (d) $BCl _3$is a Lewis acid since it can accept a pair of electrons.$\\$

42   The concentration of hydrogen ion in a sample of soft drink is $3.8 * 10 ^{–3} M$. what is its $pH?$

Given,$\\$ $[ H ^+ ] = 3.8 * 10 ^{- 3} M$$\\ \therefore pH Value of soft drink\\ =- \log[ H ^+ ]$$\\$ $=- \log ( 3.8 * 10 ^{- 3 })$$\\ =- \log 3.8 - \log10 ^{- 3}$$\\$ $=- \log 3.8 + 3$$\\ =-0.58 + 3$$\\$ $= 2.42$$\\ 43 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. ##### Solution : Given, pH\\ = 3.76\\ It is known that,\\ pH =- \log[H ^+ ]\\ = \log[H ^+ ] =- pH\\ = [H ^+ ] = antilog (-pH)\\ =antilog (-3.76)\\ =1.74 * 10 ^{- 4} M \\ Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 * 10 ^{- 4} M . 44 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 * 10 ^{–4} , 1.8* 10^{ –4} and 4.8 * 10^{ –9} respectively. Calculate the ionization constants of the corresponding conjugate base. ##### Solution : It is known that,\\ K_b=\dfrac{K_w}{K_a} \\ Given,\\ K _a of HF = 6.8 * 10^{ –4}$$\\$ Hence, $K_ b$ of its conjugate base $F^ –$$\\ =\dfrac{K_w}{K_a}\\ =\dfrac{10^{-14}}{6.8*10^{-4}}\\ =1.5*10^{-11} \\ Given,\\ K_ a of HCOOH = 1.8 * 10^{–4}$$\\$ Hence, $K_ b$of its conjugate base $HCOO^ –$$\\ \dfrac{K_w}{K_a}\\ \dfrac{10^{-14}}{1.8*10^{-4}}\\ =5.6*10^{-11} \\ Given,\\ K _a of HCN = 4.8 * 10 ^{–9}$$\\$ Hence,$K_ b$ of its conjugate base $CN ^–$$\\ =\dfrac{K_w}{K_a}\\ =\dfrac{10^{-14}}{4.8*10^{-9}}\\ =2.08*10^{-6} 45 The ionization constant of phenol is 1.0 * 10^{ –10} . What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate? ##### Solution : Ionization of phenol:\\ C _6 H _5 OH + H_2 O\leftrightarrow C_ 6 H_ 5 O ^-+H _3 O ^+$$\\$ Initialconc.$\qquad 0.05 \qquad \qquad \quad 0 \qquad 0$ $\\$ At equilibrium $\qquad 0.05-x \qquad \qquad \quad x \qquad x$ $\\$ $K_a=\dfrac{[C_6H_5O^-][H_3O^+]}{[C_6H_5OH]}\\ K_a=\dfrac{x*x}{0.05-x}$ $\\$ As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.$\\$ $\therefore x=\sqrt{1*10^{-10}*0.05}\\ =\sqrt{5*10^{-12}}\\ =2.2*10^{-6}M=[H_3O^+]$ $\\$ Since $[H_3O^+]=[C_6H_5O^-]\\ [C_6H_5O^-]=2.2*10^{-6} M$ $\\$ Now, let $\alpha$ be the degree of ionization of phenol in the presence of $0.01 M C _6 H _5 ONa.$$\\ C _6 H_ 5 ONa \to C _6 H_ 5 O ^-+ Na ^+$$\\$ Conc.0.01$\\$

Also, $\\$ $C _6 H _5 OH + H_2 O\leftrightarrow C_ 6 H_ 5 O ^-+H _3 O ^+$$\\ Conc. \quad 0.05-0.05 \alpha \quad 0.05 \alpha \quad 0.05 \alpha \\ [ C _6 H _5 OH ] = 0.05 - 0.05 \alpha ;0.05 M$$\\$ $[ C _6 H _5 O ^-] = 0.01 + 0.05 \alpha ;0.01 M$$\\ [ H_3 O ^+ ] = 0.05 \alpha$$\\$ $K_a=\dfrac{[C_6H_5O^-][H_3O^+]}{[C_6H_5OH]}$ $\\$ $K_a=\dfrac{(0.01)(0.05 \alpha )}{0.05}\\ 1.0*10^{-10}=.01\alpha \\ \alpha =1*10^{-8}$ $\\$

46   The first ionization constant of $H _2 S$ is $9.1 * 10^{ –8}$ . Calculate the concentration of $HS ^–$ion in its $0.1 M$solution. How will this concentration be affected if the solution is $0.1 M$ in $HCl$ also? If the second dissociation constant of $H _2 S$ is $1.2 * 10^{ –13}$ , calculate the concentrationof $S ^{2–}$ under both conditions.

##### Solution :

(i) To calculate the concentration of $HS^ –$ ion:$\\$ Case I (in the absence of HCl):$\\$ Let the concentration of $HS ^–$ be x M.$\\$ $H_ 2 S \leftrightarrow H ^++ HS ^-$$\\ C _l \qquad 0.1\qquad 0 \qquad 0 \\ C _f \qquad 0.1-x\qquad x \qquad x \\ Then, \\ K_{a_I}=\dfrac{[H^+][HS^-]}{[H_2S]}\\ 9.1*10^{-8}=\dfrac{(x)(x)}{0.1-x}\\ (9.1*10^{-8})(0.1-x)=x^2 \\ Taking 0.1 – xM ; 0.1 M, we have ( 9.1 * 10 ^{- 8 })( 0.1 )= x ^2 . \\ 9.1 * 10 ^{- 9 }= x^ 2\\ x=\sqrt{9.1*10^{-9}}\\ =9.54*10^{-5}M\\ =[HS^-]=9.54*10^{-5}m \\ Case II (in the presence of HCl):\\ In the presence of 0.1 M of HCl, let [ HS ^- ] be y M$$\\$ Then , $H_2S \leftrightarrow HS^-+H^+$ $\\$ $C _l \qquad 0.1\qquad 0 \qquad 0$ $\\$ $C _f \qquad 0.1-y\qquad y \qquad y$ $\\$ Now, $K_{a_I}=\dfrac{[HS^-][H^+]}{[H_2S]}\\ 9.1*10^{-8}=\dfrac{(y)(0.1+y)}{0.1-y}\\ (9.1*10^{-8})=\dfrac{y*0.1}{(0.1)}\ \ \ (\therefore 0.1-y;0.1M)\\ (and \ \ 0.1+y;0.1M)\\ (9.1*10^{-8})=x\\ =[HS^-]=9.1*10^{-8}$ $\\$ To calculate the concentration of$[ S 2 ^{2-} ]$$\\ Case I (in the absence of 0.1 M HCl):\\ HS ^- \leftrightarrow H ^++ S^{ 2 -}$$\\$ $[ HS ^- ] = 9.54 * 10 ^{- 5} M$(From first ionization, case I)$\\$ Let, $[ S^{ 2 -}] be X .$$\\ Also, [ H ^+ ] = 9.54 * 10 ^{- 5} M(From first ionization, case I)\\ K_{ a _2}=\dfrac{[H^+][S^{2-}]}{[HS^-]}\\ K_{a_2}=\dfrac{( 9.54 * 10 ^{-5} ) (X )}{9.54*10^{-5}}\\ 1.2*10^{13}=X=[S^{2-}] \\ Case II (in the presence of 0.1 M HCl):\\ Again, let the concentration of HS^ – be X' M.\\ [ HS ^- ] = 9.1 * 10 ^{- 8} M (From first ionization, case II)\\ [ H ^+] = 0.1 M (From HCl case II)\\ [ S^{ 2 -}] X '$$\\$ Then $,K_{a_2}=\dfrac{[H^+][S^{2-}]}{HS^-}\\ 1.2*10^{-13}=\dfrac{(0.1)(X')}{9.1*10^{-8}}\\ 10.92*10^{-21}=0.1X'\\ \dfrac{10.92*10^{-21}}{0.1}=X'\\ X'=\dfrac{1.092*10^{-20}}{0.1}\\ =1.092*10^{-10}M\\ =K_{a_1}=1.74*10^{-5}$ $\\$

47   The ionization constant of acetic acid is $1.74 * 10^{ –5}$ . Calculate the degree of dissociationof acetic acid in its $0.05 M$ solution. Calculate the concentration of acetate ion in the solution and its pH.

##### Solution :

Method 1$\\$ 1) $CH _3 COOH \leftrightarrow CH _3 COO ^-+ H ^- \quad K _a = 1.74 * 10 ^{- 15}$$\\ 2) H _2 O + H _2 O \leftrightarrow H _3 O ^++ OH ^- \quad K _w = 1.0 * 10 ^{- 14}$$\\$ Since K a > > K w ,:$\\$ $CH _3 COOH ^- H _2 O \leftrightarrow CH _3 COO ^-+H _3 O^+$ $\\$ $C_l=0.05 \qquad \qquad 0 \qquad 0 \\ 0.05-.05\alpha \qquad 0.05\alpha \qquad 0.05 \alpha \\ K_a=\dfrac{(.05\alpha)(.05 \alpha)}{(.05-0.05 \alpha)}\\ =\dfrac{(.05 \alpha)(0.05 \alpha)}{.05(1-\alpha)}\\ =\dfrac{.05 \alpha^2}{1-\alpha}\\ 1.74*10^{-5}=\dfrac{.05\alpha^2}{1-\alpha }\\ 1.74*10^{-5}-1.74*10^{-5}\alpha=0.05 \alpha ^2\\ 0.05 \alpha^2 +1.74*10^{-5}\alpha -1.74*10^{-5}\\ D=b^2-4ac\\ =(1.74*10^{-5})^2-4(.05)(1.74*10^{-5})\\ =3.02*10^{-25}+.348*10^{-5}\\ \alpha =\sqrt{\dfrac{K_a}{c}}\\ \alpha = \sqrt{\dfrac{1.74*10^{-5}}{.05}}\\ =\sqrt{\dfrac{34.8*10^{-5}*10}{10}}\\ =\sqrt{3.48*10^{-6}}\\ =CH_3COOH\leftrightarrow CH_3COO^-+H^+\\ \alpha \underline{1.86*10^{-3}}\\ [CH_3COO^-]=0.05*1.86*10^{-3}\\ =\dfrac{0.93*10^{-3}}{1000}\\ =\underline{.000093}$ $\\$

Method 2$\\$ Degree of dissociation,$\\$ $\alpha =\sqrt{\dfrac{K_a}{C}}\\ c=0.05 M\\ K_a=1.74*10^{-5}$ $\\$ Then, $\\$ $\alpha =\sqrt{\dfrac{1.74*10^{-5}}{.05}}\\ \alpha =\sqrt{3.48*10^{-5}}\\ \alpha= \sqrt{3.48}*10^{-4}\\ \alpha=1.8610^{-2}\\ CH_3COOH \leftrightarrow CH_3CO^-+H^+$ $\\$ Thus, concentration of $CH _3 COO ^– = c.\alpha$ $\\$ $=.05*1.86*10^{-2}\\ =.093*10^{-2}\\ =.00093M$ $\\$ Since $[ oAc ^-]=[ H ^+] ,\\ [ H ^+]=.00093=.093*10^{-2},\\ pH=-\log[H^+]\\ =-\log(.093*10^{-2}),\\ \therefore pH=3.03$ $\\$ Hence, the concentration of acetate ion in the solution is $0.00093 M$ and its pH is $3.03.$

48   It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its $pK _a$ .

Let the organic acid be HA.$\\$ $= HA \leftrightarrow H^++ A ^-$$\\ Concentration of HA = 0.01 M pH$$\\$ $= 4.15\\ - \log[ H ^+ ] = 4.15\\ [ H ^+ ] = 7.08 * 10 ^{- 5}$$\\ Now, K_a=\dfrac{[H^+][A^-]}{[HA]}\\ [H^+]=[A^-]=7.08*10^{-5}\\ [HA]=0.01 \\ Then,\\ K_a=\dfrac{(7.08*10^{-5})(7.08*10^{-5})}{0.01}\\ K_a=5.01*10^{-7}\\ pK_a=-\log \ K_a\\ =-\log \ \ K_a\\ =(5.01*10^{-7})\\ pK_a=6.3001 49 Assuming complete dissociation, calculate the pH of the following solutions:\\ (i) 0.003 M HCl\\ (ii) 0.005 M NaOH\\ (iii)0.002 M HBr\\ (iv) 0.002 M KOH ##### Solution : (i) 0.003MHCl:\\ H _2 O + HCl \leftrightarrow H _3 O ^++ Cl ^- Since HCl is completely ionized,\\ [H_3O^+]=[HCl]\\ =[H_3O^+]=0.003 \\ Now,\\ pH=-\log[H_2O^+]=-\log(.003)\\ =2.52 \\ Hence, the pH of the solution is 2.52.\\ (ii) 0.005MNaOH:\\ NaOH(aq)\leftrightarrow Na^+(aq)+HO^-(aq)\\ [HO^-]=[NaOH]\\ =[HO^-]=.005\\ pOH=-\log[HO^-]=-\log(.005)\\ pOH=2.30\\ =11.70 \\ Hence, the pH of the solution is 11.70.\\ (iii) 0.002 HBr:\\ HBr + H _2 O \leftrightarrow H_ 3 O ^++ Br ^-\\ [H _3 O ]= [ HBr ]\\ =[ H_ 3 O ^+]= .002\\ \therefore pH = \log [ H _3 O ^+]\\ =- \log ( 0.002 )\\ = 2.69$$\\$ Hence, the pH of the solution is 2.69.$\\$ (iv)0.002 M KOH: $\\$ $KOH(aq)\leftrightarrow K^+(aq)+OH^{-}(aq)\\ [OH^-]=[KOH]\\ =[OH^-]=.002$ $\\$ Now, $pOH=-\log[OH^-]\\ =2.69\\ \therefore pH=14-2.69\\ =11.31$ $\\$ Hence, the pH of the solution is $11.31.$

50   Calculate the pH of the following solutions:$\\$ a) 2 g of TlOH dissolved in water to give 2 litre of solution.$\\$ b) 0.3 g of Ca(OH) 2 dissolved in water to give 500 mL of solution.$\\$ c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.$\\$ d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.$\\$

##### Solution :

(a) For 2g of TlOH dissolved in water to give 2 L of solution:$\\$ $[TIOH(aq)]=\dfrac{2}{2}g/L\\ =\dfrac{2}{2}*\dfrac{1}{221}M\\ =\dfrac{1}{221}M\\ TIOH(aq)\to Tl^+(aq)+OH^-(aq)\\ [OH^-(aq)]=[TIOH(aq)]=\dfrac{1}{221}M\\ K_w=[H^+][OH^-]\\ 10^{-14}=[H^+](\dfrac{1}{221})\\ 221*10^{-14}=[H^+]\\ =pH=-\log[H^+]=-\log(221*10^{-14})\\ =-\log(2.21*10^{-12})\\ =11.65$ $\\$ (b) For 0.3 g of $Ca(OH) ^2$ dissolved in water to give 500 mL of solution:$\\$ $Ca(OH)_2\to Ca^2+2OH^-\\ [Ca(OH)_2]=0.3*\dfrac{1000}{500}=0.6M\\ [OH^-(aq)]=2*[Ca(OH)_2(aq)]=2*0.6=1.2M\\ [H^+]=\dfrac{K_w}{[OH^-(aq)]}\\ =\dfrac{10^{-14}}{1.2}M\\ =0.833*10^{-14}\\ pH=-\log(0.833*10^{-14})\\ =-\log(8.33*10^{-13})\\ =(-0.902+13)\\ =12.098$ $\\$

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:$\\$ $NaOH\to Na^+(aq)+OH^-{aq}\\ [NaOH]=0.3*\dfrac{1000}{200}=1.5M\\ [OH^-(aq)]=1.5M$ $\\$ Then,$[H^+]=\dfrac{10^{14}}{1.5}\\ =6.66*10^{-13}\\ pH=-\log(6.66*10^{-13})\\ 12.18$ $\\$ (d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:$\\$ $13.6*1 mL=M_2*1000mL$ $\\$ (Before dilution) (After dilution)$\\$ $13.6*10^{-3}=M_2*1L M_2\\ =1.36*10^{-2}[H^+]=1.36*10^{-2}pH=-\log(1.36*10^{-2})\\ =(-0.1335+2)=1.866 \approx 1.87$

51   The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the $pK_ a$ of bromoacetic acid.

##### Solution :

Degree of ionization, $\alpha = 0.132$$\\ Concentration, c = 0.1 M$$\\$ Thus, the concentration of $H_ 3 O + = c.\alpha $$\\ = 0.1 * 0.132\\ = 0.0132\\ pH=-\log[H^+]\\ =-\log(0.0132)\\ =1.879:1.88 \\ Now, \\ K _a = C \alpha ^ 2\\ = 0.1 *( 0.132 )^2\\ K _a = .0017\\ pK _a = 2.75$$\\$

52   The degree of ionization of a $0.1M$ bromoacetic acid solution is $0.132.$ Calculate the $pH$ ofthe solution and the $pK _a$ of bromoacetic acid.

Degree of ionization, $\alpha = 0.132$$\\ Concentration, c = 0.1 M$$\\$ Thus, the concentration of $H _3 O + = c.\alpha $$\\ = 0.1 * 0.132\\ = 0.0132\\ pH=-\log[H^+]\\ =-\log(0.0132)\\ =1.879:1.88 \\ Now,\\ K_a=C\alpha^2\\ =0.1*(0.132)^2\\ K_a=.0017\\ pK_a=2.75 53 What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline. ##### Solution : K _b = 4.27 * 10^{ –10}\\ c = 0.001M\\ pH=?\\ \alpha =?\\ k_ a = C \alpha^ 2\\ 4.27 * 10 ^{- 10 }= 0.001*\alpha^ 2\\ 4.270 * 10 ^{- 10 }= \alpha^ 2\\ 65.34*10^{-5}=\alpha =6.53*10^{-4} \\ Then , [anion]= c\alpha=0.01*65.34*10^{-5}\\ =0.65*10^{-5}\\ pOH=-\log (.065*10^{-5})\\ =6.187\\ pH=7.813 \\ Now,\\ K_a*K_b=K_w\\ \therefore 4.27*10^{-10}*K_a=K_w\\ K_a=\dfrac{10^{-14}}{4.27*10^{-10}}\\ =2.34*10^{-5} \\ Thus, the ionization constant of the conjugate acid of aniline is 2.34 * 10 ^{- 5} . 54 Calculate the degree of ionization of 0.05M acetic acid if its pK_ a value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl? ##### Solution : c= 0.05 M\\ pK _a = 4.74\\ pK _a =- \log( K a )\\ K _a = 1.82 * 10 ^{- 5}\\ K_a=c\alpha^2 \\ \alpha=\sqrt{\dfrac{K_a}{c}}\\ \alpha =\sqrt{\dfrac{1.82*10^{-5}}{5810^{-2}}}=1.908*10^{-2} \\ When HCl is added to the solution, the concentration of H ^+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.\\ Case I:When 0.01 M HCl is taken.\\ Let x be the amount of acetic acid dissociated after the addition of HCl.\\ CH_3COOH \leftrightarrow H^+ + CH_3COO^- \\ Initialconc. \qquad 0.05M \quad 0 \quad 0 \\ After dissociation \qquad 0.05 -x \quad 0.01+x \quad x \\ As the dissociation of a very small amount of acetic acid will take place, the values i.e.,\\ 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.\\ K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}\\ \therefore K_a=\dfrac{(0.01)x}{0.05}\\ x=\dfrac{1.82*10^{-5}*0.05}{0.01}\\ x=1.82 * 10 ^{- 3 }* 0.05 M$$\\$ Now, $\\$ $\alpha =\dfrac{\text{Amount of acid dissociated}}{\text{Amount of acid taken}}\\ =\dfrac{1.82*10^{-5}*0.05}{0.05}\\ =1.82*10^{-3}$ $\\$

Case II:When 0.1 M HCl is taken. Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:$\\$ $[ CH _3 COOH ] = 0.05 - X ;0.05 M\\ [ CH _3 COO ^- ] = X\\ [ H ^+ ] = 0.1 + X ;0.1 M\\ K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}\\ \therefore K_a=\dfrac{(0.1)x}{0.05}\\ x=\dfrac{1.82*10^{-5}*0.05}{0.1}\\ x=1.82*10^{-4}*0.05 M$ $\\$ Now,$\\$ $\alpha=\dfrac{\text{Amount of acid dissociated}}{\text{Amount of acid taken}}\\ =\dfrac{1.82*10^{-4}*0.05}{0.05}\\ =1.82*10^{-4}$

55   The ionization constant of dimethylamine is $5.4 * 10^{ –4}$ . Calculate its degree of ionization in its $0.02 M$ solution. What percentage of dimethylamine is ionized if the solution is also $0.1 M$ in $NaOH?$

##### Solution :

$K _b = 5.4 * 10 ^{- 4}\\ c = 0.02 M$ $\\$ Then, $\alpha =\sqrt{\dfrac{K_b}{c}}\\ =\sqrt{\dfrac{5.4*10^{-4}}{0.02}}\\ =0.1643$ $\\$ Now, if $0.1 M$ of $NaOH$ is added to the solution, then $NaOH$ (being a strong base) undergoes complete ionization.$\\$ $NaOH(aq) \leftrightarrow Na^+(aq)+OH^-(aq)\\ \qquad \qquad 0.1M \quad 0.1M$ $\\$ And,$(CH_3)_2NH+H_2O \leftrightarrow (CH_3)_2NH^+_2+OH\\ (0.02-x) \qquad \qquad x \qquad x \\ ;0.02M \qquad \qquad \qquad ;0.1M$ $\\$ Then, $[ ( CH _3 )_ 2 NH _2 ^+]= x\\ [ OH ^-]= x + 0.1;0.1\\ =K_b =\dfrac{[(CH_3)_2NH^+_2][OH^-]}{[(CH_3)_2NH]}\\ 5.4*10^{-4}=\dfrac{x*0.1}{0.02}\\ x=0.0054$ $\\$ It means that in the presence of $0.1 M NaOH, 0.54$% of dimethylamine will get dissociated.

56   Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:$\\$ (a) Human muscle-fluid, 6.83$\\$ (b) Human stomach fluid, 1.2$\\$ (c) Human blood, 7.38$\\$ (d) Human saliva, 6.4.

##### Solution :

(a) Human muscle fluid 6.83:$\\$ $pH = 6.83 pH = – \log [H ^+ ]\\ \therefore 6.83 = – \log [H ^+ ]\\ [H ^+ ] =1.48 * 10 ^{–7} M$ (b) Human stomach fluid, 1.2:$\\$ $pH =1.2\\ 1.2 = – \log [H^ + ]\\ \therefore [H ^+ ] = 0.063$ (c) Human blood, 7.38:$\\$ $pH = 7.38 = – \log [H ^+ ]\\ \therefore [H ^+ ] = 4.17 * 10 ^{–8} M$ (d) Human saliva, 6.4:$\\$ $pH = 6.4\\ 6.4 = – \log [H ^+ ]\\ [H ^+ ] = 3.98 * 10^{ –7}$

57   The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2,2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

##### Solution :

The hydrogen ion concentration in the given substances can be calculated by using the given relation:$pH = –\log [H ^+ ]$$\\ (i) pH of milk = 6.8$$\\$ Since, $pH = –\log [H ^+ ]\\ 6.8 = –\log [H^ + ] \log\\ [H^ + ] = –6.8\\ [H^ + ] = anitlog(–6.8)\\ = 1.5 * 19 ^{–7 }M$ $\\$

(ii) $pH$ of black coffee = $5.0$$\\ Since, pH = –\log [H^ + ]\\ 5.0 = –\log [H ^+ ] \log\\ [H ^+ ] = –5.0\\ [H ^+ ] = anitlog(–5.0)\\ = 10 –5 M$$\\$ (iii) $pH$ of tomato juice =$4.2$$\\ Since, pH = –\log [H^ + ]\\ 4.2 = –\log [H ^+ ] \log\\ [H ^+ ] = –4.2\\ [H^ + ] = anitlog(–4.2)\\ = 6.31 * 10 ^{–5} M \\ (iv) pH of lemon juice = 2.2$$\\$ Since, $pH = –\log [H ^+ ]\\ 2.2 = –\log [H^ + ] \log\\ [H ^+] = –2.2\\ [H^ + ] = anitlog(–2.2)\\ = 6.31 * 10^{ –3} M\\$ (v) $pH$ of egg white = $7.8$$\\ Since, pH = –\log [H^ + ]\\ 7.8 = –\log [H^ + ] \log\\ [H^ + ] = –7.8\\ [H ^+ ] = anitlog(–7.8)\\ =1.58 * 10 ^{–8} M \\ 58 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH? ##### Solution : [KOH_{aq}]=\dfrac{0.561}{\frac{1}{5}}g/L\\ =2.805 g/L\\ =2.805*\dfrac{1}{56.11}M\\ =0.5 M\\ KOH_{aq}\to K^+_{(aq)}+OH^-_{(aq)}\\ [OH^-]=.05M=[K^+]\\ [H^+][H^-]=K_w\\ [H^+]=\dfrac{K_w}{[OH^-]}\\ =\dfrac{10^{-14}}{0.05}=2*10^{-11}M\\ \therefore pH=12.70 \\ 59 The solubility of Sr(OH)_ 2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution. ##### Solution : Solubility of Sr(OH)_ 2 = 19.23 g/L$$\\$ Then, concentration of $Sr(OH) _2$$\\ \dfrac{19.23}{121.63}M\\ =0.1581 M\\ Sr ( OH ) _{2( aq )} \to Sr ^{2 +} ( aq ) + 2( OH ^- ) ( aq )\\ \therefore [Sr^{ 2+} ]= 0.1581 M\\ [ OH ^- ] = 2 * 0.1581 M = 0.3126 M$$\\$ Now,$\\$ $K _w =[ OH ^-][ H ^+]\\ \dfrac{10^{-14}}{0.3126}=[H^+]\\ =[ H ^+]= 3.2 * 10^{+ 14}\\ \therefore pH = 13.495 ;13.50$$\\ 60 The ionization constant of propanoic acid is 1.32 * 10^{ –5} . Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also? ##### Solution : Let the degree of ionization of propanoic acid be \alpha .\\ Then, representing propionic acid as HA, we have:\\ HA +H_2O \leftrightarrow H_3O^++A^-\\ (.05-0.0\alpha)\approx \quad .05 \alpha \quad .05 \alpha \\ K_a=\dfrac{[H_3O^+][A^-]}{[HA]}\\ =\dfrac{(.05 \alpha)(.05 \alpha)}{0.05}=.05 \alpha^2\\ \alpha=\sqrt{\dfrac{K_\alpha}{.05}}=1.63*10^{-2} \\ Then,[H_3O^+]=.05\alpha=.05\alpha81.63*10^{-2}=K_b.15*10^{-4}M\\ \therefore pH=3.09 \\ In the presence of 0.1M of HCl, let \alpha ' be the degree of ionization.\\ Then,[H_3O^+]=0.01\\ [A^-]=005\alpha '\\ [HA]=.05\\ K_a=\dfrac{0.01*0.5 \alpha '}{.05}\\ 1.32*10^{-5}=.01*\alpha '\\ \alpha '=1.32*10^{-5} \\ 61 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution. ##### Solution : c = 0.1 M\\ pH = 2.34\\ - \log [ H ^+] = pH\\ - \log [ H ^+]= 2.34\\ [ H^+]= 4.5 * 10 ^{- 3}$$\\$ Also,$\\$ $[ H ^+]= c \alpha \\ 4.5 * 10 ^{- 3 }= 0.1*\alpha \\ \dfrac{4.5*10^{-3}}{0.1}=\alpha\\ \alpha =4.5*10^{-3}=.045$ $\\$ Then, $\\$ $K_a=c \alpha ^2\\ =0.1*(45*10^{-3})^2\\ =202.5*10^{-6}\\ =2.02*10^{-4}$ $\\$

62   The ionization constant of nitrous acid is $4.5 * 10^{ –4}$ . Calculate the $pH$ of $0.04 M$ sodium nitrite solution and also its degree of hydrolysis.

##### Solution :

$NaNO _2$ is the salt of a strong base $(NaOH)$ and a weak acid $(HNO _2 ).$$\\ NO_^- 2 + H _2 O \leftrightarrow HNO _2 + OH ^-$$\\$ $K_b=\dfrac{[HNO_2][OH^-]}{[NO^-_2]}\\ =\dfrac{K_w}{K_a}=\dfrac{10^{-14}}{4.5*10^{-14}}=.22*10^{-10}$ $\\$ Now, If x moles of the salt undergo hydrolysis, then the concentration of various speciespresent in the solution will be:$\\$ $[NO^-_2]=.04-x;0.04\\ [HNO_2]=x\\ [OH^-]=x\\ K_b=\dfrac{x^2}{0.04}=0.22*10^{-10}\\ x^2=0.0088*10^{-10}\\ x=.093*10^{-5}\\ \therefore [OH^-]=0.093*10^{-5}M\\ [H_3O^+]=\dfrac{10^{-14}}{.093*10^{-5}}=10.75*10^{-9}M\\ =pH=-\log(10.75*10^{-9})\\ =7.96$ $\\$ Therefore, degree of hydrolysis$\\$ $=\dfrac{x}{0.04}=\dfrac{.093*106{-5}}{.04}\\ =2.325*10^{-5}$ $\\$

63   A $0.02 M$ solution of pyridinium hydrochloride has $pH = 3.44$. Calculate the ionization constant of pyridine

$pH = 3.44$$\\ We know that,\\ pH = – \log [H ^+ ]\\ \therefore [ H ^+ ] = 3.63 *10 ^{- 4} \\ Then,\\ K_b=\dfrac{(3.63810^{-4})^2}{0.02}(\text{concentration}=0.02M)\\ =K_b=6.6*10^{-6} \\ Now,K_b=\dfrac{K_w}{K_a}\\ =K_a=\dfrac{K_w}{K_a}=\dfrac{10^{-14}}{6.6*10^{-6}}\\ =1.51*10^{-9} \\ 64 Predict if the solutions of the following salts are neutral, acidic or basic:\\ NaCl, KBr, NaCN, NH _4 NO _3 , NaNO _2 and KF ##### Solution : (i) NaCl:\\ NaCl +H_2O \leftrightarrow NaOH+HCl \\ \qquad \qquad \qquad \text{Strong base}\quad \text{Strong base} \\ Therefore, it is a neutral solution.\\ (ii) KBr:\\ KBr + H_ 2 O \leftrightarrow KOH + HBr$$\\$ $\qquad \qquad \qquad \text{Strong base}\quad \text{Strong base}$ $\\$ Therefore, it is a neutral solution.$\\$ (iii) NaCN:$\\$ $NaCN + H _2 O \leftrightarrow HCN + NaOH$ $\\$ $\qquad \qquad \qquad \text{Weak acid}\quad \text{Strong base}$ $\\$ Therefore, it is a basic solution.$\\$ (iv) $NH _4 NO_ 3$$\\ NH _4 NO _3 + H _2 O \leftrightarrow NH_ 4 OH + HNO _3$$\\$ $\qquad \qquad \qquad \text{Strong base}\quad \text{Weak acid}$ $\\$ Therefore, it is a acidic solution.$\\$ (v) $NaNO_ 2$$\\ NaNO_ 2 + H_ 2 O \leftrightarrow NaOH+HNO_2 \\ \qquad \qquad \qquad \text{Strong base}\quad \text{Weak acid} \\ Therefore, it is a basic solution.\\ (vi) KF\\ KF + H _2 O \leftrightarrow KOH + HF$$\\$ $\qquad \qquad \qquad \text{Strong base}\quad \text{Weak acid}$ $\\$ Therefore, it is a basic solution.$\\$

65   The ionization constant of chloroacetic acid is $1.35 * 10^{ –3}$ . What will be the $pH$ of $0.1M$acid and its $0.1M$ sodium salt solution?