## Chemistry 9th Edition

a. $pH = 5.82$ b. $pH = 10.95$
a. 1. Analyze each ion that forms this salt: $CH_3NH_3^+$: Conjugate pair of a weak base. So, it will affect the pH of the solution. $Cl^-$: Conjugate pair of a strong acid (HCl). So, it will not affect the pH of the solution. 2. Calculate the $pH$ for a $CH_3NH_3^+$ 0.10M solution: 3. Since $CH_3NH_3^+$ is the conjugate acid of $CH_3NH_2$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $4.4\times 10^{- 4} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 4.4\times 10^{- 4}}$ $K_a = 2.3\times 10^{- 11}$ 4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $CH_3NH_3^+(aq) + H_2O(l) \lt -- \gt CH_3NH_2(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $CH_3NH_2$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [CH_3NH_2] = 0 + x = x$ -$[CH_3NH_3^+] = [CH_3NH_3^+]_{initial} - x$ For approximation, we are going to consider $[CH_3NH_3^+]_{initial} = [CH_3NH_3^+]$ 5. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3NH_2]}{ [CH_3NH_3^+]}$ $Ka = 2.3 \times 10^{- 11}= \frac{x * x}{ 0.1}$ $Ka = 2.3 \times 10^{- 11}= \frac{x^2}{ 0.1}$ $x^2 = 0.1 \times 2.3 \times 10^{-11}$ $x = \sqrt { 0.1 \times 2.3 \times 10^{-11}} = 1.5 \times 10^{-6}$ Percent dissociation: $\frac{ 1.5 \times 10^{- 6}}{ 0.1} \times 100\% = 0.0015\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3NH_2] = x = 1.5 \times 10^{- 6}M$ And, since 'x' has a very small value (compared to the initial concentration): $[CH_3NH_3^+] \approx 0.1M$ 6. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.5 \times 10^{- 6})$ $pH = 5.82$ ------------- b. 1. Analyze each ion that forms this salt: $Na^+$: Conjugate pair of a strong base (NaOH). So, it will not affect the pH of the solution. $CN^-$: Conjugate pair of a weak acid. So, it will affect the pH of the solution, because it acts as a weak base. 2. Calculate the $pH$ for a $0.050M$ $CN^-$ solution: 3. Since $CN^-$ is the conjugate base of $HCN$ , we can calculate its kb by using this equation: $K_a * K_b = K_w = 10^{-14}$ $6.2\times 10^{- 10} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 6.2\times 10^{- 10}}$ $K_b = 1.6\times 10^{- 5}$ 3. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $CN^-(aq) + H_2O(l) \lt -- \gt HCN(aq) + OH^-(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $HCN$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [HCN] = 0 + x = x$ -$[CN^-] = [CN^-]_{initial} - x$ For approximation, we are going to consider $[CN^-]_{initial} = [CN^-]$ 4. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][HCN]}{ [CN^-]}$ $Kb = 1.6 \times 10^{- 5}= \frac{x * x}{ 0.05}$ $Kb = 1.6 \times 10^{- 5}= \frac{x^2}{ 0.05}$ $x^2 = 1.6 \times 10^{-5} \times 0.05$ $x = \sqrt { 1.6 \times 10^{-5} \times 0.05} = 8.9 \times 10^{-4}$ Percent ionization: $\frac{ 8.9 \times 10^{- 4}}{ 0.05} \times 100\% = 1.8\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [HCN] = x = 8.9 \times 10^{- 4}M$ $[CN^-] \approx 0.050M$ 5. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 8.9 \times 10^{- 4})$ $pOH = 3.05$ 6. Find the pH: $pH + pOH = 14$ $pH + 3.05 = 14$ $pH = 10.95$