## Chemistry (4th Edition)

(a) $pH = 2.87$ (b) $pH = 4.56$ (c) $pH = 5.34$ (d) $pH = 8.78$ (e) $pH = 12.10$
(a) 1. We have these concentrations at equilibrium: -$[H_3O^+] = [CH_3COO^-] = x$ -$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.1 - x$ For approximation, we consider: $[CH_3COOH] = 0.1M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.1}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.1}$ $1.8 \times 10^{- 6} = x^2$ $x = 1.3 \times 10^{- 3}$ Percent dissociation: $\frac{ 1.3 \times 10^{- 3}}{ 0.1} \times 100\% = 1.3\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3COO^-] = x = 1.3 \times 10^{- 3}M$ And, since 'x' has a very small value (compared to the initial concentration): $[CH_3COOH] \approx 0.1M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.3 \times 10^{- 3})$ $pH = 2.87$ (b) 1. Find the numbers of moles: $C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.2* 0.005 = 1 \times 10^{-3}$ moles 2. Write the acid-base reaction: $CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$ - Total volume: 0.025 + 0.005 = 0.03L 3. Since the base is the limiting reactant, only $0.001$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3COOH] = 0.0025 - 0.001 = 1.5 \times 10^{-3}$ moles. Concentration: $\frac{1.5 \times 10^{-3}}{ 0.03} = 0.05M$ $[KOH] = 0.001 - 0.001 = 0$ $[KCH_3COO] = 0 + 0.001 = 0.001$ moles. Concentration: $\frac{ 0.001}{ 0.03} = 0.033M$ 4. Calculate the $pK_a$ for the acid $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.74$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.74 + log(\frac{0.033}{0.05})$ $pH = 4.74 + log(0.67)$ $pH = 4.74 + -0.176$ $pH = 4.56$ (c) 1. Find the numbers of moles: $C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.2* 0.01 = 2 \times 10^{-3}$ moles 2. Write the acid-base reaction: $CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$ - Total volume: 0.025 + 0.01 = 0.035L 3. Since the base is the limiting reactant, only $0.002$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3COOH] = 0.0025 - 0.002 = 5 \times 10^{-4}$ moles. Concentration: $\frac{5 \times 10^{-4}}{ 0.035} = 0.014M$ $[KOH] = 0.002 - 0.002 = 0$ $[KCH_3COO] = 0 + 0.002 = 0.002$ moles. Concentration: $\frac{ 0.002}{ 0.035} = 0.057M$ 4. Calculate the $pK_a$ for the acid $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.74$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.74 + log(\frac{0.057}{0.014})$ $pH = 4.74 + log(4)$ $pH = 4.74 + 0.602$ $pH = 5.34$ (d) 1. Find the numbers of moles: $C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.2* 0.0125 = 2.5 \times 10^{-3}$ moles 2. When the number of moles is equal, the reactants are totally consumed: $CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$ - Total volume: 0.025 + 0.0125 = 0.0375L 3. So, those are the final concentrations: $[CH_3COOH] = 0.0025 - 0.0025 = 0$ mol. $[KOH] = 0.0025 - 0.0025 = 0$ mol $[KCH_3COO] = 0 + 0.0025 = 0.0025$ moles. Concentration: $\frac{ 0.0025}{ 0.0375} = 0.0666666666666667M$ - Therefore, we have a weak base salt solution: - Since $CH_3COO^-$ is the conjugate base of $CH_3COOH$ , we can calculate its kb by using this equation: $K_a * K_b = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_b = 5.6\times 10^{- 10}$ 4. We have these concentrations at equilibrium: -$[OH^-] = [CH_3COOH] = x$ -$[CH_3COO^-] = [CH_3COO^-]_{initial} - x = 0.0666666666666667 - x$ For approximation, we consider: $[CH_3COO^-] = 0.0666666666666667M$ 5. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][CH_3COOH]}{ [CH_3COO^-]}$ $Kb = 5.6 \times 10^{- 10}= \frac{x * x}{ 0.067}$ $Kb = 5.6 \times 10^{- 10}= \frac{x^2}{ 0.067}$ $3.7 \times 10^{- 11} = x^2$ $x = 6.1 \times 10^{- 6}$ Percent ionization: $\frac{ 6.1 \times 10^{- 6}}{ 0.067} \times 100\% = 0.0091\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [CH_3COOH] = x = 6.1 \times 10^{- 6}M$ $[CH_3COO^-] \approx 0.067M$ 6. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 6.1 \times 10^{- 6})$ $pOH = 5.22$ 7. Find the pH: $pH + pOH = 14$ $pH + 5.22 = 14$ $pH = 8.78$ (e) 1. Find the numbers of moles: $C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.025 = 2.5 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.2* 0.015 = 3 \times 10^{-3}$ moles Write the acid-base reaction: $CH_3COOH(aq) + KOH(aq) -- \gt KCH_3COO(aq) + H_2O(l)$ - Total volume: 0.025 + 0.015 = 0.04L Since the acid is the limiting reactant, only $0.0025$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3COOH] = 0.0025 - 0.0025 = 0M$. $[KOH] = 0.003 - 0.0025 = 5 \times 10^{-4}$ mol Concentration: $\frac{5 \times 10^{-4}}{ 0.04} = 0.012M$ $[KCH_3COO] = 0 + 0.0025 = 0.0025$ moles. Concentration: $\frac{ 0.0025}{ 0.04} = 0.063M$ - We have a strong and a weak base. We can ignore the weak one and calculate the pH based only on the strong base concentration: $[OH^-] = [KOH]$ 2. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 0.012)$ $pOH = 1.90$ 3. Find the pH: $pH + pOH = 14$ $pH + 1.90 = 14$ $pH = 12.10$