Chapter 15 - Acid-Base Equilibria - Exercises - Page 753: 59

a. $pH = 2.72$ b. $pH = 4.26$ c. $pH = 4.74$ d.$pH = 5.22$ e. $pH = 8.79$ f. $pH = 12.15$

a. 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $CH_3CO_2H(aq) + H_2O(l) \lt -- \gt CH_3CO_2^-(aq) + H_3O^+(aq) $ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $CH_3CO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [CH_3CO_2^-] = 0 + x = x$ -$[CH_3CO_2H] = [CH_3CO_2H]_{initial} - x$ For approximation, we are going to consider $[CH_3CO_2H]_{initial} = [CH_3CO_2H]$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3CO_2^-]}{ [CH_3CO_2H]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.2}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.2}$ $x^2 = 0.2 \times 1.8 \times 10^{-5} $ $x = \sqrt { 0.2 \times 1.8 \times 10^{-5}} = 1.9 \times 10^{-3} $ Percent dissociation: $\frac{ 1.9 \times 10^{- 3}}{ 0.2} \times 100\% = 0.95\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3CO_2^-] = x = 1.9 \times 10^{- 3}M $ And, since 'x' has a very small value (compared to the initial concentration): $[CH_3CO_2H] \approx 0.2M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.9 \times 10^{- 3})$ $pH = 2.72$ ----------------- b. 1000ml = 1L 100ml = 0.1 L 50ml = 0.05 L 1. Find the numbers of moles: $C(CH_3CO_2H) * V(CH_3CO_2H) = 0.2* 0.1 = 0.02$ moles $C(KOH) * V(KOH) = 0.1* 0.05 = 5 \times 10^{-3}$ moles 2. Write the acid-base reaction: $CH_3CO_2H(aq) + KOH(aq) -- \gt KCH_3CO_2(aq) + H_2O(l)$ - Total volume: 0.1 + 0.05 = 0.15L 3. Since the base is the limiting reactant, only $ 0.005$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3CO_2H] = 0.02 - 0.005 = 0.015$ moles. Concentration: $\frac{0.015}{ 0.15} = 0.1M$ $[KOH] = 0.005 - 0.005 = 0$ $[KCH_3CO_2] = 0 + 0.005 = 0.005$ moles. Concentration: $\frac{ 0.005}{ 0.15} = 0.033M$ 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: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.74 + log(\frac{0.033}{0.1})$ $pH = 4.74 + log(0.33)$ $pH = 4.74 + -0.477$ $pH = 4.26$ ----------------- c. 1000ml = 1L 100ml = 0.1 L 100ml = 0.1 L 1. Find the numbers of moles: $C(CH_3CO_2H) * V(CH_3CO_2H) = 0.2* 0.1 = 0.02$ moles $C(KOH) * V(KOH) = 0.1* 0.1 = 0.01$ moles 2. Write the acid-base reaction: $CH_3CO_2H(aq) + KOH(aq) -- \gt KCH_3CO_2(aq) + H_2O(l)$ - Total volume: 0.1 + 0.1 = 0.2L 3. Since the base is the limiting reactant, only $ 0.01$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3CO_2H] = 0.02 - 0.01 = 0.01$ moles. Concentration: $\frac{0.01}{ 0.2} = 0.05M$ $[KOH] = 0.01 - 0.01 = 0$ $[KCH_3CO_2] = 0 + 0.01 = 0.01$ moles. Concentration: $\frac{ 0.01}{ 0.2} = 0.05M$ 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: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.74 + log(\frac{0.05}{0.05})$ $pH = 4.74 + log(1)$ $pH = 4.74 + 0$ $pH = 4.74$ ----------------- 1000ml = 1L 100ml = 0.1 L 250ml = 0.15 L 1. Find the numbers of moles: $C(CH_3CO_2H) * V(CH_3CO_2H) = 0.2* 0.1 = 0.02$ moles $C(KOH) * V(KOH) = 0.1* 0.15 = 0.015$ moles 2. Write the acid-base reaction: $CH_3CO_2H(aq) + KOH(aq) -- \gt KCH_3CO_2(aq) + H_2O(l)$ - Total volume: 0.1 + 0.15 = 0.25L 3. Since the base is the limiting reactant, only $ 0.015$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3CO_2H] = 0.02 - 0.015 = 5 \times 10^{-3}$ moles. Concentration: $\frac{5 \times 10^{-3}}{ 0.25} = 0.02M$ $[KOH] = 0.015 - 0.015 = 0$ $[KCH_3CO_2] = 0 + 0.015 = 0.015$ moles. Concentration: $\frac{ 0.015}{ 0.25} = 0.06M$ 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: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.74 + log(\frac{0.06}{0.02})$ $pH = 4.74 + log(3)$ $pH = 4.74 + 0.477$ $pH = 5.22$ ------------- e. 1000ml = 1L 100ml = 0.1 L 200ml = 0.2 L 1. Find the numbers of moles: $C(CH_3CO_2H) * V(CH_3CO_2H) = 0.2* 0.1 = 0.02$ moles $C(KOH) * V(KOH) = 0.1* 0.2 = 0.02$ moles 2. When the number of moles is equal, the reactants are totally consumed: $CH_3CO_2H(aq) + KOH(aq) -- \gt KCH_3CO_2(aq) + H_2O(l)$ - Total volume: 0.1 + 0.2 = 0.3L 3. So, those are the final concentrations: $[CH_3CO_2H] = 0.02 - 0.02 = 0$ mol. $[KOH] = 0.02 - 0.02 = 0$ mol $[KCH_3CO_2] = 0 + 0.02 = 0.02$ moles. Concentration: $\frac{ 0.02}{ 0.3} = 0.0667M$ - Therefore, we have a weak base salt solution: - Since $CH_3CO_2^-$ is the conjugate base of $CH_3CO_2H$ , 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}$ $K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_b = 5.56\times 10^{- 10}$ 4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $CH_3CO_2^-(aq) + H_2O(l) \lt -- \gt CH_3CO_2H(aq) + OH^-(aq) $ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $CH_3CO_2H$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [CH_3CO_2H] = 0 + x = x$ -$[CH_3CO_2^-] = [CH_3CO_2^-]_{initial} - x $ For approximation, we are going to consider $[CH_3CO_2^-]_{initial} = [CH_3CO_2^-]$ 5. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][CH_3CO_2H]}{ [CH_3CO_2^-]}$ $Kb = 5.56 \times 10^{- 10}= \frac{x * x}{ 0.0667}$ $Kb = 5.56 \times 10^{- 10}= \frac{x^2}{ 0.0667}$ $x^2 = 5.56 \times 10^{-10} \times 0.0667$ $x = \sqrt { 5.56 \times 10^{-10} \times 0.0667} = 6.1 \times 10^{-6}$ Percent ionization: $\frac{ 6.1 \times 10^{- 6}}{ 0.0667} \times 100\% = 0.0091\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [CH_3CO_2H] = x = 6.1 \times 10^{- 6}M $ $[CH_3CO_2^-] \approx 0.0667M$ 6. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 6.1 \times 10^{- 6})$ $pOH = 5.21$ 7. Find the pH: $pH + pOH = 14$ $pH + 5.21 = 14$ $pH = 8.79$ ------------------ f. 1000ml = 1L 100ml = 0.1 L 350ml = 0.25 L 1. Find the numbers of moles: $C(CH_3CO_2H) * V(CH_3CO_2H) = 0.2* 0.1 = 0.02$ moles $C(KOH) * V(KOH) = 0.1* 0.25 = 0.025$ moles Write the acid-base reaction: $CH_3CO_2H(aq) + KOH(aq) -- \gt KCH_3CO_2(aq) + H_2O(l)$ - Total volume: 0.1 + 0.25 = 0.35L Since the acid is the limiting reactant, only $ 0.02$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3CO_2H] = 0.02 - 0.02 = 0M$. $[KOH] = 0.025 - 0.02 = 5 \times 10^{-3}$ mol Concentration: $\frac{5 \times 10^{-3}}{ 0.35} = 0.014M$ $[KCH_3CO_2] = 0 + 0.02 = 0.02$ moles. Concentration: $\frac{ 0.02}{ 0.35} = 0.057M$ - 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: $pOH = -log[OH^-]$ $pOH = -log( 0.014)$ $pOH = 1.85$ 3. Find the pH: $pH + pOH = 14$ $pH + 1.85 = 14$ $pH = 12.15$