Let's solve each question step by step: --- ## **8. pH of .075 M Mandelic Acid Solution** Given: - \([HA] = .075\, M\) - \(K_a = 4. \times 10^{-4}\) Mandelic acid (\(HA\)) dissociates as: \[ HA \leftrightarrow H^+ + A^- \] Let \(x\) be the amount of \(HA\) that dissociates: - \([H^+] = x\) - \([A^-] = x\) - \([HA] = .075 - x \approx .075\) (since \(K_a\) is small, \(x\) will be small) **Ka expression:** \[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{.075} \] \[ 4. \times 10^{-4} = \frac{x^2}{.075} \] \[ x^2 = (4. \times 10^{-4}) \times .075 = 3. \times 10^{-5} \] \[ x = \sqrt{3. \times 10^{-5}} \approx 5.48 \times 10^{-3} \] \[ [H^+] = 5.48 \times 10^{-3} \text{ M} \] \[ \text{pH} = -\log(5.48 \times 10^{-3}) \approx 2.26 \] **Final Answer:** \[ \boxed{\text{pH} = 2.26} \] --- ## **9. Titration of KOH with HCl** **Given:** - \(85.\, \text{mL}\) KOH required \(65.\, \text{mL}\) of \(.250\, M\) HCl to reach equivalence. ### **a) Concentration of KOH in the starting solution** \[ \text{Moles of HCl} = .250 \times .065 = .01625\, \text{mol} \] At equivalence, moles of KOH = moles of HCl = .01625 mol \[ \text{Concentration of KOH} = \frac{.01625}{.085} = .191\, M \] **Final Answer:** \[ \boxed{.191\, M} \] --- ### **b) pH at the Equivalence Point** At equivalence, all KOH and HCl have reacted to form water and KCl (a neutral salt): \[ \text{Total volume} = 85. + 65. = 150.\, \text{mL} = .150\, \text{L} \] Since both are strong acid and base, the pH at equivalence is **7.00**. **Final Answer:** \[ \boxed{7.00} \] --- ### **c) pH after 10. mL of acid has been added** **Volume of HCl added:** \(10.\, \text{mL}\) \[ \text{Moles HCl added} = .250 \times .010 = .00250\, \text{mol} \] **Initial moles of KOH:** \(.191 \times .085 = .016235\, \text{mol}\) After addition: - HCl neutralizes an equal amount of KOH - Remaining KOH: \(.016235 - .00250 = .013735\, \text{mol}\) - Total volume: \(85. + 10. = 95.\, \text{mL} = .095\, \text{L}\) \[ [\text{KOH}] = \frac{.013735}{.095} = .1446\, M \] \[ [\text{OH}^-] = .1446\, M \] \[ \text{pOH} = -\log(.1446) = .840 \] \[ \text{pH} = 14.00 - .840 = 13.16 \] **Final Answer:** \[ \boxed{13.16} \] --- ### **d) pH after 75. mL of acid has been added** \[ \text{Moles HCl added} = .250 \times .075 = .01875\, \text{mol} \] KOH started with \(.016235\, \text{mol}\), so HCl is in **excess**: \[ \text{Excess HCl} = .01875 - .016235 = .002515\, \text{mol} \] \[ \text{Total volume} = 85. + 75. = 160.\, \text{mL} = .160\, \text{L} \] \[ [\text{H}^+] = \frac{.002515}{.160} = .01572\, M \] \[ \text{pH} = -\log(.01572) = 1.80 \] **Final Answer:** \[ \boxed{1.80} \] --- ### **e) pH after 64.9 mL of acid has been added** \[ \text{Moles HCl added} = .250 \times .0649 = .016225\, \text{mol} \] KOH started with \(.016235\, \text{mol}\), so a tiny excess of KOH remains: \[ \text{Excess KOH} = .016235 - .016225 = .000010\, \text{mol} \] \[ \text{Total volume} = 85. + 64.9 = 149.9\, \text{mL} = .1499\, \text{L} \] \[ [\text{OH}^-] = \frac{.000010}{.1499} = 6.67 \times 10^{-5}\, M \] \[ \text{pOH} = -\log(6.67 \times 10^{-5}) = 4.176 \] \[ \text{pH} = 14.00 - 4.176 = 9.82 \] **Final Answer:** \[ \boxed{9.82} \] --- ### **Summary Table** | Part | Final Answer | |------|-----------------------| | 8 | pH = **2.26** | | 9a | .191 M | | 9b | pH = **7.00** | | 9c | pH = **13.16** | | 9d | pH = **1.80** | | 9e | pH = **9.82** | Let me know if you need further explanation for any step!