QuestionJune 24, 2025

A 40.00 mL sample of 0 .1000 M diprotic malonic acid is , titrated with @.071 M KOH. What volume KOH must be added to give a pH of 5.62? K_(a1)=1.42times 10^-3 and K_(a2)=2.01times 10^-6 Show all of your work on your scratch paper and upload it as a comment to this quiz assignment.

A 40.00 mL sample of 0 .1000 M diprotic malonic acid is , titrated with @.071 M KOH. What volume KOH must be added to give a pH of 5.62? K_(a1)=1.42times 10^-3 and K_(a2)=2.01times 10^-6 Show all of your work on your scratch paper and upload it as a comment to this quiz assignment.
A 40.00 mL sample of 0 .1000 M diprotic malonic acid is , titrated with @.071 M KOH. What volume KOH must be added to give a pH of 5.62? K_(a1)=1.42times 10^-3 and K_(a2)=2.01times 10^-6 Show all of your work on your scratch paper and upload it as a comment to this quiz assignment.

Solution
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Answer

56.2 mL Explanation 1. Determine the species at pH 5.62 At pH 5.62, malonic acid is primarily in its first deprotonated form (HA^-). 2. Use Henderson-Hasselbalch equation For the first dissociation: pH = pK_{a1} + \log\left(\frac A^- HA \right). Rearrange to find \frac A^- HA : 5.62 = -\log(1.42 \times 10^{-3}) + \log\left(\frac A^- HA \right). 3. Calculate ratio of concentrations 5.62 = 2.85 + \log\left(\frac A^- HA \right). \log\left(\frac A^- HA \right) = 5.62 - 2.85 = 2.77. \frac A^- HA = 10^{2.77}. 4. Calculate moles of HA and A^- Initial moles of malonic acid (HA): 0.040 \, \text{L} \times 0.10 \, \text{M} = 0.004 \, \text{mol}. Let x be moles of KOH added: [A^-] = x, [HA] = 0.004 - x. \frac{x}{0.004-x} = 10^{2.77}. 5. Solve for x x = 10^{2.77}(0.004 - x). x = 588.84(0.004 - x). x = 2.35536 - 588.84x. 589.84x = 2.35536. x = \frac{2.35536}{589.84} \approx 0.00399 \, \text{mol}. 6. Calculate volume of KOH needed Volume of KOH: \frac{0.00399 \, \text{mol}}{0.071 \, \text{M}} \approx 0.0562 \, \text{L} or 56.2 \, \text{mL}.

Explanation

1. Determine the species at pH 5.62<br /> At pH 5.62, malonic acid is primarily in its first deprotonated form (HA^-).<br /><br />2. Use Henderson-Hasselbalch equation<br /> For the first dissociation: $pH = pK_{a1} + \log\left(\frac{[A^-]}{[HA]}\right)$.<br /> Rearrange to find $\frac{[A^-]}{[HA]}$: $5.62 = -\log(1.42 \times 10^{-3}) + \log\left(\frac{[A^-]}{[HA]}\right)$.<br /><br />3. Calculate ratio of concentrations<br /> $5.62 = 2.85 + \log\left(\frac{[A^-]}{[HA]}\right)$.<br /> $\log\left(\frac{[A^-]}{[HA]}\right) = 5.62 - 2.85 = 2.77$.<br /> $\frac{[A^-]}{[HA]} = 10^{2.77}$.<br /><br />4. Calculate moles of HA and A^-<br /> Initial moles of malonic acid (HA): $0.040 \, \text{L} \times 0.10 \, \text{M} = 0.004 \, \text{mol}$.<br /> Let $x$ be moles of KOH added: $[A^-] = x$, $[HA] = 0.004 - x$.<br /> $\frac{x}{0.004-x} = 10^{2.77}$.<br /><br />5. Solve for x<br /> $x = 10^{2.77}(0.004 - x)$.<br /> $x = 588.84(0.004 - x)$.<br /> $x = 2.35536 - 588.84x$.<br /> $589.84x = 2.35536$.<br /> $x = \frac{2.35536}{589.84} \approx 0.00399 \, \text{mol}$.<br /><br />6. Calculate volume of KOH needed<br /> Volume of KOH: $\frac{0.00399 \, \text{mol}}{0.071 \, \text{M}} \approx 0.0562 \, \text{L}$ or $56.2 \, \text{mL}$.
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