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Chapter 17: Problem 21

(a) Calculate the pH of a buffer that is \(0.150 \mathrm{M}\) in lactic acid and \(0.120 M\) in sodium lactate. (b) Calculate the pH of a buffer formed by mixing \(75 \mathrm{~mL}\) of \(0.150 \mathrm{M}\) lactic acid with \(25 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) sodium lactate.

Short Answer

Expert verified
The pH of the buffer that is 0.150 M in lactic acid and 0.120 M in sodium lactate is 3.66, and the pH of the buffer formed by mixing 75 mL of 0.150 M lactic acid with 25 mL of 0.120 M sodium lactate is 2.61.

Step by step solution

01

Find pKa for lactic acid

The \(pK_a\) value for lactic acid is 3.86.
02

Step 2(a): Use the Henderson-Hasselbalch equation for 0.150 M lactic acid and 0.120 M sodium lactate

Plug the concentrations into the equation: \(pH = 3.86 + log_{10}\dfrac{0.120}{0.150}\) Now, calculate the pH: \(pH = 3.86 - 0.204\) \(pH = 3.656\)
03

Step 3(a): Round the pH value

Round the pH value to two decimal places to get the pH of the buffer solution: \(pH = 3.66\)
04

Step 2(b): Calculate the new concentrations of lactic acid and sodium lactate by mixing the solutions

Use the given volumes and concentrations to calculate the moles of lactic acid and sodium lactate: Moles of lactic acid = 0.150 M × 0.075 L = 0.01125 mol Moles of sodium lactate = 0.120 M × 0.025 L = 0.00300 mol Now, calculate the new total volume after mixing the solutions: Total volume = 0.075 L + 0.025 L = 0.100 L Determine the new concentrations: [\(HA\)] = (0.01125 mol) / (0.100 L) = 0.1125 M [\(A^-\)] = (0.00300 mol) / (0.100 L) = 0.0300 M
05

Step 3(b): Use the new concentrations in the Henderson-Hasselbalch equation

Plug the new concentrations into the equation: \(pH = 3.86 + log_{10}\dfrac{0.0300}{0.1125}\) Now, calculate the pH: \(pH = 3.86 - 1.25\) \(pH = 2.61\)
06

Step 4(b): Round the pH value

Round the pH value to two decimal places to get the pH of the mixed buffer solution: \(pH = 2.61\) In conclusion, the pH of the buffer that is 0.150 M in lactic acid and 0.120 M in sodium lactate is 3.66, and the pH of the buffer formed by mixing 75 mL of 0.150 M lactic acid with 25 mL of 0.120 M sodium lactate is 2.61.

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Most popular questions from this chapter

(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{+}\), (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Mn}^{2+} .\) Suggest how each mixture might be separated.

For each pair of compounds, use \(K_{s p}\) values to determine which has the greater molar solubility: (a) CdS or CuS, (b) \(\mathrm{PbCO}_{3}\) or \(\mathrm{BaCrO}_{4}\), (c) \(\mathrm{Ni}(\mathrm{OH})_{2}\) or \(\mathrm{NiCO}_{3}\), (d) \(\mathrm{AgI}\) or \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\)

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to $8.5 ?(\mathbf{b})\( Will \)\mathrm{AgIO}_{3}\( precipitate when \)20 \mathrm{~mL}$ of \(0.010 \mathrm{M} \mathrm{AgIO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of $0.015 \mathrm{M} \mathrm{NaIO}_{3} ?\left(K_{s p}\right.\( of \)\mathrm{AgIO}_{3}$ is \(3.1 \times 10^{-8} .\) )

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathbf{a})\) a solution formed by adding \(30.0 \mathrm{~g}\) of furoic acid and \(25.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.300 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) $\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\( and \)30.0 \mathrm{~mL}\( of \)0.250 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$ and diluting the total volume to \(125 \mathrm{~mL},(\mathbf{c})\) a solution prepared by adding $25.0 \mathrm{~mL}\( of \)1.00 \mathrm{M} \mathrm{NaOH}\( solution to \)100.0 \mathrm{~mL}\( of \)0.100 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$
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