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Chapter 17: Problem 27

A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 \(\mathrm{L}\) (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.02 mol of \(\mathrm{KOH}\) ? (c) What is the \(\mathrm{pH}\) of the buffer after the addition of 0.02 \(\mathrm{mol}\) of \(\mathrm{HNO}_{3} ?\)

Short Answer

Expert verified
The pH of the initial buffer solution is 4.92. After adding 0.02 mol of KOH, the pH increases to 5.24. When 0.02 mol of HNO3 is added, the pH decreases to 4.67.

Step by step solution

01

Understand the components of the buffer solution

The buffer solution contains acetic acid and sodium acetate. Acetic acid, \(CH_3COOH\), is a weak acid and will create an equilibrium with its conjugate base, acetate (\(CH_3COO^-\)). Sodium acetate is the sodium salt of the acetate ion, which will increase the amount of \(CH_3COO^-\) in the solution. This forms the buffer solution, which maintains its pH despite the addition of small amounts of acidic or basic substances.
02

Calculate the pH of the buffer using the Henderson-Hasselbalch equation

The pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation: \[pH = pK_a + \log_{10}\frac{[A^-]}{[HA]}\] In this case, \(HA\) represents acetic acid (\(CH_3COOH\)) and \(A^-\) represents acetate (\(CH_3COO^-\)). The \(pK_a\) of acetic acid is 4.74. Using the given amounts of acetic acid and sodium acetate, we find the concentrations: \[[CH_3COOH] = \frac{0.10\: mol}{1.00\: L} = 0.10\: M\] \[[CH_3COO^-] = \frac{0.13\: mol}{1.00\: L} = 0.13\: M\] Now, we can plug these values into the Henderson-Hasselbalch equation to calculate the pH of the buffer: \[pH = 4.74 + \log_{10}\frac{0.13}{0.10} = 4.92\]
03

Calculate the pH after adding 0.02 mol of KOH

KOH is a strong base, which will react completely with the acetic acid, forming water and more acetate: \[CH_3COOH + OH^- \longrightarrow CH_3COO^- + H_2O\] In this case, 0.02 mol of KOH was added, meaning 0.02 mol of acetic acid will react, forming 0.02 mol of acetate. After adding KOH: \[[CH_3COOH] = 0.10\:mol - 0.02\:mol = 0.08\:mol\] \[[CH_3COO^-] = 0.13\: mol + 0.02\:mol = 0.15\:mol\] Now, plug these values into the Henderson-Hasselbalch equation: \[pH = 4.74 + \log_{10}\frac{0.15}{0.08} = 5.24\]
04

Calculate the pH after adding 0.02 mol of HNO3

HNO3 is a strong acid, which will react completely with the acetate, forming water and acetic acid: \[CH_3COO^- + H^+ \longrightarrow CH_3COOH\] In this case, 0.02 mol of HNO3 was added, meaning 0.02 mol of acetate will react, forming 0.02 mol of acetic acid. After adding HNO3: \[[CH_3COOH] = 0.10\:mol + 0.02\:mol = 0.12\:mol\] \[[CH_3COO^-] = 0.13\:mol - 0.02\:mol = 0.11\:mol\] Now, plug these values into the Henderson-Hasselbalch equation: \[pH = 4.74 + \log_{10}\frac{0.11}{0.12} = 4.67\] In conclusion, the pH of the initial buffer is 4.92, the pH after adding 0.02 mol of KOH is 5.24, and the pH after adding 0.02 mol of HNO3 is 4.67.

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Most popular questions from this chapter

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to p \(K_{a}\) for the acid.

Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either M \(^{+}\) or \(A^{-} .(\mathbf{b})\) Common ions alter the equilibrium constant for the reaction of an ionic solid with water. (c) The common-ion effect does not apply to unusual ions like \(\mathrm{SO}_{3}^{2-}\) . (d) The solubility of a salt MA is affected equally by the addition of either \(\mathrm{A}^{-}\) or a noncommon ion.

The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20}\) . The \(\mathrm{AsO}_{4}^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{pK}_{a 1}=\right.\) \(2.22 ; \mathrm{p} K_{a 2}=6.98 ; \mathrm{p} K_{a 3}=11.50 )\) . (a) Calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water. (b) Calculate the pH of a saturated solution of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water.

(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: MnCO \(_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2} .\)

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating 0.200 \(\mathrm{M}\) solutions of each of the following bases with 0.200 \(M \mathrm{HBr} :(\mathbf{a})\) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\)
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