Buffer Chemistry: Calculating pH Changes in a Buffer

In buffer chemistry, a buffer is made by mixing 1.00 L of 0.20 M acetic acid (CH\(_3\)COOH) with 1.00 L of 0.20 M sodium acetate (CH\(_3\)COONa); given \(K_a=1.8\times 10^{-5}\), find the initial pH and the pH after adding 0.010 mol HCl and after adding 0.010 mol NaOH (assume volume change is negligible).

Accepted answer Answer included

Buffer chemistry: problem

A buffer is prepared by combining a weak acid and its conjugate base: 1.00 L of 0.20 M acetic acid, CH₃COOH (denote as \(HA\)), and 1.00 L of 0.20 M sodium acetate, CH₃COO⁻ (denote as \(A^-\)). Given \(K_a=1.8\times 10^{-5}\) for acetic acid, determine:

the initial pH of the buffer,
the pH after adding 0.010 mol HCl,
the pH after adding 0.010 mol NaOH.

Assumption: the added strong acid/base changes moles but not the total volume appreciably, so ratios of concentrations can be taken as ratios of moles.

Key ideas

1) What makes a buffer?

A buffer contains a conjugate pair \(HA/A^-\). Added \(H^+\) is consumed mainly by \(A^-\) (\(A^- + H^+ \rightarrow HA\)), and added \(OH^-\) is consumed mainly by \(HA\) (\(HA + OH^- \rightarrow A^- + H_2O\)). The result is a relatively small change in the ratio \(\frac{[A^-]}{[HA]}\), so the pH changes only modestly.

2) Henderson–Hasselbalch equation

For \(HA \rightleftharpoons H^+ + A^-\),

\[ K_a=\frac{[H^+]\cdot [A^-]}{[HA]} \quad \Rightarrow \quad [H^+]=K_a\cdot \frac{[HA]}{[A^-]} \]

\[ -\log[H^+] = -\log K_a - \log\!\left(\frac{[HA]}{[A^-]}\right) \quad \Rightarrow \quad pH=pK_a+\log\!\left(\frac{[A^-]}{[HA]}\right) \]

Step-by-step solution

Step 1: Compute \(pK_a\)

\[ pK_a = -\log(1.8\times 10^{-5}) \approx 4.745 \]

Step 2: Initial pH

Initial moles in each 1.00 L portion:

\[ n(HA)=0.20\cdot 1.00 = 0.200\ \text{mol}, \quad n(A^-)=0.20\cdot 1.00 = 0.200\ \text{mol} \]

The ratio is 1, so:

\[ pH = pK_a + \log(1) = pK_a \approx 4.745 \]

Step 3: Add 0.010 mol HCl (strong acid)

HCl provides \(0.010\) mol \(H^+\), which reacts essentially completely with \(A^-\):

\[ A^- + H^+ \rightarrow HA \]

Update moles:

\[ n(A^-) = 0.200 - 0.010 = 0.190\ \text{mol} \quad ; \quad n(HA) = 0.200 + 0.010 = 0.210\ \text{mol} \]

\[ pH = pK_a + \log\!\left(\frac{0.190}{0.210}\right) \approx 4.745 + \log(0.9048) \approx 4.701 \]

Step 4: Add 0.010 mol NaOH (strong base)

NaOH provides \(0.010\) mol \(OH^-\), which reacts essentially completely with \(HA\):

\[ HA + OH^- \rightarrow A^- + H_2O \]

Update moles:

\[ n(HA) = 0.200 - 0.010 = 0.190\ \text{mol} \quad ; \quad n(A^-) = 0.200 + 0.010 = 0.210\ \text{mol} \]

\[ pH = pK_a + \log\!\left(\frac{0.210}{0.190}\right) \approx 4.745 + \log(1.1053) \approx 4.788 \]

Summary table

Case	\(n(A^-)\) (mol)	\(n(HA)\) (mol)	\(\dfrac{[A^-]}{[HA]}\approx \dfrac{n(A^-)}{n(HA)}\)	pH
Initial buffer	0.200	0.200	1.000	\(\approx 4.745\)
After 0.010 mol HCl	0.190	0.210	\(\approx 0.9048\)	\(\approx 4.701\)
After 0.010 mol NaOH	0.210	0.190	\(\approx 1.1053\)	\(\approx 4.788\)

Visualization: how pH depends on the buffer ratio

Buffer chemistry is often summarized by \(pH=pK_a+\log\!\left(\frac{[A^-]}{[HA]}\right)\). The plot below uses \(x=\log\!\left(\frac{[A^-]}{[HA]}\right)\) (base 10), so the relationship becomes a straight line \(pH=pK_a+x\). The three marked points correspond to the initial buffer and the two small additions.

Near the center (\([A^-]\approx[HA]\), so \(x\approx 0\)), the two small additions shift \(x\) by about \(\pm 0.043\), producing pH changes of the same magnitude around \(pK_a\).

Final answers

\(pK_a\approx 4.745\), initial pH \(\approx 4.745\); after adding 0.010 mol HCl, pH \(\approx 4.701\); after adding 0.010 mol NaOH, pH \(\approx 4.788\).

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