Ionic Equilibrium: Speed-Solving Techniques (2)

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Tags Ionic Equilibrium

Question

Calculate the pH of a 0.010.01 M acetic acid solution. Given Ka=1.8×10−5K_a = 1.8 \times 10^{-5}.

Solution — Step by Step

CH3_3COOH ⇌\rightleftharpoons CH3_3COO−^- + H+^+. Initially, [HA]=0.01[\text{HA}] = 0.01 M, [A−]=[H+]=0[\text{A}^-] = [\text{H}^+] = 0. At equilibrium, [H+]=x[\text{H}^+] = x, [A−]=x[\text{A}^-] = x, [HA]=0.01−x[\text{HA}] = 0.01 - x.

 Ka=[H+][A−][HA]=x20.01−xK_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{x^2}{0.01 - x}

Since KaK_a is small, x≪0.01x \ll 0.01, so 0.01−x≈0.010.01 - x \approx 0.01. Then:

x2≈Ka⋅C=1.8×10−5×0.01=1.8×10−7x^2 \approx K_a \cdot C = 1.8 \times 10^{-5} \times 0.01 = 1.8 \times 10^{-7} x≈1.8×10−7≈4.24×10−4 Mx \approx \sqrt{1.8 \times 10^{-7}} \approx 4.24 \times 10^{-4} \text{ M} pH=−log⁡[H+]=−log⁡(4.24×10−4)≈3.37\text{pH} = -\log[\text{H}^+] = -\log(4.24 \times 10^{-4}) \approx 3.37

Final Answer: pH ≈3.37\approx 3.37.

Why This Works

For a weak acid with Ka≪CK_a \ll C, the approximation [H+]≈Ka⋅C[\text{H}^+] \approx \sqrt{K_a \cdot C} saves you from solving a quadratic. The shortcut breaks if the degree of dissociation α=x/C\alpha = x/C exceeds about 5%5\% — here α≈4.24%\alpha \approx 4.24\%, so we’re fine.

The shortcut formula in pH form: pH=12(pKa−log⁡C)\text{pH} = \dfrac{1}{2}(\text{p}K_a - \log C). With pKa=4.74K_a = 4.74 and log⁡C=−2\log C = -2, pH=(4.74+2)/2=3.37\text{pH} = (4.74 + 2)/2 = 3.37. Same answer, one line.

Alternative Method

Solve the quadratic exactly: x2+Kax−KaC=0x^2 + K_a x - K_a C = 0. Substituting numbers gives x=4.16×10−4x = 4.16 \times 10^{-4} M and pH ≈3.38\approx 3.38. Marginally different, confirming the shortcut works at this concentration.

Using the strong-acid formula pH =−log⁡C= -\log C on a weak acid gives pH =2= 2 here, which is half a pH unit too acidic. Always check whether the acid is strong (HCl, HNO3_3, H2_2SO4_4) or weak before choosing the formula.

Memorize: pH=12(pKa−log⁡C)\text{pH} = \frac{1}{2}(\text{p}K_a - \log C) for weak acids, pOH=12(pKb−log⁡C)\text{pOH} = \frac{1}{2}(\text{p}K_b - \log C) for weak bases. NEET asks one such calculation every year — recognize the pattern in 5 seconds.

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