Weak-Acid Dilution and Buffer Neutralization
A monoprotic weak acid HA is dissolved in pure water at 25°C. A 0.100 M solution has pH 2.90. A student calculates Ka using the approximation Ka≈[H+]²/C and obtains 1.6×10−5. The student then dilutes the solution tenfold to 0.0100 M and predicts pH=3.40 by adding 0.50 to the original pH, because the acid is diluted by a factor of ten. The instructor asks whether the approximation is valid at both concentrations, whether the pH shift is exactly 0.50, and how the buffer region behaves after partial neutralization. The acid is monoprotic, no other solutes are present, activities are approximated by concentrations, and water autoionization may be neglected where justified. The temperature remains 25°C throughout. A second sample of 100.0 mL of the 0.100 M HA solution is treated with 0.0050 mol of solid NaOH; the volume change is negligible.
1. Calculate Ka without assuming x≪C and assess the approximation at 0.100 M.
Read the worked answer
[H+]=10^−2.90=1.26×10^−3 M. Exact: Ka=[H+]²/(C−[H+])=(1.26×10^−3)²/(0.100−1.26×10^−3)=1.59×10^−6/0.09874=1.61×10^−5. Approximation: Ka≈(1.26×10^−3)²/0.100=1.59×10^−5. The approximation error is about 1.3%; [H+]/C≈1.3%, so x≪C is acceptable at 0.100 M.
2. Calculate the pH at 0.0100 M using Ka=1.61×10−5, solve the quadratic, and compare with the naive prediction 3.40.
Read the worked answer
Let x=[H+] at 0.0100 M. x²/(0.0100−x)=1.61×10^−5, so x²+1.61×10^−5x−1.61×10^−7=0. x=[−1.61×10^−5+√((1.61×10^−5)²+4×1.61×10^−7)]/2. The discriminant is 6.44×10^−7, square root 8.03×10^−4; x=(8.03×10^−4−1.61×10^−5)/2=3.93×10^−4 M. pH=−log(3.93×10^−4)=3.41. The naive 3.40 is close but not exact; the shift is 0.51, not exactly 0.50, because dissociation is slightly more complete at the lower concentration.
3. For the second sample, after adding 0.0050 mol NaOH to 100.0 mL of 0.100 M HA, calculate the new pH.
Read the worked answer
Moles HA initially=0.100 L×0.100 M=0.0100 mol. Added NaOH=0.0050 mol, which converts 0.0050 mol HA to A−. After reaction: HA=0.0050 mol, A−=0.0050 mol in 0.100 L (volume change neglected). pH=pKa+log([A−]/[HA]); pKa=−log(1.61×10^−5)=4.79. Since [A−]=[HA], the log term is zero, so pH=4.79. The result is the half-neutralization point; the quadratic formula is unnecessary because the buffer ratio is exact.