Question 10.14: What is the pH of 1.00 L of the 0.100 M hydrofluoric acid -0...
What is the pH of 1.00 L of the 0.100 M hydrofluoric acid -0.120 M fluoride ion buffer system described in Worked Example 10.13 after 0.020 mol of NaOH is added?
ANALYSIS Initially, the 0.100 M HF -0.120 M NaF buffer has pH = 3.54, as calculated in Worked Example 10.13. The added base will react with the acid as indicated in the neutralization reaction,
which means [HF] decreases and [F^-] increases. With the pK_a and the concentrations of HF and F^- known, pH can be calculated using the Henderson-Hasselbalch equation.
BALLPARK ESTIMATE After the neutralization reaction, there is more conjugate base (F^-) and less conjugate acid (HF), and so we expect the pH to increase slightly from the initial value of 3.54.
When 0.020 mol of NaOH is added to 1.00 L of the buffer, the HF concentration decreases from 0.100 M to 0.080 M as a result of an acid-base reaction. At the same time, the F^- concentration increases from 0.120 M to 0.140 M because additional F^- is produced by the neutralization. Using these new values gives
pH = 3.46 + \log \left(\frac{(0.140)}{(0.080)} \right) = 3.46 + 0.24 = 3.70The addition of 0.020 mol of base causes the pH of the buffer to rise only from 3.54 to 3.70.
BALLPARK CHECK The final pH, 3.70, is slightly more basic than the initial pH of 3.54, consistent with our prediction.