Please help, specially for (c) and (d).?
Calculate the pH of a buffer consisting of 0.50 M HF and 0.45 M KF (a) before and (b) after the addition of 0.40 g of NaOH to 1.0 L of the buffer
(Ka of HF = 6.8E-4). For sake of comparison,
calculate the pH before and after adding the same amount of NaOH to (c) 1.0 L of pure water and (d) 1.0 L of 0.50 M HCl.
Discuss any differences in the magnitude of change in the pH between the three scenarios.
- hcbiochemLv 75 months ago
HF <--> H+ + F-
Ka = 6.8X10^-4 = [H+][F-]/[HF]
(a) 6.8X10^-4 = [H+](0.45) / 0.50
[H+] = 7.56X10^-4 M
pH = 3.12
(a) Initial moles HF = 0.50 mol
Initial moles F- = 0.45 mol
Moles NaOH added = 0.40 g / 40 g/mol = 0.01
Added NaOH quantitatively neutralizes 0.01 mol HF forming an additional 0.01 mol F-. So, after addition, [HF] = 0.49 and [F-] = 0.46
6.8X10^-4 = [H+] (0.46)/(0.49)
[H+] = 7.24X10^-4
pH = 3.14
Adding 0.01 mol NaOH to 1 L water (which initially has a pH = 7.0) results in [OH-] = 0.01 M
pOH = 2.0
pH = 12.0
Adding 0.01 mol NaOH to 0.50 M HCl decreases [H+] from 0.50 M to 0.49 M
pH before addition = 0.301
pH after addition = 0.309