Question

Determine the pH change when 0.068 mol HCl is added to 1.00 L of a buffer solution that is 0.438 M in HF and 0.237 M in F^-.

pH after addition − pH before addition = pH change = _____________

Answer 1

Ka of HF = 6.6*10^-4

Lets calculate the initial pH

Ka = 6.6*10^-4

pKa = - log (Ka)
= - log(6.6*10^-4)
= 3.18

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.18+ log {0.237/0.438}
= 2.913


Lets calculate the final pH

mol of HCl added = 0.068 mol

F- will react with H+ to form HF

Before Reaction:
mol of F- = 0.237 M *1.0 L
mol of F- = 0.237 mol

mol of HF = 0.438 M *1.0 L
mol of HF = 0.438 mol

after reaction,
mol of F- = mol present initially - mol added
mol of F- = (0.237 - 0.068) mol
mol of F- = 0.169 mol

mol of HF = mol present initially + mol added
mol of HF = (0.438 + 0.068) mol
mol of HF = 0.506 mol


Ka = 6.6*10^-4

pKa = - log (Ka)
= - log(6.6*10^-4)
= 3.18

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.18+ log {0.169/0.506}
= 2.701


pH change = |final pH - initial pH|
= |2.701 - 2.913|
= 0.212
Answer: 0.212