Question

A 1.50 L buffer solution is 0.250 mol L-1 in HF and 0.250 mol L-1 in NaF. Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH. Assume no volume change upon the addition of the base. The Ka for HF is 3.5 × 10-4.

Thanks in advance!

Answer 1

millimols HF = mL x M = approx 375
mmols NaF = 1500 x 0.25 = 375
mmols NaOH added = 50

........HF + OH^- ==> H2O + F^-
I......375...0........0.....375
add..........50................
C.....-50...-50.......50.....50
E......325...0...............425

Substitute the E line into the HH equation and solve for pH.
By the way the pH of the initial solution is just pH = pKa since the acid and base are equal.