A buffer is made by dissolving 0.10 mol of NaF in 1.00 L of 0.20 M HF. What is the pH of this buffer?
K _a(HF) = 6.3×10^{- 4}

Question

A buffer is made by dissolving 0.10 mol of NaF in 1.00 L of 0.20 M HF. What is the pH of this buffer? K _a(HF) = 6.3×10^{- 4}

Quizplus · Accepted Answer

First, we need to find the amount of HF left after addition of NaF.

Initial moles of HF = 0.20 mol/L x 1.00 L = 0.20 mol 
Moles of NaF added = 0.10 mol 
Moles of HF left after addition of NaF = 0.20 - 0.10 = 0.10 mol

Next, we can use the Henderson-Hasselbalch equation to find the pH of the buffer:

pH = pKa + log([base]/[acid])

where pKa is the dissociation constant of HF (6.3 x 10^-4), [base] is the concentration of NaF (0.10 mol/L), and [acid] is the concentration of HF (0.10 mol/L).

pH = 6.4 - log(0.10/0.10) 
pH = 3.46

Therefore, the answer is D, 3.46.

A Buffer Is Made by Dissolving 0