Question

What is the pH of 1.00 L of the buffer solution that contains 0.100 M HF and 0.120 M NaF after 0.020 mol of HNO₃ is added?

Answer 1

Answer:

pH = 3,38

Explanation:

The reaction of HNO₃ with NaF is:

HNO₃ + NaF → HF + NaNO₃

If you have 1L of 0,120M NaF, moles of NaF are 0,120 moles. These moles reacts with 0,020 mol of HNO₃ to produce 0,020 moles of HF. That means the final concentrations of HF and NaF are:

HF: 0,100mol + 0,020 mol = 0,120 mol

NaF: 0,120mol - 0,020 mol = 0,100 mol

Now, to obtain the pH of this buffer (Weak acid + conjugate base) you can use  Henderson-Hasselblach formula:

pH = pka + log₁₀ [NaF] / [HF]

Replacing:

pH = 3.46 + log₁₀ [0,100] / [0,120]

pH = 3.38

I hope it helps!