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?
the of hf is 3.5×10⁻⁴, and so p = 3.46.
express your answer using two decimal places.

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!

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