Question

What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.150 mol of HCl were added

Answer 1

Answer:

pH = 2.21

Explanation:

Hello there!

In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:

$NaF+HCl\rightarrow NaCl+HF$

It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:

$n_{HF}^{new}=0.300mol+0.150mol=0.450mol\\\\n_{NaF}^{new}=0.200mol-0.150mol=0.050mol$

Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:

$[HF]=\frac{0.450mol}{1.0L} =0.450M$

$[NaF]=\frac{0.050mol}{1.0L} =0.050M$

Now, calculated the pKa of HF:

$pKa=-log(6.8x10^{-4})=3.17$

We can proceed to the HH equation:

$pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21$

Best regards!