40 mL of 0.10 mol L$^{-1}$ NaOH is mixed with 60 mL of 0.10 mol L$^{-1}$ HCl - HSC - SSCE Chemistry - Question 18 - 2016 - Paper 1

For HCl:

$\text{moles of HCl} = 0.10 \: \text{mol L}^{-1} \times 0.060 \: \text{L} = 0.006 \: \text{mol}$

In this reaction, both NaOH and HCl react in a 1:1 molar ratio:

$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Here, NaOH is the limiting reagent with 0.004 mol, while we have 0.006 mol of HCl.

After the reaction, there will be:

$\text{Excess HCl} = \text{Initial moles of HCl} - \text{moles of NaOH} = 0.006 \: ext{mol} - 0.004 \: ext{mol} = 0.002 \: ext{mol}$

The total volume after mixing is:

$\text{Total volume} = 40 \: \text{mL} + 60 \: \text{mL} = 100 \: \text{mL} = 0.1 \: \text{L}$

To find the concentration of excess HCl:

$\text{Concentration of HCl} = \frac{0.002 \: \text{mol}}{0.1 \: \text{L}} = 0.02 \: \text{mol L}^{-1}$

The pH of a strong acid can be calculated using:

$\text{pH} = -\log[\text{H}^+]$

Substituting the concentration:

$\text{pH} = -\log(0.02) \approx 1.7$