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

To find the number of moles of HCl, we use the same formula:

For HCl:

$\text{Moles} = 0.10 \text{ mol L}^{-1} \times 0.060 \text{ L} = 0.006 \text{ mol}$

When NaOH and HCl are mixed, they will react in a 1:1 ratio:

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

Since we have 0.004 mol of NaOH and 0.006 mol of HCl, NaOH is the limiting reagent.

After the reaction, the moles of HCl remaining can be calculated as follows:

$\text{Remaining HCl} = 0.006 \text{ mol} - 0.004 \text{ mol} = 0.002 \text{ mol}$

The total volume of the mixed solution is:

$\text{Total Volume} = 40 \text{ mL} + 60 \text{ mL} = 100 \text{ mL} = 0.100 \text{ L}$

The concentration of the remaining HCl is given by:

$\text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{0.002 \text{ mol}}{0.100 \text{ L}} = 0.020 \text{ mol L}^{-1}$

The pH of the solution can be calculated using:

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

Where

$[\text{H}^+] = 0.020 \text{ mol L}^{-1}$

Thus,

$\text{pH} = -\log(0.020) \approx 1.7$