What is the pH of the resultant solution after 20.0 mL of 0.20 mol L$^{-1}$ HCl(aq) is mixed with 20.0 mL of 0.50 mol L$^{-1}$ NaOH(aq)? - HSC - SSCE Chemistry - Question 15 - 2021 - Paper 1

Likewise, calculate the moles of NaOH:

$\text{Moles of NaOH} = 0.50 \: \text{mol L}^{-1} \times 0.020 \: \text{L} = 0.010 \: \text{mol}$

When HCl and NaOH react, they do so in a 1:1 ratio.

After mixing, the moles of HCl (0.004 mol) are less than the moles of NaOH (0.010 mol), making HCl the limiting reagent.

Subtract the moles of HCl from NaOH:

$\text{Remaining moles of NaOH} = 0.010 \: \text{mol} - 0.004 \: \text{mol} = 0.006 \: \text{mol}$

The total volume after mixing is:

$\text{Total Volume} = 20.0 \: \text{mL} + 20.0 \: \text{mL} = 40.0 \: \text{mL} = 0.040 \: \text{L}$

The concentration of NaOH is given by:

$\text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{0.006 \: \text{mol}}{0.040 \: \text{L}} = 0.15 \: \text{mol L}^{-1}$

To find pOH:

$\text{pOH} = -\log[0.15] \approx 0.823$

Now convert to pH:

$\text{pH} + \text{pOH} = 14 \\ \text{pH} = 14 - 0.823 \approx 13.177$

Thus, rounding to one decimal place, the resultant pH is approximately 13.2.