# pH Calculation After Mixing Acid and Base

This is part of HSC Chemistry course under the topic Using Brønsted-Lowry Theory.

### HSC Chemistry Syllabus

• Calculate the pH of the resultant solution when solutions of acids and/or bases are diluted or mixed

### How to Approach These Questions

The pH of the final solution after mixing an acid and a base depends on the substance in excess.

• When an acid is in excess, the pH of the final mixture will be < 7.0
• When a base is in excess, the pH of the final mixture will be > 7.0

Therefore, the objective of this type of questions is to calculate the number of moles of acid and base, and determine whether the acid or the base is in excess.

### Example 1

Calculate the pH after 30 mL of 0.10 mol L–1 sodium hydroxide is added to 50 mL of 0.30 mol L­–1 hydrochloric acid solution. Include a balanced equation in your answer.

Solution:

$$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$$

$$n(NaOH) = c \times V = 0.10 \times 0.030 = 0.0030 \hspace{0.1cm} mol$$

Moles of HCl present:

$$n(HCl) = c \times V = 0.30 \times 0.050 = 0.015 \hspace{0.1cm} mol$$

Since 1 mol HCl reacts with 1 mol NaOH, then all 0.0030 mol of NaOH will be neutralised by the HCl

Moles of NaOH remaining:

$$n(NaOH)_{excess} = 0.015 - 0.0030 = 0.012 \hspace{0.1cm} mol$$

Volume solution =

$$V = 30 + 50 = 80 \hspace{0.1cm} mL$$

Remaining [HCl]:

$$c(HCl) = \frac{0.012}{0.08} = 0.15 \hspace{0.1cm} mol \hspace{0.1cm} L^{-1}$$

Since HCl is a strong acid, [H+] = [HCl] = 0.15 mol/L

pH = –log(0.15) = 0.82 (2 s.f.)