**Problem Statement: – Calculate the pH of the following solutions:**

- 01M HCl
- 05M H
_{2}SO_{4} - 001M KOH

**Solution**:

**1. pH of 0.01M HCl solution**

Since HCl is a strong acid, it completely dissociates in aqueous solution

Therefore, the concentration of H^{+}, [H^{+}] will be 0.01M

Now, pH = – log[H^{+}]

= – log (0.01)

= 2

**2. pH of 0.05M H _{2}SO_{4} solution**

Since H_{2}SO_{4} is a strong acid, it completely dissociates in aqueous solution

Therefore, [H^{+}] = 0.01M

Now, pH = – log[H^{+}]

= – log (0.1)

= 1

**3. pH of 0.001M KOH solution**

Since KOH is a strong base, it completely dissociates in aqueous solution

Therefore, [OH^{–}] = 0.01M

Now, pOH = – log[OH^{–}]

= – log (0.001)

= 3

Hence, pH = 14 – pOH = 14 – 3 = 11