Calculate the pH of the following aqueous solutions:
(i) `5 xx 10^(-8)"M HCl"`
(ii) `5 xx 10^(-10)" M HCl "`
(iii) `10^(-8) "M NaOH"`
(iv) `10^(-10)" M NaOH"`

Answer

To solve the problem of calculating the pH of the given aqueous solutions, we will follow these steps for each solution:

### (i) Calculate the pH of `5 x 10^(-8) M HCl`

1. **Identify the concentration of H⁺ ions from HCl**:
- HCl is a strong acid and dissociates completely in water. Therefore, the concentration of H⁺ from HCl is `5 x 10^(-8) M`.

2. **Identify the concentration of H⁺ ions from water**:
- Pure water contributes H⁺ ions at a concentration of `1 x 10^(-7) M`.

3. **Calculate the total concentration of H⁺ ions**:
- Total H⁺ = Concentration from HCl + Concentration from water
- Total H⁺ = `5 x 10^(-8) + 1 x 10^(-7) = 1.5 x 10^(-7) M`.

4. **Calculate the pH**:
- pH = -log[H⁺]
- pH = -log(1.5 x 10^(-7)).
- pH ≈ 6.82.

### (ii) Calculate the pH of `5 x 10^(-10) M HCl`

1. **Identify the concentration of H⁺ ions from HCl**:
- HCl concentration = `5 x 10^(-10) M`.

2. **Identify the concentration of H⁺ ions from water**:
- Water contributes H⁺ at `1 x 10^(-7) M`.

3. **Calculate the total concentration of H⁺ ions**:
- Since `5 x 10^(-10) M` is negligible compared to `1 x 10^(-7) M`, we can approximate:
- Total H⁺ ≈ `1 x 10^(-7) M`.

4. **Calculate the pH**:
- pH = -log(1 x 10^(-7)).
- pH = 7.

### (iii) Calculate the pH of `10^(-8) M NaOH`

1. **Identify the concentration of OH⁻ ions from NaOH**:
- NaOH concentration = `10^(-8) M`.

2. **Identify the concentration of OH⁻ ions from water**:
- Water contributes OH⁻ at `1 x 10^(-7) M`.

3. **Calculate the total concentration of OH⁻ ions**:
- Total OH⁻ = `10^(-8) + 1 x 10^(-7) = 1.1 x 10^(-7) M`.

4. **Calculate the pOH**:
- pOH = -log[OH⁻]
- pOH = -log(1.1 x 10^(-7)).
- pOH ≈ 6.96.

5. **Calculate the pH**:
- pH = 14 - pOH = 14 - 6.96 = 7.04.

### (iv) Calculate the pH of `10^(-10) M NaOH`

1. **Identify the concentration of OH⁻ ions from NaOH**:
- NaOH concentration = `10^(-10) M`.

2. **Identify the concentration of OH⁻ ions from water**:
- Water contributes OH⁻ at `1 x 10^(-7) M`.

3. **Calculate the total concentration of OH⁻ ions**:
- Since `10^(-10) M` is negligible compared to `1 x 10^(-7) M`, we can approximate:
- Total OH⁻ ≈ `1 x 10^(-7) M`.

4. **Calculate the pOH**:
- pOH = -log(1 x 10^(-7)).
- pOH = 7.

5. **Calculate the pH**:
- pH = 14 - pOH = 14 - 7 = 7.

### Summary of Results:
- (i) pH of `5 x 10^(-8) M HCl` = 6.82
- (ii) pH of `5 x 10^(-10) M HCl` = 7
- (iii) pH of `10^(-8) M NaOH` = 7.04
- (iv) pH of `10^(-10) M NaOH` = 7