Assuming complete dissociation, calculate the `pH` of the following solutions,
a. `0.003 M HCl, b. 0.005 M NaOH`,
c. `0.002 M HBr, d. 0.002 M KOH`

Explanation

To calculate the pH of the given solutions, we will follow these steps for each solution:

### a. For 0.003 M HCl

1. **Identify the dissociation**: HCl dissociates completely into H⁺ and Cl⁻ ions. Therefore, the concentration of H⁺ ions is the same as the concentration of HCl, which is 0.003 M.

\[ \text{[H⁺]} = 0.003 \, \text{M} \]

2. **Calculate pH**: The pH is calculated using the formula:

\[ \text{pH} = -\log[\text{H⁺}] \]

Substituting the values:

\[ \text{pH} = -\log(0.003) \]

3. **Solve the logarithm**:
\[ \text{pH} = -\log(3 \times 10^{-3}) \]
\[ = -(\log(3) + \log(10^{-3})) \]
\[ = -\log(3) + 3 \]
\[ \approx 2.52 \]

### b. For 0.005 M NaOH

1. **Identify the dissociation**: NaOH dissociates completely into Na⁺ and OH⁻ ions. Therefore, the concentration of OH⁻ ions is 0.005 M.

\[ \text{[OH⁻]} = 0.005 \, \text{M} \]

2. **Calculate [H⁺] using Kw**: The ion product of water (Kw) at 25°C is \( 1.0 \times 10^{-14} \).

\[ \text{[H⁺]} = \frac{K_w}{[\text{OH⁻}]} = \frac{1.0 \times 10^{-14}}{0.005} \]
\[ = 2.0 \times 10^{-12} \, \text{M} \]

3. **Calculate pH**:

\[ \text{pH} = -\log(2.0 \times 10^{-12}) \]
\[ = -(\log(2) + \log(10^{-12})) \]
\[ = -\log(2) + 12 \]
\[ \approx 11.70 \]

### c. For 0.002 M HBr

1. **Identify the dissociation**: HBr dissociates completely into H⁺ and Br⁻ ions. Therefore, the concentration of H⁺ ions is 0.002 M.

\[ \text{[H⁺]} = 0.002 \, \text{M} \]

2. **Calculate pH**:

\[ \text{pH} = -\log(0.002) \]
\[ = -\log(2 \times 10^{-3}) \]
\[ = -(\log(2) + \log(10^{-3})) \]
\[ = -\log(2) + 3 \]
\[ \approx 2.70 \]

### d. For 0.002 M KOH

1. **Identify the dissociation**: KOH dissociates completely into K⁺ and OH⁻ ions. Therefore, the concentration of OH⁻ ions is 0.002 M.

\[ \text{[OH⁻]} = 0.002 \, \text{M} \]

2. **Calculate [H⁺] using Kw**:

\[ \text{[H⁺]} = \frac{K_w}{[\text{OH⁻}]} = \frac{1.0 \times 10^{-14}}{0.002} \]
\[ = 5.0 \times 10^{-12} \, \text{M} \]

3. **Calculate pH**:

\[ \text{pH} = -\log(5.0 \times 10^{-12}) \]
\[ = -(\log(5) + \log(10^{-12})) \]
\[ = -\log(5) + 12 \]
\[ \approx 11.30 \]

### Summary of pH Values:
- **pH of 0.003 M HCl**: 2.52
- **pH of 0.005 M NaOH**: 11.70
- **pH of 0.002 M HBr**: 2.70
- **pH of 0.002 M KOH**: 11.30