Calculate pH of basic solution having `OH^(-)` as (i) `10^(-2)` M and ( ii) `10^(-8)` M

Answer

To calculate the pH of a basic solution given the concentration of hydroxide ions (OH⁻), we can follow these steps:

### Step 1: Calculate pOH from OH⁻ concentration
The formula to calculate pOH is:
\[ \text{pOH} = -\log[\text{OH}^-] \]

#### (i) For \([\text{OH}^-] = 10^{-2} \, \text{M}\):
\[ \text{pOH} = -\log(10^{-2}) = 2 \]

#### (ii) For \([\text{OH}^-] = 10^{-8} \, \text{M}\):
\[ \text{pOH} = -\log(10^{-8}) = 8 \]

### Step 2: Calculate pH using the relationship between pH and pOH
We know that at 25°C:
\[ \text{pH} + \text{pOH} = 14 \]

#### (i) For \(\text{pOH} = 2\):
\[ \text{pH} = 14 - \text{pOH} = 14 - 2 = 12 \]

#### (ii) For \(\text{pOH} = 8\):
\[ \text{pH} = 14 - \text{pOH} = 14 - 8 = 6 \]

### Step 3: Adjust for the second case
In the second case, since the concentration of OH⁻ is \(10^{-8} \, \text{M}\), we need to consider the contribution of water's autoionization, which produces \(10^{-7} \, \text{M}\) of OH⁻. Therefore, the total concentration of OH⁻ becomes:
\[ [\text{OH}^-] = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \, \text{M} \]

Now recalculate pOH:
\[ \text{pOH} = -\log(1.1 \times 10^{-7}) \approx 7 - 0.041 = 6.959 \]

Then calculate pH:
\[ \text{pH} = 14 - 6.959 \approx 7.041 \]

### Final Answers:
1. For \([\text{OH}^-] = 10^{-2} \, \text{M}\), \(\text{pH} = 12\)
2. For \([\text{OH}^-] = 10^{-8} \, \text{M}\), \(\text{pH} \approx 7.041\)