In a buffer solution consisting of a mixture of weak base and its salt, the ratio of salt to base is increases 10 times , the pOH of the solution will

Options

A. Increase by one units
B. Decrease by one units
C. Increase by ten units
D. Decrease by ten units

Answer

To solve the problem, we need to analyze how the pOH of a buffer solution changes when the ratio of salt to base is increased by 10 times.

### Step-by-Step Solution:

1. **Understanding the Buffer Solution**:
A buffer solution consists of a weak base and its corresponding salt. The pOH of a basic buffer can be expressed using the formula:
\[ \text{pOH} = \text{pK}_B + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]

2. **Initial Condition**:
Initially, we assume that the concentrations of salt and base are equal. Therefore, the ratio of salt to base is:
\[ \frac{[\text{Salt}]}{[\text{Base}]} = 1 \]
Substituting this into the pOH formula gives:
\[ \text{pOH} = \text{pK}_B + \log(1) = \text{pK}_B + 0 = \text{pK}_B \]

3. **Change in Ratio**:
Now, the problem states that the ratio of salt to base is increased 10 times. This means:
\[ \frac{[\text{Salt}]}{[\text{Base}]} = 10 \]

4. **Calculating New pOH**:
Substituting the new ratio into the pOH formula:
\[ \text{pOH} = \text{pK}_B + \log(10) \]
Since \(\log(10) = 1\), we have:
\[ \text{pOH} = \text{pK}_B + 1 \]

5. **Change in pOH**:
Initially, the pOH was \(\text{pK}_B\). After increasing the ratio of salt to base, the new pOH is \(\text{pK}_B + 1\). This indicates that the pOH has increased by 1 unit.

### Conclusion:
Thus, the pOH of the solution will increase by 1 unit.