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 understand how the pOH of a buffer solution changes when the ratio of salt to base is altered. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Buffer Solution
A buffer solution consists of a weak base and its salt. The pOH of a basic buffer can be calculated using the Henderson-Hasselbalch equation, which for a basic buffer is given by:

\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]

### Step 2: Initial Ratio of Salt to Base
Let’s denote the initial concentration of the salt as \([\text{Salt}]\) and the concentration of the base as \([\text{Base}]\). The initial ratio of salt to base is:

\[ \text{Ratio}_1 = \frac{[\text{Salt}]}{[\text{Base}]} = X \]

Thus, the initial pOH can be expressed as:

\[ \text{pOH}_1 = \text{pK}_b + \log(X) \]

### Step 3: Increase the Ratio of Salt to Base
According to the problem, the ratio of salt to base is increased 10 times. Therefore, the new ratio becomes:

\[ \text{Ratio}_2 = \frac{[\text{Salt}]}{[\text{Base}]} = 10X \]

### Step 4: Calculate the New pOH
Now, we can calculate the new pOH using the new ratio:

\[ \text{pOH}_2 = \text{pK}_b + \log(10X) \]

Using the properties of logarithms, we can break this down:

\[ \text{pOH}_2 = \text{pK}_b + \log(10) + \log(X) \]

Since \(\log(10) = 1\), we can simplify this to:

\[ \text{pOH}_2 = \text{pK}_b + 1 + \log(X) \]

### Step 5: Relate New pOH to Initial pOH
Now, we can relate \(\text{pOH}_2\) to \(\text{pOH}_1\):

\[ \text{pOH}_2 = \text{pOH}_1 + 1 \]

### Conclusion
Thus, when the ratio of salt to base is increased 10 times, the pOH of the solution increases by 1 unit.

### Final Answer
The pOH of the solution will increase by 1 unit.

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