Calculate the pH of a solution obtained by diluting 1 mL of 0.10 M weak monoacidic base to 100 mL at constant temperature if Kb of the base is `1 xx 10^(-5)`.

Explanation

To calculate the pH of the solution obtained by diluting 1 mL of 0.10 M weak monoacidic base to 100 mL, we will follow these steps:

### Step 1: Calculate the concentration after dilution
We can use the dilution formula:
\[ M_1 V_1 = M_2 V_2 \]
Where:
- \( M_1 = 0.10 \, M \) (initial concentration)
- \( V_1 = 1 \, mL \) (initial volume)
- \( V_2 = 100 \, mL \) (final volume)

Rearranging the formula to find \( M_2 \):
\[ M_2 = \frac{M_1 V_1}{V_2} \]

Substituting the values:
\[ M_2 = \frac{0.10 \, M \times 1 \, mL}{100 \, mL} \]
\[ M_2 = \frac{0.10}{100} \]
\[ M_2 = 0.001 \, M \text{ or } 1 \times 10^{-3} \, M \]

### Step 2: Set up the equilibrium expression
For a weak base \( B \), the dissociation can be represented as:
\[ B + H_2O \rightleftharpoons BH^+ + OH^- \]

The equilibrium expression for the base dissociation constant \( K_b \) is:
\[ K_b = \frac{[BH^+][OH^-]}{[B]} \]

Given \( K_b = 1 \times 10^{-5} \).

### Step 3: Define the concentrations at equilibrium
Let \( C \) be the initial concentration of the base after dilution, which we found to be \( 1 \times 10^{-3} \, M \).

At equilibrium:
- \([BH^+] = x\)
- \([OH^-] = x\)
- \([B] = C - x \approx C\) (since \( x \) is small compared to \( C \))

Thus, we can write:
\[ K_b = \frac{x^2}{C} \]

Substituting the known values:
\[ 1 \times 10^{-5} = \frac{x^2}{1 \times 10^{-3}} \]

### Step 4: Solve for \( x \)
Rearranging gives:
\[ x^2 = 1 \times 10^{-5} \times 1 \times 10^{-3} \]
\[ x^2 = 1 \times 10^{-8} \]
\[ x = \sqrt{1 \times 10^{-8}} \]
\[ x = 1 \times 10^{-4} \]

### Step 5: Calculate the pOH
Since \( x \) represents the concentration of \( OH^- \):
\[ [OH^-] = 1 \times 10^{-4} \, M \]

Now, we can calculate pOH:
\[ pOH = -\log[OH^-] \]
\[ pOH = -\log(1 \times 10^{-4}) \]
\[ pOH = 4 \]

### Step 6: Calculate the pH
Using the relationship:
\[ pH + pOH = 14 \]
We can find pH:
\[ pH = 14 - pOH \]
\[ pH = 14 - 4 \]
\[ pH = 10 \]

### Final Answer
The pH of the solution is **10**.

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