`2.0 g`of diborane `(B_(2)H_(6))` reacts with water to product `100mL` solution. If `K_(a)` for `H_(3)BO_(3)` is `7.3 xx 10^(-10)`, calculated the `pH` of solution.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation.
Diborane \( B_2H_6 \) reacts with water to produce boric acid \( H_3BO_3 \) and hydrogen gas \( H_2 \). The balanced equation is:

\[ B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 3H_2 \]

### Step 2: Calculate the number of moles of diborane \( B_2H_6 \).
The molar mass of diborane \( B_2H_6 \) can be calculated as follows:
- Atomic mass of Boron (B) = 10.8 g/mol
- Atomic mass of Hydrogen (H) = 1 g/mol

\[ \text{Molar mass of } B_2H_6 = 2 \times 10.8 + 6 \times 1 = 21.6 + 6 = 27.6 \text{ g/mol} \]

Now, calculate the moles of \( B_2H_6 \) in 2 g:

\[ \text{Moles of } B_2H_6 = \frac{2 \text{ g}}{27.6 \text{ g/mol}} \approx 0.0725 \text{ moles} \]

### Step 3: Determine the moles of \( H_3BO_3 \) produced.
From the balanced equation, 1 mole of \( B_2H_6 \) produces 2 moles of \( H_3BO_3 \). Therefore, the moles of \( H_3BO_3 \) produced are:

\[ \text{Moles of } H_3BO_3 = 2 \times 0.0725 \approx 0.145 \text{ moles} \]

### Step 4: Calculate the concentration of \( H_3BO_3 \).
The volume of the solution is given as 100 mL, which is 0.1 L. The concentration \( C \) of \( H_3BO_3 \) can be calculated as:

\[ C = \frac{\text{Moles of } H_3BO_3}{\text{Volume in L}} = \frac{0.145 \text{ moles}}{0.1 \text{ L}} = 1.45 \text{ M} \]

### Step 5: Set up the equilibrium expression for \( H_3BO_3 \).
The dissociation of \( H_3BO_3 \) can be represented as:

\[ H_3BO_3 \rightleftharpoons H^+ + BO_3^{3-} \]

Let \( \alpha \) be the degree of dissociation. The equilibrium concentrations will be:
- \( [H_3BO_3] = C(1 - \alpha) \)
- \( [H^+] = C\alpha \)
- \( [BO_3^{3-}] = C\alpha \)

The expression for \( K_a \) is:

\[ K_a = \frac{[H^+][BO_3^{3-}]}{[H_3BO_3]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \approx \frac{C\alpha^2}{C} = C\alpha^2 \quad (\text{assuming } \alpha \text{ is small}) \]

### Step 6: Substitute the values into the \( K_a \) expression.
Given \( K_a = 7.3 \times 10^{-10} \) and \( C = 1.45 \):

\[ 7.3 \times 10^{-10} = 1.45 \alpha^2 \]

### Step 7: Solve for \( \alpha \).
Rearranging gives:

\[ \alpha^2 = \frac{7.3 \times 10^{-10}}{1.45} \approx 5.034 \times 10^{-10} \]

Taking the square root:

\[ \alpha \approx \sqrt{5.034 \times 10^{-10}} \approx 2.24 \times 10^{-5} \]

### Step 8: Calculate the concentration of \( H^+ \).
The concentration of \( H^+ \) ions is:

\[ [H^+] = C\alpha = 1.45 \times 2.24 \times 10^{-5} \approx 3.25 \times 10^{-5} \text{ M} \]

### Step 9: Calculate the pH.
The pH is given by:

\[ pH = -\log[H^+] = -\log(3.25 \times 10^{-5}) \approx 4.49 \]

Thus, the final pH of the solution is approximately **4.49**.

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