Diborane, `B_(2) H_(6)`, reacts with water to from boric acid and hydrogen , What is the `pH` of the solution which results when `1.104 g` of `B_(2)H_(6)` reacts with `100 mL` water ? Assume the final volume to be `100 mL`.
Given : `K_(a)` of `H_(3) BO_(3) = 8 xx 10^(-10) pK_(a)` , Atomic weight of `B = 10,8 g , MW` of `B_(2) H_(6) = 27.6 g mol^(-1)`.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the moles of diborane (B₂H₆)
Given:
- Mass of diborane (B₂H₆) = 1.104 g
- Molecular weight (MW) of B₂H₆ = 27.6 g/mol

Using the formula for moles:
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