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

`B_(2)H_(6) + 6H_(3) O rarr 2H_(3) BO_(3) + 2H_(2)`
`{:"1mole" ,"2 mole":}`
`27.6 g`
`27.6 g` of `B_(2) H_(6)` gives `2 `moles of `H_(3) BO_(3)`
:. `1.104 g of B_(2) H_(6)` gives `= (2xx 1.104)/(27.6) = 0.08 mol`
Since `0.08 mol` of `H_(3) BO_(3)` is present in `100 mL` (i.e. `0.1 L`), resulting molarity of `H_(3) BO_(3)` solution is
`M of H_(3) BO_(3) = (0.08 mol)/(0.1 L) = 0.8 M`
Since `H_(3) BO_(3)` is a weak acid, thereore `pH` of weak acid (i.e. `pH_(WA)`) is given by :
`pH_(WA) = (1)/(2) (pK_(a) - log C)`
=`(1)/(2) (9.1 - log 0.8)`
=`(1)/(2) [9.1 - log 0.8)`
=`(1)/(2) [9.1 -(0.1)] = 4.6`
:. `pH` of `H_(3) BO_(3) = 4.6`.
`[(log 0.8=log8xx10^(-1)),(=3 log2-1),(=3xx0.3-1),(=0.9-1),(=-0.1)]`.