`2.0 g`of dibrona `(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.

Answer

A, D

Explanation

`B_(2)H_(6) + 6H_(2)O rarr 2H_(3)BO_(3) + 3H_(2)`
`1mol (=27.6g) of B_(2)H_(6) = 2 mol H_(3)BO_(3)`
`:. 2.0 g of B_(2)H_(6) = (2xx2)/(27.6) = 0.145 "mole" H_(3)BO_(3)`
`:. [H_(3)BO_(3)] = (0.145 xx 1000)/(100) = 1.45M`
Now `H_(3)BO_(3) + H_(2)O hArr B(OH)_(4)^(Theta) + H^(o+)`
`K_(a) = (C alpha^(2))/(1-alpha) ( :' 1 - alpha =1)`
or `7.3 xx 10^(-10) = 1.45 xx alpha^(2) :. alpha = 2.24 xx 10^(-5)`
`[H^(o+)] = C alpha = 1.45 xx 2.24 xx 10^(-5) = 3.25 xx 10^(-5)`
`pH = 4.4881`