Boric acid `Br(OH)_(3)` is monobasic Lewis acid if gives `H^(+)` ions in aquoeus soluions as follows `B(OH)_(3) + H_(2)O hArr [B(OH)_(4)]^(-) + H^(+)`
`K_(a) = 5.9 xx 10^(-10)` The `P^(H)` of `0.025M` aqueous solution is `[log 5.9 = 0.771]`

Options

A. `5.42`
B. `4.52`
C. `2.45`
D. `4.675`

Answer

A

Explanation

`P^(H) = - 1/2 [Log K_(a) + logC]`
`= - 1/2["log"(5.9 xx 10^(-10)) + "log" (0.025)]`
`= - 1/2[(-10 + "log"5.9) + (-2 + "log" 2.85)]`
`= - 1/2[(-10 + 0.771 )+ (-2 + 0.3979)]`
`= - 1/2[(-9229) + (-16021)]`
`= - 1/2 [-9.229 - 1.6021]`
`=- 1/2[-10.3811] = 5.4255`