The pH of a solution of `B(OH)_2` is 10.6. Calculate its solubility and solubility product.

Explanation

pH=10.6
pH+pOH=14
or pOH=14-10.6 =3.4
-log `[OH^-]` =3.4
or log `[OH^-]=-3.4=4.6`
`[OH^-]=3.98xx10^(-4) "mol L"^(-1)`
Now, `B(OH)_2` dissociates as :
`B(OH)_2 hArr B^(+)+2OH^-`
It is clear that 1 mol of `B(OH)_2` gives 2 mol of `OH^-`
Solubility of `B(OH)_2=1/2[OH^-]`
`=1/2xx(3.98xx10^(-4))` `=1.99xx10^(-4) "mol L"^(-1)`
Now, `K_(sp)=[B^+][OH^-]^2`
`therefore [B^+]=1.99xx10^(-4) "mol L"^(-1)`
`[OH^-]=3.98xx10^(-4) "mol L"^(-1)`
`therefore K_(sp)=(1.99xx10^(-4))xx(3.98xx10^(-4))^2`
`=3.15xx10^(-11)`