Question

Calculate : $K_b$ for $B{\left({OH}_4\right)}^{-}$ given $K_a\left(B{\left(OH\right)}_3\right)=6\times {10}^{-10}$

• A. $8.3\times {10}^{-6}$
• B. $1.66\times {10}^{-5}$
• C. $3.3\times {10}^{-5}$
• D. None of these

Answer 1

The correct option is B $1.66\times {10}^{-5}$

As we know,

$K_a\times K_b={10}^{-14}$ and

Given, $K_a\left(B{\left(OH\right)}_3\right)=6\times {10}^{-10}$

Therefore, $K_b$ = $1.66\times {10}^{-5}$