The pH of 0.01 MBOH solution is 10 . What is its degree of dissociation? (Given K_b of BOH is 1 10^-6 )

Question

Accepted Answer

Correct answer: D. (a) BOH ⇌B^++ OH^- K_b= (B^+ ) ( OH^- ) [ BOH ]=1 10^-6 pH=10 ⇒ pOH=14-10=4 ⇒ ( OH^- )=10^-4M . Now, K_b= C^21- = ( OH^- ) ^21- ⇒ K_b ( OH^- )= ^21- ≅ ^2= 10^-610^-4=10^-2 ⇒ =10^-1 or0.1 or10 \%

$The pH of 0.01 MBOH solution is 10 . What is its degree of dissociation? (Given K_{b} of BOH is 1 \times 10^{- 6})$

Solution

More Ionic Equilibrium PYQs