Question

If the equilibrium constant of BOH $\rightleftarrows$B⁺ + OH^– at 25ºC is 2.5 × 10^–6 , therefore equilibrium constant for neutralisation of BOH against strong acid (HCl) is-

• A. 4 × 10^–9
• B. 2.5 × 10⁸
• C. 4 × 10⁹
• D. 2.5 × 10^–8

Answer 1

Answer: (B)

2.5 × 10⁸

Answer 2

BOH + H⁺ $\xrightarrow { \mathrm { Keq } }$ B⁺ + H₂O

K_eq = $\frac { \left[ \mathrm { B } ^ { + } \right] } { [ \mathrm { BOH } ] \left[ \mathrm { H } ^ { + } \right] }$ = $\frac { \left[ \mathrm { B } ^ { + } \right] \left[ \mathrm { OH } ^ { - } \right] } { [ \mathrm { BOH } ] }$

× $\frac { 1 } { \left[ \mathrm { H } ^ { + } \right] \left[ \mathrm { OH } ^ { - } \right] }$ =

K_eq = $\frac { 2.5 \times 10 ^ { - 6 } } { 10 ^ { 14 } }$ = 2.5 × 10⁸