Question

When 0.1 mole of $N{H}_3$ is dissolved in water to make 1L of solution, the $[OH{]}^{-}$ of solution is $1.34\times {10}^{-3}$M.Calculate $K_b$ for $N{H}_3$

Answer 1

Reaction involved-

$N{H}_3+H_{2}O\to N{H}_4++O{H}^{-}$

Given-

$4N{H}_4^{+}=O{H}^{-}=1.34\times {10}^{-3}$ M

$N{H}_3$ reacted $=(0.1-1.34\times {10}^{-3})$ M

Condition at equilibrium-

$=(0.1-1.34\times {10}^{-3})$ M

$=0.1-0.00134$ M

= 0.09866 M

$K_b=\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]/\left[N{H}_3\right]$

$=(1.34\times {10}^{-3})\times 1.34\times {10}^{-3})/0.09866$

$=1.8\times {10}^{-5}$