Question

What is $\left[N{H}_4^{+}\right]$ in a solution that contain $0.02MN{H}_3(K_b=1.8\times {10}^{-5})$ and $0.01MKOH$?

• A. $9\times {10}^{-6}$
• B. $1.8\times {10}^{-3}$
• C. $3.6\times {10}^{-5}$
• D. None of these

Answer 1

The correct option is C $3.6\times {10}^{-5}$

Given: $K_b=1.8\times {10}^{-5},N{H}_3=0.02,KOH=0.01$

Solution: $KOH\iff K^{+}+O{H}^{-}$ (KOH is a strong electrolyte)

At eq 0.01 0 0

(Ammonia is a weak electrolyte)

$N{H}_3+H_2{O}_{\left(l\right)}\iff N{H}_{4\left(aq\right)}^{+}+O{H}_{\left(aq\right)}^{-}$

(0.02-x) x (0.01+ x) (at equillibrium)

Here the value of x is very less than 0.01. So we can neglect the value of x.

Then, (0.01-x) is changed to 0.01 and (0.02-x) is changed to 0.02

$K_b=\frac{0.01\times x}{0.02}$

$x=1.8\times {10}^{-5}\times 2=3.6\times {10}^{-5}$

Hence the correct option is C.