Question

If the dissociation constant of $N{H}_{4}OH$ is $1.8\times {10}^{-5}$, the concentration of $\stackrel{⊝}{O}H$ ions, in mol $L^{-1}$ of 0.1 M $N{H}_{4}OH$ is:

• A. $1.8\times {10}^{-6}$
• B. $1.34\times {10}^{-3}$
• C. $4.20\times {10}^{-2}$
• D. $5.0\times {10}^{-2}$

Answer 1

The correct option is B $1.34\times {10}^{-3}$

As we know, conc. of $\left[O{H}^{-}\right]=C\alpha =\sqrt{K_b\times C}$

So, by putting given values,

$\left[O{H}^{-}\right]=\sqrt{1.8\times {10}^{-5}\times 0.1}=1.34\times {10}^{-3}$

Option B is correct.