Question

When $0.100$ mole of ammonia, $NH_3$ is dissolved in sufficient water to make $1.0 \,L$ of solution, the solution is found to have a hydroxide ion concentration of $1.34 \times 10^{-3}\, M$. Calculate $K_b$ for ammonia.

• A. $1.56 \times 10^3$
• B. $1.82 \times 10^{-5}$
• C. $1.37 \times 10^{-3}$
• D. $1.28 \times 10^5$

Answer 1

Answer: (B)

$NH_{3}+H_{2}O { \rightleftharpoons} NH_{4}^{+}+OH^{-}$
At equi. $(0.100-1.34 \times 10^{-3})M 1.34\times 10^{-3}\,M 1.34 \times 10^{-3}\,M = 0.09866\, M$
$K_{b}=\frac{\left[NH_{4}^{+}\left[OH^{-}\right]\right]}{\left[NH_{3}\right]}=\frac{1.34\times10^{-3}\times1.34\times10^{-3}}{0.09866}$
$=1.8199 \times 10^{-5}\approx1.82 \times10^{-5}$