Question

The $O{H}^{-}$ concentration in a mixture of 5.0mL of 0.0504 M$N{H}_{4}Cl$ and 2 mL of $0.0210MN{H}_3$ solution is $x\times {10}^{-6}M$. The value of x is .......(Nearest integer) [Given $K_W=1\times {10}^{-14}andKb=1.8\times {10}^{-5}$]

Answer 1

Answer is 3

$\left[N{H}_4^{+}\right]=0.0504\&\left[N{H}_3\right]=0.0210$

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

$\left[H{O}^{-}\right]=\frac{K_b\times \left[N{H}_3\right]}{\left[N{H}_4^{+}\right]}=1.8\times {10}^{-5}\times \frac{2}{5}\times \frac{210}{504}$

$=3\times {10}^{-6}$