Question

The ph of aqueous ammonia is 11. Find molarity of solution kb of nh4oh is 10^-5

Answer 1

Answer: Molarity of solution is 0.1 M.

Explanation:

The pH of the aqueous ammonia is 11. Then pOH of the aqueous ammonia will be :

Since, pH + pOH = 14 the pOH will be 3.

$pOH=-log[OH^-]$

$3=-log[OH^-]$

$[OH^-]=0.001 M$

$NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)$

Concentration of hydroxide ion will be equal to the concentration of the ammonium ions formed in the aqueous solution of ammonia.

$[OH^-]=[NH_4^+]$

Expression of $K_b$ is written as:

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

On substituting the value in above expression we will get the concentration of undissociated ammonia present in the solution.

$10^{-5}=\frac{0.001\times 0.001}{[NH_3]}$

$[NH_3]=0.1 M$

Molarity of solution is 0.1 M.