Question

What is $[NH_4^ + ]$ in a solution that contain $0.02M$ $N{H_3}, ({K_b} = 1.8 \times {10^{ - 5}})$ and $0.01M\;KOH$?
A. $9 \times {10^{ - 6}}$
B. $1.8 \times {10^{ - 3}}$
C. $3.6 \times {10^{ - 5}}$
D. None of these

Answer 1

Hint:As we know that base constant of a solution is the dissociation or ionisation of a base. Also we know that when in the solution of weak electrolytes another weak electrolyte is added then the degree of dissociation of both weak electrolytes decreases due to the common ion effect formed in solution.

Formula used: ${K_b} = \dfrac{{[NH_4^ + ][O{H^ - }]}}{{[N{H_3}][{H_2}O]}}$

Complete answer:
Now we know that when in the solution of weak electrolytes another strong electrolyte is added then the degree of dissociation of weak electrolytes decreases due to the common ion effect formed in solution.
So we are given with potassium hydroxide which is a strong base so it will completely dissociate into solution and it will be given as:
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;KOH \rightleftharpoons {K^ + } + O{H^ - }$

Initially:	$0.01$	$0$	$0$
At equilibrium:	$0$	$0.01$	$0.01$

Now if we talk about ammonium ions further, we know that it is obtained from weak acid so:
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;N{H_3}(aq) + {H_2}O(l) \rightleftharpoons NH_4^ + (aq) + O{H^ - }(aq)$

Initially:	$0.02$	-	$0$	$0$
At equilibrium:	$0.02 - x$	-	$x$	$x + 0.01$

We can say that due to the low value of base constant ${K_b}$ and common ion effect of hydroxide ion we can neglect the value of $x$ with respect to $0.01$ because the value of $x$ would be less after solving so we can neglect it.
Now using the formula:
${K_b} = \dfrac{{[NH_4^ + ][O{H^ - }]}}{{[N{H_3}][{H_2}O]}}$
We can solve for the concentration of ammonium ions by substituting all the values:
${\Rightarrow K_b} = \dfrac{{[x][x + 0.01]}}{{[0.02 - x]}}$
$\Rightarrow 1.8 \times {10^{ - 5}} = \dfrac{{x \times 0.01}}{{0.02}}$
$\Rightarrow x = 2 \times 1.8 \times {10^{ - 5}}$
$\Rightarrow x = 3.6 \times {10^{ - 5}}$
Therefore, the concentration of ammonium ions in the solution is $3.6 \times {10^{ - 5}}$.

Hence the correct answer is (C).

Note:
Remember higher the value of base constant, higher will be the dissociation suggesting that the base is stronger and when the value of base constant is low, lower will be the dissociation suggesting that the base is weak and thus do not completely dissociate into the solution. Dissociation also depends on temperature, if dissociation is Endothermic then it will increase with increase in temperature and if it is exothermic, it will decrease with increase in temperature.