JEE MAIN - Chemistry (2021 - 26th August Morning Shift - No. 16)
The OH$$-$$ concentration in a mixture of 5.0 mL of 0.0504 M NH4Cl and 2 mL of 0.0210 M NH3 solution is x $$\times$$ 10$$-$$6 M. The value of x is ___________. (Nearest integer)
[Given Kw = 1 $$\times$$ 10$$-$$14 and Kb = 1.8 $$\times$$ 10$$-$$5]
[Given Kw = 1 $$\times$$ 10$$-$$14 and Kb = 1.8 $$\times$$ 10$$-$$5]
Answer
3
Explanation
$$\left[ {NH_4^ + } \right]$$ = 0.0504 & [NH3] = 0.0210
So, $${K_b} = {{[NH_4^ + ][H{O^ - }]} \over {[N{H_3}]}}$$
$$[H{O^ - }] = {{{K_b} \times [N{H_3}]} \over {[NH_4^ + ]}} = 1.8 \times {10^{ - 5}} \times {2 \over 5} \times {{210} \over {504}} = 3 \times {10^{ - 6}}$$
So, $${K_b} = {{[NH_4^ + ][H{O^ - }]} \over {[N{H_3}]}}$$
$$[H{O^ - }] = {{{K_b} \times [N{H_3}]} \over {[NH_4^ + ]}} = 1.8 \times {10^{ - 5}} \times {2 \over 5} \times {{210} \over {504}} = 3 \times {10^{ - 6}}$$
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