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Question

A solution contains $$0.2$$M $$NH_4OH$$ and $$0.2$$M $$NH_4Cl$$. If $$1.0$$ mL of $$0.001$$ M HCl is added to it. What will be the $$[OH^-]$$ of the resulting solution $$[K_b=2\times 10^{-5}]$$?


A
2×105
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B
5×1010
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C
2×103
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D
None of these
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Solution

The correct option is A $$2\times 10^{-5}$$
Concentration oF $$NH_4OH=0.2\,M$$

Concentration of $$NH_4Cl=0.2\,M$$

Concentration OF $$HCl=0.001\,M$$

$${K_b} = 2 \times {10^{ - 5}}$$
Applying Henderson Equation

$${POH} = {P^{{K_a}}} + \log \left[ {\frac{{salt}}{{base}}} \right]$$

$${POH} = 5 - \log 2 + \log \frac{{0.2}}{{0.2}}$$

           $$ = 5 - \log 2$$
Hence,
$$\left[ {O{H^ - }} \right] = 2 \times {10^{ - 5}}$$
so, the option $$(A)$$ is the correct answer.

Chemistry

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