Single Choice

If $$0.00050$$ mol $$NaH{CO}_{3}$$ is added to $$1$$ litre of a buffered solution of $$pH\ 8$$, then how much material will exist in each of the three forms $${H}_{2}{CO}_{3},H{CO}_{3}^{-}$$ and $${CO}_{3}^{2-}$$? For $${H}_{2}{CO}_{3}$$, $${K}_{1}=5\times {10}^{-7}$$; $${K}_{2}=5\times {10}^{-13}$$

A$$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$$
Correct Answer
B$$[{H}_{2}{CO}_{3}]=4.9\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=9.85\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$$
C$$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=2.45\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=4.9\times {10}^{-8}M$$
D$$[{H}_{2}{CO}_{3}]=92.45\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=9.85\times {10}^{-8}M$$

Solution

Given, $$pH=8, [{H}^{+}]={10}^{-8}, [{OH}^{-}={10}^{-6}$$

$$H{CO}_{3}^{-}\rightleftharpoons {H}^{+}+{CO}_{3}^{-2}$$ $$K=5\times {10}^{-13}$$
$$0.0005$$
$$0.0005-y-z$$ $${10}^{-8}$$ $$y$$
$$H{CO}_{3}^{-}+{H}_{2}O\rightleftharpoons {H}_{2}{CO}_{3}+{OH}^{-}$$ $$K=\cfrac{{K}_{w}}{{K}_{1}}=2\times {10}^{-8}$$
$$0.0005$$
$$0.0005-y-z$$ $$z$$ $${10}^{-6}$$
Since equilibrium constant for first reaction is very less, $$y<$$2\times {10}^{-8}=\cfrac{z({10}^{-6})}{0.0005-z}$$
$$51z=0.0005,\Rightarrow z=9.8\times {10}^{-6}$$
$$[{H}_{2}{CO}_{3}]=9.8\times {10}^{-6}M$$
$$[H{CO}_{3}^{-}]=0.0005-9.8\times {10}^{-6}=4.9\times {10}^{-4}M$$
$$5\times {10}^{-3}=\cfrac{{10}^{-8}\times y}{4.9\times {10}^{-4}}$$
$$[{CO}_{3}^{2-}]=y=2.45\times {10}^{-8}M$$

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The total number of different kind of buffers obtained during the titration of $$H_3PO_4$$ with $$NaOH$$ are:

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