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Chemistry Question on Equilibrium

0.01 M solution of H2AH_2A has pH equal to 4. If Ka1K_{a1} for the acid is 4.45×10−74.45 \times 10^{-7}, the concentration of HA−HA^- ion in solution would be

A

0.01 M

B

4.45×10−54.45 \times 10^{-5}

C

8.0×10−58.0 \times 10^{-5}

D

unpredictable

Answer

4.45×10−54.45 \times 10^{-5}

Explanation

Solution

H2A<=>HA−+H+H_{2}A {<=>} HA^{-}+H^{+} Now, [HA−]=Ka×[H2A][H+]\left[HA^{-}\right] = \frac{K_{a} \times \left[H_{2}A\right]}{\left[H^{+}\right] } =4.45×10−7×1×10−21×10−4=\frac{4.45\times10^{-7}\times1\times10^{-2}}{1\times10^{-4}}