Question

$0.01\, M$ solution of $ H_{2}A $ has pH equal to $4$, if $ K_{a_{1}} $ for the acid is $ 4.45\times 10^{-7}, $ the concentration of $ HA^{-} $ ion in solution would be

• A. $0.01 \,N$
• B. $ 4.45\times 10^{-5} $
• C. $ 8.0\times 10^{-5} $
• D. unpredictable

Answer 1

Answer: (B)

$H_{2} A H^{+}+H A^{-}$as
$p H=4\left[H^{+}\right]=10^{-4} M$
$K_{a_{1}}=\frac{\left[H^{+}\right]\left[H A^{-}\right]}{\left[H_{2} A\right]}$
Or $\frac{10^{-4} \times\left[H A^{-}\right]}{10^{-2}}=4.45 \times 10^{-7}$
Or $\left[H A^{-}\right]=4.45 \times 10^{-5}$