Question

What $\%$ of the carbon in the $H_{2}C{O}_3-HC{O}_3^{-}$ buffer should be in the form of $HC{O}_3^{-}$ so as to have a neutral solution? $(K_a=4\times {10}^{-7})$

• A. $20\%$
• B. $40\%$
• C. $60\%$
• D. $80\%$

Answer 1

The correct option is D $80\%$

Using Henderson Hasselbalch Equation for buffer, $pH=p{K}_a+log\frac{A^{-}}{HA}$

$pH=p{K}_a+log\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}$ --(1)

$p{K}_a=-log\left(K_a\right)=-log(4\times {10}^{-7})=-log4+7$

pH = 7 (Neutral solution)

Substitute in 1:

$7=-log4+7+log\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}$

$log\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=log4$

$\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=\frac{4}{1}$

Thus, percentage of buffer in the form $HC{O}_3^{-}=\frac{4}{5}\times 100=80$ %