Question

The $pH$ of a blood stream is maintained by a proper balance of $H_{2}C{O}_3$ and $NaHC{O}_3.$ What volume of 5 M $NaHC{O}_3$ solution should be mixed with 10 mL of a sample of solution which is 2.5 M in $H_{2}C{O}_3,$ in order to maintain a $pH=\mathrm{7.4.}$ (Take $p{K}_{a_1}$ for $H_{2}C{O}_3=6.7,log2=0.3,log5=0.7)$

• A. 10 mL
• B. 25 mL
• C. 40 mL
• D. 50 mL

Answer 1

The correct option is B 25 mL

Given, $p{K}_a=6.7,pH=7.4,\left[NaHC{O}_3\right]=5M,\left[H_{2}C{O}_3\right]=2.5M,V=10mL={10}^{-3}L$
Using Henderson-Hasselbalch equation:
$pH=p{K}_a+log(\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}$
$pH=p{K}_{a_1}+log\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}$
$\Rightarrow 7.4=6.7+log\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}$
$\Rightarrow log\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=0.7\Rightarrow \frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=5$
$\therefore 5\times$ moles of $H_{2}C{O}_3=$ moles of $HC{O}_3^{-}$ (since both are components of same solution, volume of solution would be same for both)
$\Rightarrow 5\times 2.5\times 10\times {10}^{-3}=5\times V$.
Required volume = 0.025 L = 25 mL.