Question

The pH of blood stream is maintained by a proper balance of $H_{2}C{O}_3$ and $NaHC{O}_3$ concentration. What volume of $5M$ $NaHC{O}_3$ solution should be mixed with a $10$ mL sample of blood which is $2M$ in $H_{2}C{O}_3$ in order to maintain its pH?

• A. $40$ mL
• B. $38$ mL
• C. $50$ mL
• D. $78$ mL

Answer 1

The correct option is D $78$ mL

Let $V$ mL of $5M$ $NaHC{O}_3$ solution is added.
Total volume will be $V+10$ mL.
The concentration of salt, $NaHC{O}_3$ will be $\frac{5M\times VmL}{V+10mL}$

The concentration of acid $H_{2}C{O}_3$ will be $\frac{2M\times 10mL}{V+10mL}$

Hence, the ratio $\frac{\left[Salt\right]}{\left[Acid\right]}$ will be $\frac{\frac{5M\times VmL}{V+10mL}}{\frac{2M\times 10ml}{V+10mL}}=\frac{5M\times VmL}{2M\times 10mL}$
The expression for the pH of the solution is $pH=p{K}_a+log\frac{\left[Salt\right]}{\left[Acid\right]}$
Substituting values in the above expression, we get
$7.4=6.11+log\frac{5M\times VmL}{2M\times 10mL}$
$1.29=log\frac{V}{4}$
$19.5=\frac{V}{4}$ $⟹V=78$
Hence, $78$ mL of $5M$ $NaHC{O}_3$ solution is added.