Question

To maintain the $pH$ of $7.4$ for blood at normal condition which is $2M$ in $H_{2}C{O}_3$ (at equilibrium), what volume of $7.8MNaHC{O}_3$ solution is required to mix with $10\text{mL}$ of blood?
$\text{Given}:K\left(H_{2}C{O}_3\right)=7.8\times {10}^{-7};{10}^{7.4}=2.511\times {10}^7$

• A. $25.11\text{mL}$
• B. $100.44\text{mL}$
• C. $50.22\text{mL}$
• D. $5.022\text{mL}$

Answer 1

The correct option is C $50.22\text{mL}$

$\left[H_{2}C{O}_3\right]=2M{V}_{\text{blood}}=10\text{mL}\left[NaHC{O}_3\right]=7.8M\text{Let, V mL 7.8 M}NaHC{O}_3\text{is mixed}\therefore \text{Total}V=(10+V)\text{mL}\therefore [H_{2}C{O}_3{]}_{\text{mix}}=\frac{2\times 10}{(V+10)}[NaHC{O}_3]=\frac{7.8\times V}{(V+10)}\therefore pH=pKa+\log \frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}\Rightarrow 7.4=-\log (7.8\times {10}^{-7})+\log \frac{7.8V}{20}\Rightarrow 7.4=\log (\frac{7.8V}{20}\times \frac{1}{7.8\times {10}^{-7}})\Rightarrow (10{)}^{7.4}=\frac{V}{20}\times {10}^7\Rightarrow 2.511\times {10}^7=\frac{V}{20}\times {10}^7\Rightarrow V=50.22\text{mL}$