The pH of blood stream is maintained by a proper balance of `H_(2)CO_(3)` and `NaHCO_(3)` concentrations. What volume of 5 M `NaHCO_(3)` solution, shnould be mixed with 10 mL sample of blood, which is 2 M in `H_(2)CO_(3)` in order to maintain a pH of `7.4 (K_(a)for H_(2)CO_(3)`in blood =`7.8xx10^(-7)) `

Options

A. 40 mL
B. 38 mL
C. 50 mL
D. 79 mL

Answer

To solve the problem, we need to determine the volume of a 5 M NaHCO₃ solution that should be mixed with a 10 mL sample of blood containing 2 M H₂CO₃ in order to maintain a pH of 7.4.

### Step-by-Step Solution:

1. **Identify Given Values:**
- pH = 7.4
- Concentration of H₂CO₃ = 2 M
- Volume of blood sample = 10 mL
- Concentration of NaHCO₃ = 5 M
- \( K_a \) for H₂CO₃ = \( 7.8 \times 10^{-7} \)

2. **Calculate \( pK_a \):**
\[ pK_a = -\log(K_a) = -\log(7.8 \times 10^{-7}) \approx 6.1 \]

3. **Set Up the Henderson-Hasselbalch Equation:**
The Henderson-Hasselbalch equation is given by:
\[ pH = pK_a + \log\left(\frac{[NaHCO₃]}{[H₂CO₃]}\right) \]

4. **Substitute Known Values into the Equation:**
\[ 7.4 = 6.1 + \log\left(\frac{[NaHCO₃]}{[H₂CO₃]}\right) \]

5. **Rearranging the Equation:**
\[ \log\left(\frac{[NaHCO₃]}{[H₂CO₃]}\right) = 7.4 - 6.1 = 1.3 \]

6. **Convert Logarithmic Form to Exponential Form:**
\[ \frac{[NaHCO₃]}{[H₂CO₃]} = 10^{1.3} \approx 19.95 \]

7. **Calculate the Concentration of H₂CO₃:**
The concentration of H₂CO₃ in the final solution can be calculated as follows:
\[ [H₂CO₃] = \frac{2 \, \text{mol/L} \times 10 \, \text{mL}}{V + 10 \, \text{mL}} = \frac{20}{V + 10} \, \text{mol/L} \]

8. **Calculate the Concentration of NaHCO₃:**
The concentration of NaHCO₃ can be calculated as:
\[ [NaHCO₃] = \frac{5 \, \text{mol/L} \times V \, \text{mL}}{V + 10 \, \text{mL}} = \frac{5V}{V + 10} \, \text{mol/L} \]

9. **Set Up the Ratio:**
Using the ratio from step 6:
\[ \frac{5V}{V + 10} = 19.95 \times \frac{20}{V + 10} \]

10. **Cross-Multiply and Simplify:**
\[ 5V = 19.95 \times 20 \]
\[ 5V = 399 \]
\[ V = \frac{399}{5} \approx 79.8 \, \text{mL} \]

11. **Final Answer:**
The volume of 5 M NaHCO₃ solution required is approximately **79 mL**.