The pH of blood stream is maintained by a proper balance of `H_(2)CO_(3)` and `NaHCO_(3)` concentration. What volume of 5M concentration. `NaHCO_(3)` solution should 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 is "4.0xx10^(-7))`?

Options

A. 40 ml
B. 35 ml
C. 25 ml
D. 38 ml

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Volume of blood sample (V_blood) = 10 ml
- Concentration of H₂CO₃ (C_H₂CO₃) = 2 M
- Desired pH = 7.4
- \( K_a \) for H₂CO₃ = \( 4.0 \times 10^{-7} \)
- Concentration of NaHCO₃ (C_NaHCO₃) = 5 M

2. **Calculate the Number of Moles of H₂CO₃:**
\[ \text{Moles of H₂CO₃} = C_{H₂CO₃} \times V_{blood} = 2 \, \text{mol/L} \times 0.01 \, \text{L} = 0.02 \, \text{mol} \]

3. **Calculate \( pK_a \):**
\[ pK_a = -\log(K_a) = -\log(4.0 \times 10^{-7}) \approx 6.4 \]

4. **Use the Henderson-Hasselbalch Equation:**
The Henderson-Hasselbalch equation is given by:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Here, [A⁻] is the concentration of NaHCO₃ and [HA] is the concentration of H₂CO₃.

5. **Set Up the Equation:**
\[ 7.4 = 6.4 + \log\left(\frac{C_{NaHCO₃}}{C_{H₂CO₃}}\right) \]
Rearranging gives:
\[ \log\left(\frac{C_{NaHCO₃}}{C_{H₂CO₃}}\right) = 7.4 - 6.4 = 1 \]
Therefore:
\[ \frac{C_{NaHCO₃}}{C_{H₂CO₃}} = 10^1 = 10 \]

6. **Express Concentrations in Terms of Volume:**
Let the volume of NaHCO₃ solution added be \( V \) ml. The total volume after adding NaHCO₃ will be \( 10 + V \) ml.

- Concentration of H₂CO₃:
\[ C_{H₂CO₃} = \frac{0.02 \, \text{mol}}{0.01 + \frac{V}{1000}} \quad \text{(in L)} \]
- Concentration of NaHCO₃:
\[ C_{NaHCO₃} = \frac{5V \, \text{mol/L}}{0.01 + \frac{V}{1000}} \quad \text{(in L)} \]

7. **Substituting into the Ratio:**
\[ \frac{5V}{0.02} = 10 \]
Rearranging gives:
\[ 5V = 0.2 \quad \Rightarrow \quad V = \frac{0.2}{5} = 0.04 \, \text{L} = 40 \, \text{ml} \]

### Final Answer:
The volume of 5M NaHCO₃ solution required is **40 ml**.