The `pH` of blood is `7,4`. If the buffer in blood constitute `CO_(2)` and `HCO_(3)^(Theta)` ions, calculate the ratio of conjugate base of acid `(H_(2)CO_(3))` to maintain the `pH` of blood. Given `K_(1)` of `H_(2)CO_(3) = 4.5 xx 10^(-7)`.

Options

A. `11.25`
B. `10.0`
C. `8.5`
D. None

Explanation

To solve the problem, we need to calculate the ratio of the conjugate base (HCO3^-) to the acid (H2CO3) in the blood buffer system, given the pH and the dissociation constant (K1) for carbonic acid (H2CO3).

### Step-by-Step Solution:

1. **Understand the Relationship**:
The dissociation of carbonic acid (H2CO3) can be represented as:
\[ H2CO3 \rightleftharpoons H^+ + HCO3^- \]
The equilibrium constant (K1) for this reaction is given by:
\[ K1 = \frac{[H^+][HCO3^-]}{[H2CO3]} \]

2. **Calculate the Concentration of H+**:
The pH of the blood is given as 7.4. We can find the concentration of hydrogen ions (H+) using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-7.4} \]
Calculating this gives:
\[ [H^+] \approx 4 \times 10^{-8} \, \text{M} \]

3. **Substitute Known Values into K1 Expression**:
We know K1 for H2CO3 is \(4.5 \times 10^{-7}\). We can rearrange the K1 expression to find the ratio of the conjugate base to the acid:
\[ K1 = \frac{[H^+][HCO3^-]}{[H2CO3]} \]
Rearranging gives:
\[ \frac{[HCO3^-]}{[H2CO3]} = \frac{K1}{[H^+]} \]

4. **Plug in the Values**:
Substituting the known values into the equation:
\[ \frac{[HCO3^-]}{[H2CO3]} = \frac{4.5 \times 10^{-7}}{4 \times 10^{-8}} \]

5. **Calculate the Ratio**:
Performing the division:
\[ \frac{[HCO3^-]}{[H2CO3]} = \frac{4.5}{4} \times 10^{1} = 1.125 \times 10 = 11.25 \]

6. **Conclusion**:
The ratio of the conjugate base (HCO3^-) to the acid (H2CO3) required to maintain the pH of blood is approximately 11.25.

### Final Answer:
The ratio of HCO3^- to H2CO3 is **11.25**.