Calculate the pH of `0.010M NaHCO_(3)` solution.
`K_(1)=4.5xx10^(-7) , K_(2)= 4.7xx10^(-11)` for carbonic acid.

Answer

To calculate the pH of a `0.010 M NaHCO₃` solution, we need to consider the dissociation of carbonic acid (H₂CO₃) and its bicarbonate ion (HCO₃⁻). The relevant dissociation constants for carbonic acid are given as:

- \( K_1 = 4.5 \times 10^{-7} \)
- \( K_2 = 4.7 \times 10^{-11} \)

### Step-by-Step Solution:

1. **Understanding the Dissociation**:
Sodium bicarbonate (NaHCO₃) is a salt that will dissociate in solution to give bicarbonate ions (HCO₃⁻). The bicarbonate ion can act as a weak acid and can further dissociate to produce hydrogen ions (H⁺) and carbonate ions (CO₃²⁻). The relevant equilibria are:
- \( H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \) with \( K_1 \)
- \( HCO_3^- \rightleftharpoons H^+ + CO_3^{2-} \) with \( K_2 \)

2. **Using the Formula for H⁺ Concentration**:
For a weak diprotic acid like carbonic acid, the concentration of H⁺ ions can be approximated using the formula:
\[ [H^+] = \sqrt{K_1 \cdot K_2} \]

3. **Calculating H⁺ Concentration**:
Substitute the values of \( K_1 \) and \( K_2 \):
\[ [H^+] = \sqrt{(4.5 \times 10^{-7}) \cdot (4.7 \times 10^{-11})} \]
First, calculate the product:
\[ 4.5 \times 4.7 = 21.15 \]
Then add the exponents:
\[ 10^{-7} \times 10^{-11} = 10^{-18} \]
So,
\[ [H^+] = \sqrt{21.15 \times 10^{-18}} = \sqrt{21.15} \times 10^{-9} \]
Calculating \( \sqrt{21.15} \):
\[ \sqrt{21.15} \approx 4.59 \]
Thus,
\[ [H^+] \approx 4.59 \times 10^{-9} \, \text{M} \]

4. **Calculating pH**:
Now, we can calculate the pH using the formula:
\[ pH = -\log[H^+] \]
Substitute the value of \( [H^+] \):
\[ pH = -\log(4.59 \times 10^{-9}) \]
Using the logarithmic property:
\[ pH = -\log(4.59) - (-9) = 9 - \log(4.59) \]
Calculate \( \log(4.59) \):
\[ \log(4.59) \approx 0.694 \]
Therefore,
\[ pH = 9 - 0.694 = 8.306 \]

### Final Answer:
The pH of the `0.010 M NaHCO₃` solution is approximately **8.31**.