Find pH of 0.1 M `NaHCO_(3)`?K1=5*10^-7, K2=5*10^-11 for carbonic acids.

Options

A. 8.68
B. 9.68
C. 7.68
D. 5.58

Answer

To find the pH of a 0.1 M NaHCO₃ solution, we can follow these steps:

### Step 1: Identify the relevant equilibrium reactions
NaHCO₃ (sodium bicarbonate) can act as a weak base in water. The relevant equilibrium reaction is:
\[ \text{NaHCO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 + \text{Na}^+ + \text{OH}^- \]

### Step 2: Determine the Kb for NaHCO₃
We know the dissociation constants for carbonic acid:
- \( K_1 = 5 \times 10^{-7} \) (for H₂CO₃ dissociating to HCO₃⁻)
- \( K_2 = 5 \times 10^{-11} \) (for HCO₃⁻ dissociating to CO₃²⁻)

Using the relationship between Kb and Ka:
\[ K_b = \frac{K_w}{K_a} \]
Where \( K_w = 1 \times 10^{-14} \) at 25°C.

For the bicarbonate ion (HCO₃⁻), we can use \( K_1 \) as \( K_a \):
\[ K_b = \frac{1 \times 10^{-14}}{5 \times 10^{-7}} = 2.0 \times 10^{-8} \]

### Step 3: Set up the equilibrium expression
Let \( x \) be the concentration of OH⁻ produced at equilibrium. The equilibrium expression can be written as:
\[ K_b = \frac{[OH^-]^2}{[NaHCO_3] - x} \]
Assuming \( x \) is small compared to the initial concentration (0.1 M), we can approximate:
\[ K_b \approx \frac{x^2}{0.1} \]

Substituting the value of \( K_b \):
\[ 2.0 \times 10^{-8} = \frac{x^2}{0.1} \]

### Step 4: Solve for x
Rearranging the equation gives:
\[ x^2 = 2.0 \times 10^{-8} \times 0.1 \]
\[ x^2 = 2.0 \times 10^{-9} \]
\[ x = \sqrt{2.0 \times 10^{-9}} \]
\[ x \approx 4.47 \times 10^{-5} \, \text{M} \]

### Step 5: Calculate pOH
Now we can find the pOH:
\[ pOH = -\log[OH^-] \]
\[ pOH = -\log(4.47 \times 10^{-5}) \]
\[ pOH \approx 4.35 \]

### Step 6: Calculate pH
Finally, we can find the pH using the relationship:
\[ pH + pOH = 14 \]
\[ pH = 14 - pOH \]
\[ pH = 14 - 4.35 \]
\[ pH \approx 9.65 \]

### Final Answer
The pH of a 0.1 M NaHCO₃ solution is approximately **9.65**.

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