Find the pH of `" 0.1 M NaHCO_(3)"`.
Use data `(K_(1)=4xx10^(-7),K_(2)=4xx10^(-11)` for `H_(2)CO_(3), log 4=0.6) :`

Options

A. 3.7
B. 8.4
C. 9.6
D. None of these

Explanation

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

### Step 1: Identify the relevant equilibrium reaction
NaHCO₃ (sodium bicarbonate) can act as a weak base in water. It can react with water to form H₂CO₃ (carbonic acid) and OH⁻ (hydroxide ions):
\[ \text{NaHCO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 + \text{NaOH} \]

### Step 2: Determine the dissociation constant (Kb)
Since NaHCO₃ is a weak base, we can calculate its Kb using the relation:
\[ K_b = \frac{K_w}{K_a} \]
Where:
- \( K_w = 1 \times 10^{-14} \) (the ion product of water)
- \( K_a = K_1 = 4 \times 10^{-7} \) (given for H₂CO₃)

Calculating Kb:
\[ K_b = \frac{1 \times 10^{-14}}{4 \times 10^{-7}} = 2.5 \times 10^{-8} \]

### Step 3: Set up the equilibrium expression
For the reaction:
\[ \text{NaHCO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 + \text{OH}^- \]
Let \( x \) be the concentration of OH⁻ produced at equilibrium. The equilibrium expression for Kb is:
\[ K_b = \frac{[\text{H}_2\text{CO}_3][\text{OH}^-]}{[\text{NaHCO}_3]} \]
Substituting the concentrations:
\[ K_b = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \]
(since \( x \) will be very small compared to 0.1 M)

### Step 4: Solve for x (OH⁻ concentration)
Setting the equation:
\[ 2.5 \times 10^{-8} = \frac{x^2}{0.1} \]
\[ x^2 = 2.5 \times 10^{-8} \times 0.1 \]
\[ x^2 = 2.5 \times 10^{-9} \]
\[ x = \sqrt{2.5 \times 10^{-9}} = 4.47 \times 10^{-5} \, \text{M} \]

### Step 5: Calculate pOH
Using the concentration of OH⁻:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(4.47 \times 10^{-5}) \]
Using a calculator:
\[ \text{pOH} \approx 4.35 \]

### Step 6: Calculate pH
Using the relation:
\[ \text{pH} + \text{pOH} = 14 \]
Thus:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4.35 = 9.65 \]

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

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