50 mL of 0.05 M `Na_(2)CO_(3)` is titrated against 0.1 M HCl. On adding 40 mL of HCl, pH of the solution will be `[Given : For H_(2)CO_(3),pK_(a1)=6.35,pK_(a2)=10.33,log 3=0.477,log 2=0.30]`

Options

A. `6.35`
B. `6.526`
C. `8.34`
D. `6.173`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the number of milliequivalents of Na₂CO₃
The formula to calculate milliequivalents (mEq) is:
\[ \text{mEq} = \text{Volume (mL)} \times \text{Molarity (M)} \]
For Na₂CO₃:
- Volume = 50 mL
- Molarity = 0.05 M

Calculating:
\[ \text{mEq of Na₂CO₃} = 50 \, \text{mL} \times 0.05 \, \text{M} = 2.5 \, \text{mEq} \]

### Step 2: Calculate the number of milliequivalents of HCl
Using the same formula:
For HCl:
- Volume = 40 mL
- Molarity = 0.1 M

Calculating:
\[ \text{mEq of HCl} = 40 \, \text{mL} \times 0.1 \, \text{M} = 4 \, \text{mEq} \]

### Step 3: Write the reaction equations
The reactions that occur are:
1. Na₂CO₃ + HCl → NaCl + NaHCO₃
2. NaHCO₃ + HCl → NaCl + H₂CO₃

### Step 4: Determine the resulting species after the reaction
From the reactions:
- 2.5 mEq of Na₂CO₃ reacts with HCl.
- 4 mEq of HCl is added.

Since Na₂CO₃ can react with HCl in a 1:1 ratio, we can see how much of each species remains:
- After 2.5 mEq of Na₂CO₃ reacts with 2.5 mEq of HCl, we have:
- 1.5 mEq of HCl remaining (4 - 2.5 = 1.5)
- 2.5 mEq of NaHCO₃ formed (from the first reaction)
- 1 mEq of H₂CO₃ formed (from the second reaction)

### Step 5: Identify the buffer solution
The resulting solution contains:
- 1 mEq of NaHCO₃ (bicarbonate)
- 1.5 mEq of H₂CO₃ (carbonic acid)

### Step 6: Calculate the pH using the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Here:
- pK₁ (for H₂CO₃) = 6.35
- Base = NaHCO₃ (1 mEq)
- Acid = H₂CO₃ (1.5 mEq)

Substituting the values:
\[ \text{pH} = 6.35 + \log\left(\frac{1}{1.5}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{1}{1.5}\right) = \log(1) - \log(1.5) = 0 - 0.1761 = -0.1761 \]
So,
\[ \text{pH} = 6.35 - 0.1761 = 6.1739 \approx 6.17 \]

### Final Answer
The pH of the solution after adding 40 mL of HCl is approximately **6.17**.

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