The pH of a buffer solution containing 0.1 M `CH_3COOH and 0.1 M CH_3COONa` is 4.74. What will be the pH if 0.05 mole of HCl is added to one litre of this buffer solution? `K_a(CH_3COOH) = 1.8xx10^(-5).`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the pKa of acetic acid (CH₃COOH)
Given that the pH of the buffer solution is 4.74, we can use this information to find the pKa of acetic acid.

We know that:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

In this case, both the salt (CH₃COONa) and the acid (CH₃COOH) are at the same concentration of 0.1 M, so:
\[ \log\left(\frac{0.1}{0.1}\right) = \log(1) = 0 \]

Thus, we have:
\[ \text{pH} = \text{pKa} \]
So,
\[ \text{pKa} = 4.74 \]

### Step 2: Calculate the new concentrations after adding HCl
When we add 0.05 moles of HCl to the buffer solution, HCl will react with the acetate ion (CH₃COO⁻) to form more acetic acid (CH₃COOH).

The initial concentrations before adding HCl are:
- \([CH₃COOH] = 0.1 \, \text{M}\)
- \([CH₃COO⁻] = 0.1 \, \text{M}\)

After adding 0.05 moles of HCl to 1 liter of the buffer solution:
- The amount of acetate ion decreases by 0.05 moles.
- The amount of acetic acid increases by 0.05 moles.

New concentrations:
- \([CH₃COOH] = 0.1 + 0.05 = 0.15 \, \text{M}\)
- \([CH₃COO⁻] = 0.1 - 0.05 = 0.05 \, \text{M}\)

### Step 3: Calculate the new pH using the Henderson-Hasselbalch equation
Now we can use the Henderson-Hasselbalch equation again to find the new pH:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Substituting the values we have:
\[ \text{pH} = 4.74 + \log\left(\frac{0.05}{0.15}\right) \]

Calculating the logarithm:
\[ \log\left(\frac{0.05}{0.15}\right) = \log\left(\frac{1}{3}\right) \approx -0.477 \]

Now substituting this back into the equation:
\[ \text{pH} = 4.74 - 0.477 \]
\[ \text{pH} \approx 4.263 \]

### Final Answer
The new pH of the buffer solution after adding 0.05 moles of HCl is approximately **4.26**.

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