Pure water is netural in nature `[H^(+)]=[OH^(-)].` When this condition is disturbed by changing the concentration of `H^(+)or OH^(-),` the natural solution changes to acidic `{[H^(+)]gt[OH^(-)]}` or basic `{[H^(+)]lt[OH^(-)]}.` This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation
(i) Salt of weak acid and strong base
`pH =1/2[pK_(w)+pK_(a)+logC]`
(ii) Salt of weak base and strong acid
`pH=1/2[pK_(w)-pK_(b)-logC]`
(iii) For salt of weak base and weak acid
`pH=1/2[pK_(w)+pK_(a)-pK_(b)]`
The pH of buffer can be calculated using t he following formula
`pH=pK_(a)+log""(["Salt"])/(["Acid"])`
`pOH=pK_(b)=log""(["Salt"])/(["Base"])`
Answer t he following questions when
`pK_(a)=4.7447`
`pK_(b)=4.7447` ltb rgt `pK_(w)=14`
When 50 ml of 0.1 m NaOH is added of 50 ml of `0.1 MCH_(3)COOH` solution the pH will be

Options

A. `4.7447`
B. `9.2553`
C. `8.7218`
D. `1.6020`

Answer

To solve the problem of calculating the pH when 50 ml of 0.1 M NaOH is added to 50 ml of 0.1 M CH₃COOH (acetic acid), we can follow these steps:

### Step 1: Calculate the moles of NaOH and CH₃COOH
First, we need to calculate the number of moles of NaOH and CH₃COOH.

- **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \, (5 \, \text{mmol}) \]

- **Moles of CH₃COOH**:
\[ \text{Moles of CH₃COOH} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \, (5 \, \text{mmol}) \]

### Step 2: Determine the reaction
When NaOH (a strong base) reacts with CH₃COOH (a weak acid), they will neutralize each other to form sodium acetate (CH₃COONa) and water.

\[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

Since both reactants are in equal amounts (5 mmol each), they will completely react with each other, producing 5 mmol of sodium acetate.

### Step 3: Calculate the concentration of the resulting solution
After the reaction, the total volume of the solution will be:
\[ \text{Total Volume} = 50 \, \text{ml} + 50 \, \text{ml} = 100 \, \text{ml} = 0.1 \, \text{L} \]

The concentration of sodium acetate (C) will be:
\[ C = \frac{\text{Moles of CH}_3\text{COONa}}{\text{Total Volume}} = \frac{0.005 \, \text{moles}}{0.1 \, \text{L}} = 0.05 \, \text{M} \]

### Step 4: Use the formula for pH of a salt formed from a weak acid and a strong base
Since sodium acetate is the salt of a weak acid (acetic acid) and a strong base (NaOH), we use the formula:
\[ \text{pH} = \frac{1}{2} \left( pK_w + pK_a + \log C \right) \]

Given:
- \( pK_w = 14 \)
- \( pK_a = 4.7447 \)
- \( C = 0.05 \, \text{M} \)

### Step 5: Calculate the pH
First, calculate \( \log C \):
\[ \log(0.05) \approx -1.3 \]

Now substitute the values into the pH formula:
\[ \text{pH} = \frac{1}{2} \left( 14 + 4.7447 + (-1.3) \right) \]
\[ = \frac{1}{2} \left( 14 + 4.7447 - 1.3 \right) \]
\[ = \frac{1}{2} \left( 17.4447 \right) \]
\[ = 8.72235 \approx 8.722 \]

### Conclusion
Thus, the pH of the solution after adding 50 ml of 0.1 M NaOH to 50 ml of 0.1 M CH₃COOH is approximately **8.722**.