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 strong 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)=4.75` lt rgt `pK_(w)=14`
When 50 ml of 0.1 M `NH_(4)OH` is added to 50 ml of 0.05 M HCl solution, the pH is nearly

Options

A. `1.60`
B. `12.40`
C. `4.75`
D. `9.25`

Answer

To solve the problem, we need to calculate the pH of the solution when 50 mL of 0.1 M NH₄OH is mixed with 50 mL of 0.05 M HCl. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the moles of NH₄OH and HCl
- **Moles of NH₄OH:**
\[ \text{Concentration} = 0.1 \, \text{M} \quad \text{and} \quad \text{Volume} = 50 \, \text{mL} = 0.050 \, \text{L} \]
\[ \text{Moles of NH₄OH} = \text{Concentration} \times \text{Volume} = 0.1 \times 0.050 = 0.005 \, \text{mol} = 5 \, \text{mmol} \]

- **Moles of HCl:**
\[ \text{Concentration} = 0.05 \, \text{M} \quad \text{and} \quad \text{Volume} = 50 \, \text{mL} = 0.050 \, \text{L} \]
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.05 \times 0.050 = 0.0025 \, \text{mol} = 2.5 \, \text{mmol} \]

### Step 2: Determine the reaction between NH₄OH and HCl
The reaction is:
\[ \text{NH₄OH} + \text{HCl} \rightarrow \text{NH₄Cl} + \text{H₂O} \]
From the stoichiometry of the reaction, 1 mole of NH₄OH reacts with 1 mole of HCl.

### Step 3: Calculate the remaining moles after the reaction
- **Moles of NH₄OH remaining:**
\[ \text{Initial moles of NH₄OH} - \text{Moles of HCl} = 5 \, \text{mmol} - 2.5 \, \text{mmol} = 2.5 \, \text{mmol} \]

- **Moles of HCl remaining:**
\[ 2.5 \, \text{mmol} - 2.5 \, \text{mmol} = 0 \, \text{mmol} \]

- **Moles of NH₄Cl formed:**
\[ 2.5 \, \text{mmol} \, (\text{from the reaction with HCl}) \]

### Step 4: Identify the resulting solution
After the reaction, we have:
- 2.5 mmol of NH₄OH (weak base)
- 2.5 mmol of NH₄Cl (conjugate acid)

This creates a buffer solution consisting of a weak base (NH₄OH) and its conjugate acid (NH₄Cl).

### Step 5: Calculate the pOH using the buffer formula
Using the formula for pOH of a buffer:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Here, both the salt (NH₄Cl) and the base (NH₄OH) have the same moles (2.5 mmol), so their concentrations will be equal, leading to:
\[ \frac{[\text{Salt}]}{[\text{Base}]} = 1 \]
Thus, \(\log(1) = 0\).

Now substituting the values:
\[ \text{pOH} = 4.75 + 0 = 4.75 \]

### Step 6: Calculate the pH
Using the relation:
\[ \text{pH} + \text{pOH} = 14 \]
We can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4.75 = 9.25 \]

### Final Answer
The pH of the resulting solution is **9.25**.

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