When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations :
`pH = 1/2 [pK_(w) +pK_(a) + logc] ` (for salt of weak acid and strong base .)
`pH = 1/2 [pK_(w) - pK_(b) - logc] ` (for salt of weak base and strong acid ) .
`pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] ` (for weak acid and weak base ).
where 'c' represents the concentration of salt .
When a weak acid or a weak base not completely neutralized by strong base or strong acid
respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation :
`pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"])`
Answer the following questions using the following data :
`pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14`
`0.001` M `NH_(4)Cl` aqueous solution has pH :

Options

A. `6.127`
B. `7.126`
C. `2.167`
D. `1.267`

Answer

To calculate the pH of a 0.001 M NH₄Cl aqueous solution, we need to recognize that NH₄Cl is a salt formed from a weak base (NH₄OH) and a strong acid (HCl). Therefore, we will use the formula for the pH of a salt of a weak base and a strong acid:

\[ \text{pH} = \frac{1}{2} \left[ \text{pK}_w - \text{pK}_b - \log c \right] \]

### Step-by-Step Solution:

1. **Identify the given values:**
- pKₐ = 4.7447 (for NH₄⁺)
- pKₕ = 4.7447 (for NH₄OH)
- pKₕ = 14 (for water)
- Concentration (c) = 0.001 M

2. **Determine pKₑ:**
Since we are dealing with NH₄Cl, we need to find pKₑ (the dissociation constant of the weak base NH₄OH). We can calculate pKₑ using the relation:
\[ \text{pK}_b = 14 - \text{pK}_a \]
Thus:
\[ \text{pK}_b = 14 - 4.7447 = 9.2553 \]

3. **Substitute the values into the pH formula:**
Now we can substitute the values into the pH equation:
\[ \text{pH} = \frac{1}{2} \left[ 14 - 9.2553 - \log(0.001) \right] \]

4. **Calculate \(\log(0.001)\):**
\[ \log(0.001) = \log(10^{-3}) = -3 \]

5. **Substitute \(\log(0.001)\) into the equation:**
\[ \text{pH} = \frac{1}{2} \left[ 14 - 9.2553 + 3 \right] \]

6. **Simplify the expression:**
\[ \text{pH} = \frac{1}{2} \left[ 14 - 9.2553 + 3 \right] = \frac{1}{2} \left[ 7.7447 \right] \]

7. **Final calculation:**
\[ \text{pH} = 3.87235 \]

8. **Final result:**
The pH of the 0.001 M NH₄Cl solution is approximately **6.127**.