When a salt reacts with water resulting into formation of acidic or basic solution, the process is referred to as salt hydrolysis. The pH of salt solution can be calculated using the following equations.
`pH = ( 1)/(2) ( p K_(w) + pK_(a) + log C)` for salt of weak acid and strong base.
`pH = (1)/(2) ( pK_(w)- pK_(b) - log C ) ` for salt of weak base and strong acid.
`pH = (1)/(2) ( pK_(w) + pK_(a) - pK_(b))` for salt of weak acid and weak base
The pH of 1M `PO_(4(aq))^(3-)` soluiton will be ( given `pK_(b) ` of `PO_(4)^(3-)` = 1.62 )

Options

A. 1.62
B. 12.38
C. 13.19
D. 7

Answer

To find the pH of a 1M phosphate ion solution (PO4^(3-)), we will use the provided equations for salt hydrolysis. Since PO4^(3-) is the salt of a weak base (HPO4^(2-)) and a strong acid (HCl), we will use the appropriate formula for calculating pH.

### Step-by-Step Solution:

1. **Identify the Type of Salt**:
- PO4^(3-) is the conjugate base of the weak acid H3PO4. Therefore, it behaves as a weak base in solution.

2. **Use the Correct pH Formula**:
- For the salt of a weak base and strong acid, the formula to calculate pH is:
\[ pH = \frac{1}{2} (pK_w - pK_b - \log C) \]
- Here, \(C\) is the concentration of the salt solution, which is given as 1M.

3. **Find \(pK_b\)**:
- We are given \(pK_b\) of PO4^(3-) as 1.62.

4. **Calculate \(pK_w\)**:
- The value of \(pK_w\) at 25°C is 14.

5. **Substitute Values into the Formula**:
- Substitute \(pK_w = 14\), \(pK_b = 1.62\), and \(C = 1\) into the formula:
\[ pH = \frac{1}{2} (14 - 1.62 - \log 1) \]
- Since \(\log 1 = 0\), the equation simplifies to:
\[ pH = \frac{1}{2} (14 - 1.62) \]

6. **Calculate the Result**:
- Calculate \(14 - 1.62\):
\[ 14 - 1.62 = 12.38 \]
- Now divide by 2:
\[ pH = \frac{12.38}{2} = 6.19 \]

7. **Final Calculation**:
- The final pH of the 1M phosphate ion solution is approximately:
\[ pH \approx 6.19 \]

### Summary of the Solution:
The pH of a 1M phosphate ion solution (PO4^(3-)) is approximately 6.19.