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-)` solution 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 \( \text{PO}_4^{3-} \) solution, we will follow these steps:

### Step 1: Identify the nature of the salt
The ion \( \text{PO}_4^{3-} \) is the conjugate base of a weak acid (phosphoric acid, \( \text{H}_3\text{PO}_4 \)). Therefore, the salt will undergo hydrolysis to produce a basic solution.

### Step 2: Use the appropriate formula
Since \( \text{PO}_4^{3-} \) is a salt of a weak acid and a strong base, we will use the formula:
\[ \text{pH} = \frac{1}{2} \left( \text{pK}_w + \text{pK}_a + \log C \right) \]
However, we need to find \( \text{pK}_a \) first.

### Step 3: Calculate \( \text{pK}_b \) and \( \text{K}_b \)
Given \( \text{pK}_b \) of \( \text{PO}_4^{3-} = 1.62 \):
\[ \text{K}_b = 10^{-\text{pK}_b} = 10^{-1.62} \approx 2.39 \times 10^{-2} \]

### Step 4: Calculate \( \text{pK}_a \)
To find \( \text{pK}_a \), we can use the relationship:
\[ \text{pK}_a + \text{pK}_b = \text{pK}_w \]
At 25°C, \( \text{pK}_w = 14 \). Thus,
\[ \text{pK}_a = 14 - \text{pK}_b = 14 - 1.62 = 12.38 \]

### Step 5: Substitute values into the pH formula
Now, substituting \( \text{pK}_w = 14 \), \( \text{pK}_a = 12.38 \), and \( C = 1 \) into the pH formula:
\[ \text{pH} = \frac{1}{2} \left( 14 + 12.38 + \log(1) \right) \]
Since \( \log(1) = 0 \), we simplify:
\[ \text{pH} = \frac{1}{2} \left( 14 + 12.38 \right) = \frac{26.38}{2} = 13.19 \]

### Final Answer
The pH of the 1M \( \text{PO}_4^{3-} \) solution is **13.19**.

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