A 0.02 M solution of `CH_3COONa` in water at `25^@C` is found to have an `H^+` concentration of `3 xx 10^(-9) g` ionic weight per litre. What is the hydrolytic constant of the salt? `K_w= 1.01 xx 10^(-14), K_a (CH_3COOH)= 1.75 xx 10^(-5)`

Answer

To find the hydrolytic constant of the salt \( CH_3COONa \), we can follow these steps:

### Step 1: Determine the concentration of \( H^+ \) ions
The problem states that the \( H^+ \) concentration is given as \( 3 \times 10^{-9} \) grams ionic weight per liter. First, we need to convert this to molarity. The molar mass of \( H^+ \) is approximately \( 1 \, g/mol \), so:

\[ \text{Concentration of } H^+ = 3 \times 10^{-9} \, \text{mol/L} \]

### Step 2: Use the relationship between \( K_w \), \( K_a \), and \( K_h \)
The hydrolytic constant \( K_h \) for the salt can be calculated using the formula:

\[ K_h = \frac{K_w}{K_a} \]

Where:
- \( K_w = 1.01 \times 10^{-14} \)
- \( K_a = 1.75 \times 10^{-5} \)

### Step 3: Calculate \( K_h \)
Substituting the values into the equation:

\[ K_h = \frac{1.01 \times 10^{-14}}{1.75 \times 10^{-5}} \]

Calculating this gives:

\[ K_h = 5.77 \times 10^{-10} \]

### Step 4: Confirm the calculation with \( H^+ \) concentration
We can also use the \( H^+ \) concentration to find \( K_h \):

\[ K_h = \frac{[H^+]^2}{C - [H^+]} \]

Where \( C \) is the concentration of the salt \( CH_3COONa \) which is \( 0.02 \, M \):

\[ K_h = \frac{(3 \times 10^{-9})^2}{0.02 - 3 \times 10^{-9}} \]

Calculating this gives:

\[ K_h \approx \frac{9 \times 10^{-18}}{0.02} \approx 4.5 \times 10^{-16} \]

### Step 5: Finalize the hydrolytic constant
The hydrolytic constant of the salt \( CH_3COONa \) is approximately:

\[ K_h \approx 5.77 \times 10^{-10} \]

### Summary
Thus, the hydrolytic constant \( K_h \) for the salt \( CH_3COONa \) is \( 5.77 \times 10^{-10} \).

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