Hydrolysis is an acid-basedreaction of a cation or anion or both ions of a salt with water, Resultan solution of hydrolysis may be acidic, basic or netural. The anion `A^(-)` which is a weakeer base than `OH^(-)` and which his its conjugate acid HA stronger then water but weaker than `H_(3)O` shown the phenomenon of hydrolysis Ex : `CH_(3)COO^(-),CN^(-),NO_(2)^(-)` etc.
The contion `B^(+)` which is a weaker acid than `H_(3)^(+)` which is a weaker acid then `H_(3)^(+)` and which has its conjugate base BOH stronger than water but weak than `OH^(-)` shown the phenmenon of hydrolysis Ex : `NH_(4)^(+)C_(6)H_(5)NH^(+),N_(2)H_(5)^(+)` etc.
The hydrolysis constant of anion and cation are given by
`A^(-)(aq.)+H_(2)O(l)hArrHA(aq.)+OH^(-)(aq)`
`" " K_(h)=(K_(w))/(K_(a))rArr([HA(aq.)][OH^(-)(aq.)])/([A^(-)(aq.)])`
`B^(+)(aq.)+H_(2)O(l)hArrBOH(aq.)+H^(+)(aq.)`
`" " K_(h)=(K_(w))/(K_(b))rArr([BOH(aq.)][H^(+)(aq.)])/([B^(-)(aq.)])`
Calculate percentage degreeof hydrolysis in a 0.1 M solution of `CH_(3)COONa.(K_(a)"of"CH_(3)COOH=10^(-5))`

Options

A. 0.1
B. 0.01
C. `10^(-4)`
D. None of these

Answer

To solve the problem of calculating the percentage degree of hydrolysis in a 0.1 M solution of CH₃COONa, we will follow these steps:

### Step 1: Understand the Hydrolysis Reaction
The hydrolysis of the acetate ion (CH₃COO⁻) can be represented as:
\[ \text{CH}_3\text{COO}^- (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{CH}_3\text{COOH} (aq) + \text{OH}^- (aq) \]

### Step 2: Write the Expression for Hydrolysis Constant (K_h)
The hydrolysis constant \( K_h \) for the reaction can be expressed as:
\[ K_h = \frac{[HA][OH^-]}{[A^-]} \]
Where:
- \( [HA] \) is the concentration of acetic acid,
- \( [OH^-] \) is the concentration of hydroxide ions,
- \( [A^-] \) is the concentration of acetate ions.

### Step 3: Calculate \( K_b \) for the Acetate Ion
Given that:
- \( K_w = 10^{-14} \)
- \( K_a \) for acetic acid (CH₃COOH) = \( 10^{-5} \)

We can calculate \( K_b \) for the acetate ion using the relationship:
\[ K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-5}} = 10^{-9} \]

### Step 4: Set Up the Equilibrium Expression
Let \( h \) be the degree of hydrolysis. The initial concentration of CH₃COO⁻ is 0.1 M. At equilibrium:
- Concentration of CH₃COO⁻ = \( 0.1 - h \)
- Concentration of CH₃COOH = \( h \)
- Concentration of OH⁻ = \( h \)

Substituting these into the \( K_h \) expression:
\[ K_h = \frac{[h][h]}{[0.1 - h]} = \frac{h^2}{0.1 - h} \]

### Step 5: Substitute \( K_h \) with \( K_b \)
Since \( K_h = K_b \):
\[ 10^{-9} = \frac{h^2}{0.1 - h} \]

### Step 6: Make an Assumption
Assuming \( h \) is small compared to 0.1 M, we can simplify:
\[ 0.1 - h \approx 0.1 \]
Thus, the equation simplifies to:
\[ 10^{-9} = \frac{h^2}{0.1} \]
\[ h^2 = 10^{-9} \times 0.1 = 10^{-10} \]

### Step 7: Solve for \( h \)
Taking the square root:
\[ h = \sqrt{10^{-10}} = 10^{-5} \]

### Step 8: Calculate Percentage Degree of Hydrolysis
The percentage degree of hydrolysis is given by:
\[ \text{Percentage Degree of Hydrolysis} = \left( \frac{h}{\text{Initial Concentration}} \right) \times 100 = \left( \frac{10^{-5}}{0.1} \right) \times 100 = 0.01\% \]

### Conclusion
The percentage degree of hydrolysis in a 0.1 M solution of CH₃COONa is 0.01%.

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