The pH of 0.1 M solution of a weak acid is 3. What is the value of the ionisation constant for the acid?

Options

A. 0.1
B. `10^(-3)`
C. `10^(-5)`
D. `10^(-7)`

Answer

To find the ionization constant (Ka) for the weak acid given the pH of its solution, we can follow these steps:

### Step 1: Calculate the concentration of hydrogen ions [H⁺]
The pH of the solution is given as 3. The concentration of hydrogen ions can be calculated using the formula:

\[ [H^+] = 10^{-\text{pH}} \]

Substituting the given pH:

\[ [H^+] = 10^{-3} = 0.001 \, \text{M} \]

### Step 2: Set up the equilibrium expression
For a weak acid (HA) that ionizes in water, the equilibrium can be represented as:

\[ HA \rightleftharpoons H^+ + A^- \]

Let the initial concentration of the weak acid (HA) be \(C = 0.1 \, \text{M}\). At equilibrium, the concentration of \(H^+\) and \(A^-\) will both be equal to \(x\), which we found to be \(0.001 \, \text{M}\). The concentration of the undissociated acid (HA) at equilibrium will be:

\[ [HA] = C - x = 0.1 - 0.001 = 0.099 \, \text{M} \]

### Step 3: Write the expression for the ionization constant (Ka)
The ionization constant (Ka) for the weak acid can be expressed as:

\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Substituting the equilibrium concentrations into the expression:

\[ K_a = \frac{(0.001)(0.001)}{0.099} \]

### Step 4: Calculate Ka
Now, we can calculate \(K_a\):

\[ K_a = \frac{0.000001}{0.099} \approx 0.0000101 \, \text{or} \, 1.01 \times 10^{-5} \]

### Final Answer
The value of the ionization constant (Ka) for the weak acid is approximately:

\[ K_a \approx 1.01 \times 10^{-5} \]

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