The solution of weak monoprotic acid which is 0.01 M has pH= 3. Calculate `K_a` of weak acid

Answer

To solve the problem of finding the acid dissociation constant \( K_a \) of a weak monoprotic acid given its concentration and pH, we can follow these steps:

### Step 1: Understand the given information
We have a weak monoprotic acid with:
- Initial concentration \( [HA] = 0.01 \, M \)
- pH = 3

### Step 2: Calculate the concentration of hydrogen ions \([H^+]\)
Using the pH value, we can find the concentration of hydrogen ions:
\[ \text{pH} = -\log[H^+] \]
Given that pH = 3,
\[ [H^+] = 10^{-3} \, M = 0.001 \, M \]

### Step 3: Set up the equilibrium expression
For a weak monoprotic acid \( HA \) dissociating in water:
\[ HA \rightleftharpoons H^+ + A^- \]
At equilibrium, we can denote:
- Initial concentration of \( HA \) = 0.01 M
- Change in concentration = \( -x \) for \( HA \) and \( +x \) for \( H^+ \) and \( A^- \)

At equilibrium:
- \( [HA] = 0.01 - x \)
- \( [H^+] = x \)
- \( [A^-] = x \)

Since we calculated \( [H^+] = 0.001 \, M \), we have \( x = 0.001 \, M \).

### Step 4: Substitute the equilibrium concentrations into the expression for \( K_a \)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the values we have:
\[ K_a = \frac{(0.001)(0.001)}{0.01 - 0.001} \]
\[ K_a = \frac{(0.001)^2}{0.009} \]

### Step 5: Calculate \( K_a \)
Now, calculate \( K_a \):
\[ K_a = \frac{0.000001}{0.009} = \frac{1 \times 10^{-6}}{9 \times 10^{-3}} = \frac{1}{9} \times 10^{-3} \approx 1.11 \times 10^{-4} \]

### Final Result
Thus, the \( K_a \) of the weak monoprotic acid is approximately:
\[ K_a \approx 1.1 \times 10^{-4} \]

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