Question

If a weak acid \(HA\) is \(0.1\%\) ionized (dissociated) in a \(0.2M\) solution, its pH will be _ _ _. (round off to one decimal place)

Hint:

For weak acids: \[ [H^+]=C\alpha \] where \(C\) is concentration and \(\alpha\) is degree of ionization in fractional form.

Answer 1

Correct Answer: 3.7

Step 1: Write the given data.
Concentration of weak acid:
\[ C=0.2M \] Percentage ionization:
\[ 0.1\% \]

Step 2: Convert percentage ionization into fractional ionization.
\[ \alpha=\frac{0.1}{100} \] \[ =0.001 \]

Step 3: Calculate hydrogen ion concentration.
For a monoprotic weak acid:
\[ [H^+]=C\alpha \] Thus,
\[ [H^+]=0.2\times0.001 \] \[ =2\times10^{-4}\ M \]

Step 4: Use the pH formula.
\[ \text{pH}=-\log[H^+] \] \[ =-\log(2\times10^{-4}) \]

Step 5: Simplify the logarithm.
\[ \text{pH} = -\left(\log2+\log10^{-4}\right) \] \[ = -(0.3010-4) \] \[ =4-0.3010 \] \[ =3.699 \]

Step 6: Round off to one decimal place.
\[ 3.699 \approx 3.7 \]

Step 7: Final conclusion.
Therefore, the pH of the solution is
\[ \boxed{3.7} \]