A weak monoprotic acid is dissolved in water. A solution that is 0.10 M is found to be 1.2% ionized. Calculate K _a for the acid.

Question

Quizplus · Accepted Answer

The ionization of a weak monoprotic acid in water can be represented as HA ⇌ H+ + A-. If the acid is 1.2% ionized, it means that 1.2% of the original 0.10 M concentration is converted into H+ and A- ions. Therefore, the concentration of H+ and A- ions each is 0.10 M * 0.012 = 0.0012 M. The equilibrium expression for the ionization of the acid is Ka = [H+][A-]/[HA], where [HA] is the concentration of the non-ionized acid. Since 1.2% is ionized, 98.8% remains non-ionized, so [HA] = 0.10 M * 0.988 = 0.0988 M. Plugging these values into the expression gives Ka = (0.0012)(0.0012)/(0.0988) = 1.4×10^-5, making choice A correct.

A Weak Monoprotic Acid Is Dissolved in Water