1.0 mole of a weak acid HA was dissolved in 2.0 L of solution. After the system had come to equilibrium, the concentration of HA was found to be 0.34 M. Calculate Ka for HA.

To solve this problem, we first need to understand the dissociation of a weak acid. A weak acid HA dissociates in water as follows:

HA ⇌ H+ + A-

At the beginning, we have 1.0 mole of HA in a 2.0 L solution, so the initial concentration of HA is 1.0 mol / 2.0 L = 0.5 M.

At equilibrium, the concentration of HA is given as 0.34 M. This means that 0.5 M - 0.34 M = 0.16 M of HA has dissociated.

Since for every HA that dissociates we get one H+ and one A-, the concentrations of H+ and A- at equilibrium are also 0.16 M.

The equilibrium constant Ka for the weak acid is then given by:

Ka = [H+][A-] / [HA]

Substituting the equilibrium concentrations into this expression gives:

Ka = (0.16 M)(0.16 M) / (0.34 M) = 0.0075

So, the Ka for the weak acid HA is 0.0075.