2-chloropropanoic acid has a Ka of 1.48E-3. Write an expression for Ka and hence or otherwise, calculate the pH of a 0.35M solution of 2-chloropropanoic acid

In these multiple part questions, it is often the case that the previous part of the question will provide necessary information for answering the following parts, as hinted by the use of 'hence'. It's very important to consider how your earlier answers could be used to complete the next part of the question.

The acid disassociation constant, Ka, for the general acid HA has the form HA -> H⁺ + A^- . All that is necessary to answer the first part of the question is to determine the molecular formula of 2-chloropropanoic acid (which we can work out to be C₃H₅O₂Cl) and from there determine what the conjugate base is. The conjugate base is the chemical species produced after loss of a proton, i.e. A^- in the general expression for K_a and we can work out that it’s C₃H₄O₂Cl by removing a hydrogen from the molecular formula.

The general form for a disassociation constant K_a for the reaction is [H⁺][A^-­]/[HA].  Thus, the K_a for this reaction is [H⁺][ C₃H₄O₂Cl^-]/[ C₃H₅O₂Cl].

The second part asks us to calculate the pH of this solution. As we know, pH = -log₁₀[H⁺]. The key thing to note here is that [H⁺] = [A^-]; this makes intuitive sense as the only sources of H⁺ and A^- are from the disassociation of HA and so they are produced in equal amounts. An assumption is made here that the concentration of HA is left relatively unchanged. This is a good approximation as 2-chloropropanoic acid is a weak acid that only undergoes partial disassociation; such a small amount of HA disassociates that we can approximate it still as 0.35M.

We can now write it as K_a = [H⁺]²/[ C₃H₅O₂Cl]. As we are given the value of K_a, we can rearrange the equation to give us [H⁺] = ([C₃H₅O₂Cl]*K_a)^0.5. Plugging the values in at this stage gives us a H⁺ concentration of 0.023M. We can then insert this into our equation to give us a pH of 1.64