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 Ka and we can work out that it’s C3H4O2Cl by removing a hydrogen from the molecular formula.
The general form for a disassociation constant Ka for the reaction is [H+][A-]/[HA]. Thus, the Ka 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 Ka = [H+]2/[ C3H5O2Cl]. As we are given the value of Ka, we can rearrange the equation to give us [H+] = ([C3H5O2Cl]*Ka)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
