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According to a previous post, Dr. Lavelle said that we would not have to calculate one from the other, but it is important to know that when you put an acid in a solution with a pKa less than the pH of the solution the acid will dissociate.
Since pH = -log[H3O+], you could use antilog to go from pKa to Ka, and then use Ka = [H3O+][A−]/[HA] to solve for the H3O+ that you need. Unfortunately, you will need to be given more information in order to solve for pH this way.
There are some questions in 6B that ask you to find pH. For example, you can go from the given concentration of an acid to pH by using -log([H3O+]=pH. You could also be given an M1V1=M2V2 type question like 6.B #5.d. The last question form I can think of is being asked to find pOH, and converting that to pH by subtracting pOH from 14.
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