Question

A solid acid is dissolved in enough water to make 200 ml of a solution. 40.0ml of the solution is titrated to a phenolphthalein en point with an NaOH solution. The neutralized solution and the remainder of the acid solution are then mixed and the PH of the resulting solution is found to be 3.35. Find Ka of the acid

Answer 1

Looks like not enough information is there but it can be done. The easy way to understand it is to assume some convenient concn for the weak acid and the NaOH and work it out. The easy way to solve it is to do it by reasoning. You can go through the numbers if you wish but the bottom line is that this is a buffered solution problem solved with the Henderson-Hasselbalch equation.
If we start with a weak acid, HA, and we add NaOH, all of the HA titrated will form the A^- and when mixed with the original HA, forms the buffer.How much of the anion is formed? That's 40 parts/200 parts. How much acid is left? That's 160 parts/200 parts. So the HH equation becomes
3.35 = pKa + log (anion)/(acid)
3.35 = pKa + log (40/160)
Solve for pKa and convert to Ka by pKa = -log Ka.
Interesting problem. eh?

Answer 2

consider a 50 ml solution of 0.55m cacl2 (aq). A solid piece of NaCl(s) with a volume of 14 cm3 is added to the aqueous calcium chloride solution. Calculate the chloride ion concentration in the mixture after the solid completely dissolves. Hint: You have to look up information to solve this problem