Hi! I saw that you did analytical chemistry and I was wondering if you could help me? A 25 mL aliquot of 0.0900 M acetic acid is titrated with a 0.100 M KOH. Calculate the pH when 25 mL of base is added. Ka = 1.75 x 10^-5. My professor keeps telling me that the pH is 11.7 but I keep receiving 5.7 and I don't know what I am missing and it's really stressing me out.
Alright, so this is weak acid/strong base and pretty straight forward. The base will completely dissociate and run the reaction to completion.
First, determine the number of mols you have of each reactant using dimensional analysis ( convert from mL of M to mols).
Second, determine which, if any, reactant is in excess. In this case, it’s the strong base. Thus only the strong base is contributing to the pH.
Third, determine the concentration of the excess OH-, since KOH will dissociate. This is your excess mols divided by the total amount of liquid used.
Fourth, find the pOH. since pOH + pH = 14 in water, this is useful. pOH=-log(concentration of OH-)
Finally, subtract pOH from 14. I ran the numbers and it worked fine for me.
I hope that helps, anon. There are probably other ways to solve it too, but this is how I was taught. I never actually used Ka in this example
Also, thinking about it physically can help. The pH above 7 indicates a basic solution, which makes sense with the excess OH- ions. That might be frustrating if you know you’re wrong but not why or how, though.
