0.5 L of a 0.30 M HCl solution is titrated with a solution of 0.6 M KOH.

Question

a)What is the pH before addition of KOH?
b)What is the total number of moles of acid?
c)What is the pH after addition of 125 mL of KOH solution?
d)What volume of the KOH solution is required to reach the equivalence point?
e)What is the pH at the equivalence point?
f)What is the pH after addition of 375 mL (total) of KOH solution?

Answer 1

The secret to working these problems is to know where you are on the titration curve and solve accordingly.
a) At the beginning, you have pure HCl.
b) moles = M x L
c) moles KOH = M x L. Subtract from moles HCl initially. The difference is the moles HCl left.
d) moles HCl = moles KOH
e) What salt do you have at the equivalence point? That's NaCl, the salt of a strong acid and strong base. The pH is determined by the hydrolysis of the salt.
f) moles KOH = M x L. Subtract moles HCl. The difference is the moles KOH in excess (note the difference between this and part (c).

Answer 2

For part a, would the pH be .523?

Answer 3

I still don't understand part e.

Answer 4

Im lost, I don't think Im doing any of it right :/

Answer 5

9.8