Question

Calculate the pH of the titration of 100.00 mL of 0.200 M HOCI with 0.400 M KOH after 0, 25.00, 50.00, & 75.00 mL of KOH have been added. The Ka for HOCI is .000000035...Please Help!!!!

Answer 1

First write the equation and balance it.

KOH + HOCl ==> KOCl + H2O

moles HOCl = M x L = ??

To do these you simply need to figure out what you have in the solution, then apply the appropriate chemical properties and calculate. I will help you get started.
The second thing you need to do is to calculate the mL for the equivalence point.
mL x M = mL x M and I get 50.00 mL KOH.

a. 0 mL of 0.4 M KOH. Here you have 100 mL of 0.2 M HOCl. You have Ka, calculate H^+ and pH.

b. 25 mL. At 25 mL you aren't to the equivalence point; therefore, you have a mixture of HOCl and KOCl. Use the buffer equation

c. 50.0 mL. You are at the equivalence point; therefore, the pH will be determine by the hydrolysis of the salt. Calculate concn of salt and hydrolyze it (NaOCl).
OCl^- + HOH ==>HOCl + OH^-
Set up ICE chart and substitute into the Kb expression
Kb = (Kw/Ka) = (HOCl)(OH^-)/(OCl^-), solve for OH^-, convert to pOH, then to pH.

d. 75 mL. This is past the equivalence point (by 25 mL), therefore, you have straight OH^- from the KOH (but diluted) so (OH^-) = 25 mL x 0.4M moles and that is in a volume of 175 mL.

Post your work and explain what you don't understand if you get stuck.