# Chemistry

posted by Judith

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.240 M HClO(aq) with 0.240 M KOH(aq). The ionization constant for HClO can be found here.

1. DrBob222

Divide the titration curve into 4 parts.
a. beginning
b. between beginning and eq pt.
c. eq pt
d. after eq pt.

First, determine mL to eq pt.
mLacid x M aci = mL base x M base and that determination tells you how to divide into the four part.
a.......... HClO ==> H^+ + ClO^-
I..........0.240M.....0.....0
C............-x.......x......x
E.........0.240-x.....x......x

Substitute the E line into the Ka expression for HClO and solve for x = (H^+), then convert to pH.

b. mmols to start = mL x M = 50*0.240 = about 12.
mmols KOH added. You don't give a value but lets say 25 mL. So 25 x 0.240 = about 6. Therefore, you have 6 acid left and you formed 6 of the salt from
.......HClO + KOH ==> KClO + H2O
I.......12.....0........0......0
C.......-6.....-6........6......6
E........6.......0.......6.......6

Use the Henderson-Hasselbalch equation to solve for the pH.
pH = pKa for HClO + log (6/6)
?

c. The pH is determined by the hydrolysis of the salt.
........ClO^- + HOH ==>HClO + OH^-
I......0.12............0......0
C........-x............x......x
E........-0.12-x........x.....x

Kb for ClO^- = (Kw/Ka for HClO) = (x*x)/(0.12-x).
Substitute and solve for x = OH^- and convert to pH.
NOTE: The concn of the salt at the eq.pt is just 1/2 of 0.240. Since they are the same molarity you have added 50 mL KOH so the M salt is just 1/2 of what acid or base started.

d. After the eq. pt is just excess KOH.

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