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
add............6.............
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.
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.
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