Posted by Anonymous on Wednesday, October 23, 2013 at 11:03pm.
A 30 mL sample of .165 M propanoic acid is titrated with .300M KOH. Calculate the pH at each volume of added base: 0 mL, 5 mL, 10, equivalence point, onehalf equivalence point, 20 mL, 25 mL. Use calculations to make a sketch of the titration curve.

chemistry  DrBob222, Wednesday, October 23, 2013 at 11:28pm
eq pt.
mL acid x M acid = mL KOH x M KOH
Solve for mL KOH.
0 mL:
.........HPr ==> H^+ + Pr^
I......0.165.....0.....0
C.........x......x....x
E.......0.165x...x....x
Ka = (H^+)(Pr^)/(HPr).
Substitute from the E line and solve for x = (H^+), then convert to pH.
For additions up to but not including eq pt use the HendersonHasselbalch equation.
For the eq pt the pH is determined by the hydrolysis of the salt. (salt) = (Pr^) = (0.165 x 30 mL)/(total mL) where total mL = 30 + mL to eq pt. I will call this 0.1 M but you need to do it more accurately
........Pr^ + HOH ==> HPr + OH^
I......0.1..............0.....0
C........x.............x.....x
E......0.1x............x.....x
Kb for Pr^ = (Kw/Ka for acid) = (x)(x)/(0.1x). Solve for x = (OH^) and convert to pH.
For all points past the eq pt it will be excess OH^.
mols base = M x L
mols acid to eq pt = ?
excess OH^ = difference.
(OH^) = mols OH excess/total L.
Then convert to pH.

chemistry  Sam Dowen, Tuesday, November 22, 2016 at 8:09pm
2.8?
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