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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO is 4.0*10^-8

b)after addition of 25.0 mL of KOH
c)after addition of 40.0 mL of KOH
d)after addition of 50.0 mL of KOH
e)after addition of 60.0 mL of KOH

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The secret to these problems is to identify what you have in solution. That will tell you how to solve the problem. This is the titration of a weak acid with a strong bsse.
a. The beginning point is just a soln of 0.150M HA (in this case HClO). I can show you how to do that if you don't remember how but I expect you know how to do it.
HA ==> H^+ A^-
0.150...0...0
-x......x...x
0.150-x..x...x

Then substitute into the Ka expression solve for H^+ and convert to pH.

c. 50.00 mL obviously is the equialence point so this is done by the hydrolysis of the salt.
............A^- + HOH ==> HA + OH^-
I........0.0750M............0....0
C...........-x..............x....x
E........0.0750-x............x....x

Kb for the A^- = (Kw/Ka for HA) = (x)(x)/(0.0750-x) and solve for x = OH^-, then convert to pH.
Where did I get the 0.075M for the salt. That's 1/2 initial concn HA and when you titrate equal molarities of monoprotic acid and base it always works out to be 1/2 the initial.

b. All point between a and c are buffer problems.
d. All points after c are just excess OH.
I shall be happy to check your numbers or explain further.

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