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.
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
a)before addition of any KOH
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
1 answer
7 answers