# Chemistry

A 25.0 mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 ✕ 10-5) is titrated with 0.100 M KOH solution. Calculate the pH after the addition of the following amounts of KOH.
a) 20.0 mL
b) 25.0 mL
c) 30.0 mL

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1. Let's call propanoic acid something simple like HPr to make typing simple.Then
millimols HPr = mL x M = 25.00 x 0.1 M = 2.500.
millimols KOH added = 20 x 0.1 = 2.0
or 25 x 0.1 = 2.50 or 30 x 0.1 = 3.0.
In order that will be as follows.
.............HPr + KOH ==> KPr + H2O
I.............2.50......0..............0.........0
C............-2.0.....-2.0...........+2.0..................
E...........0.50.........0..............2.0................
So for the 20 mL portion you have a buffer solution; i.e., you have some of the weak acid (the proponoic acid) + the salt. So use the Henderson-Hasslebalch equation.
pH = pKa + log (base)/(acid). The base is the salt. The acid is HPr.

For the 25 mL you can construct an ICE chart as I've done above and you can see that the HPr exactly equals the KOH so that is the equivalence point. Here then you have just the salt; i.e., just KPr. The pH is determined by the hydrolysis of the salt like this.The concn of the salt is millimoles/mL = 2.50/50 = 0.05 M
..................Pr^- + HOH ==> HPr + OH^-
I................0.05......................0...........0
C................-x.........................x...........x
E..................0.05-x.................x...........x
Kb for Pr^- = (Kw/Ka for HPr) = (x)(x)/(0.05-x)
Substitute and solve for x. x = (OH^-) so convert that to H^+ and calculate pH.

For the 30 mL addition, construct and ICE chart and you will see that all of the HPr has been neutralized and you now have extra KOH added after the equivalence point; therefore, you have just a solution of the salt and the KOH So the (OH^-) = millimoles extra KOH/total mL of solution. Convert to H^+ and pH.

Post your work if you get stuck on any of this.

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DrBob222

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