# chemistry

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A 20.1 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.060 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 4.90. What is the Ka of HX?

• chemistry -

There are a couple of ways to do this; one is quite straight forward, the other requires a little reasoning power. Here is the straight forward way.
mLNaOH x MNaOH = mLHX x MHX
50.0 x 0.060 = 20.1 x MHX
Solve for Molarity HX. For simplicity I shall call this about 0.15M.
So millimiles HX to start = 20.1*0.15 = about 3.0
millimoles NaOH at 30 mL = 30 mL x 0.06 = 1.8 mmoles.

Set up an ICE chart.
...........HX + NaOH ==> NaX + H2O
initial....3.0...0........0......0
react......-1.8..-1.8....1.8.....+1.8
equilibrium..1.2...0.....1.8......1.8

Substitute into Ka expression for HX (convert pH to H^+ first), and solve for Ka.
I worked it out to be about 1.9E-5 for Ka but check my work.
The more esoteric way of doing it is to use the Henderson-Hasselbalch equation.
You know, from the equivalence point, that if you add 3.0 millimoles NaOH that you must have started with 3.0 millimoles HX. So you add 1.8 millimoles NaOH at the 30 mL mark which means 1.2 HX is left and 1.8 mmoles were converted to X^-. Then substitute into pH = pKa + log (X^-)/(HX) and solve for pKa, then convert to Ka. I have 4.72 for pKa.