Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution: a solution that is 0.170 M in HC2H3O2 and 0.105 M in KC2H3O2

Answer 1

You don't need an ICE table to do this.
Use the Henderson-Hasselbalch equation.

Answer 2

The book says to use ICE

Answer 3

OK. The book wants you to do it the hard way.
Here are the two equilibria.
......CH3COOH ==> H^+ + CH3COO^- for which
I.....0.170.......0......0
C......-x.........+x.....+x
E.....0.170-x.......x.......x

..........CH3COOK ==> K^+ + CH3COO^-
I.........0.105.......0.......0
C.........-0.105.....0.105...0.105
E.........0..........0.105.....0.105

Now write the Ka for CH3COOH.
Ka = (H^+)(CH3COO^-)/(CH3COOH)
So (H^+) = x
(CH3COO^- = x from CH3COOH and 0.105 from CH3COOK for a total of x+0.105
(CH3COOH) = 0.170-x, now plug all of that in
Ka = 1.8E-5 (but use the value in your text) = (x)(x+0.105)/(0.170-x)
We can avoid a quadratic equation by making a couple of simplifying assumptions.
x + 0.105 = 0.105 (since x will be small)
0.170-x = 0.170 (since x will be small) so now it looks this way.
1.8E-5=(x)(0.105)/0.170
Solve for x. I get 2.04E-5 for pH = 4.69. You shouldn't take my word for anything. Confirm all of this for yourself.
But look how much easier the HH equation makes things.
The HH equation is
pH = pKa + log [(base)/(acid)]
pKa = 4.74
base = 0.105
acid = 0.170
pH = 4.74+log(0.150/0.170 =
pH = 4.68
The small difference is in rounding errors.

Answer 4

I agree that the Henderson-Hasslebalch equation is easier, but how do you find the pka without doing an ice table first?

Answer 5

pKa = -log(Ka)