I dont understand this Acid constant (Ka) question.?

0 ⤊

I dont understand this Acid constant (Ka) question.?

What is the pH of 0,10 M sodium acetate (NaAc) solution? The Ka of acetic acid (HAc) is 0,000018.

2006-07-17 19:34:07 · 1 answers · asked by yveswang1998 1 in Science & Mathematics ➔ Chemistry

1 answers

The ka of acetic acid is therefore 1.8*10^-5. However, you need the Kb for sodium acetate, which is the eqilibrium constant for the reaction:

CH3COO- + H20 ---> CH3COOH + OH-

Ka*Kb = Kw and Kw, at 25 degrees, is 1*10^-14. So, solving for Kb, you get 5.7*10^-10

Now it's a simple disassocation problem. Set up an ice table:

CH3COO- ----> CH3COOH + OH-
Initial 0.1 M 0 0
Change -x +x +x
Equilibrium 0.1 -x +x +x

So the equilibrium expression becomes:

x^2/0.1-x = 5.7*10^-10

Given that x (the amount of material that disassociates) is going to be very small, you can assume that 0.1 >> x, so it eliminates x from the equation (unless you like the quadratic formula)

So x^2/0.1 = 5.7*10-10
x = 5.75 * 10 -11
pH = 10.24

2006-07-17 19:44:14 · answer #1 · answered by niuchemist 6 · 0⤊ 0⤋