ManzRDH said:
what is the conc of H+ in a 2.0M solution of acetic acid CH3COOH? Ka+1.8 * 10^5 at 25 degrees C
a- 36* 10^-6
b 6*10^-3
c 3*10^-3
d 36 *10^-3
e- 6*10^-6
to do this problem, you have to recognize it's a weak acid- therefore only partially dissociates. TO do this, we set up an ICE table (I=initial concentration, C=change in conc., E=equilibrium conc.)
1) write disaccociation equation:
CH3COOH ---> CH3COO- + H+
A=CH3COOH
B=CH3COO-
C=H+
2) ICE table
A --> B + C
I 2.0...0...0 *
C -x...+x...+x **
E (2-x).+x..+x ***
-----I had to use "...." to seperate the variables...because they dont' line up right...
* initial conc. of acetic acid was given and I didn't start with any of the conj. base or H+..therefore, their initial conc. are zero.
** according to stoichiometry, for every 1 CH3COOH, I get 1 CH3COO and 1 H...so I let x represent the amount disaccotiated. And there's a negative sign under the acetic acid, because I'm losing the acid to make the c. base and H+.
*** you just simply add the colume together... 2+ -x = 2-x, 0 + x = x...
3) Looking at the ICE table, x represents the conc. of H+ and CH3COO-....so, all you have to do is solve for x...by setting up the equalibrium constant
K = [reactants] / [products]
so--
K = {[H+][CH3COO-] }/ [CH3COOH].....replace with equilibrum concentrations from ICE to get:
x^2 / (2-x) = Ka (use Ka because acetic acid is an acid..if they gave you a base, you'd have to use Kb...and get Kb from equation Kw=Ka*Kb, where Kw=1x10^-14.)
so.. x^2/(2-x) = 1.8 * 10^5
solve for x and you'll have it!
