A 0.0560-g sample of acetic acid is added to enough water to make 50.00 mL of solution. How do you calculate [H], [CH3COO-], [CH3COOH] and the pH?The pH is at equilibrium and Ka for acetic acid is...
A 0.0560-g sample of acetic acid is added to enough water to make 50.00 mL of solution. How do you calculate [H], [CH3COO-], [CH3COOH] and the pH?
The pH is at equilibrium and Ka for acetic acid is 1.8x10^-5
1 Answer
sanjeetmanna | College Teacher | (Level 3) Assistant Educator
Given
mass if CH3COOH = 0.0560 gm
Volume = 50ml = 0.05L
We have to find the concentration of acetic acid, for that first we have to find the moles of acetic acid.
Moles = mass/molar mass.
Moles = 0.0560/60.05 = 0.000932
Molarity = moles/volume = 0.000932/0.05 = 0.01864M
Now we have to follow the RICE table.
R CH3COOH → H+ + CH3COO-
I 0.01864 → 0 0
C -x → +x +x
E 0.01864-x → x x
Ka = [H+][CH3COO-]/[CH3COOH]
NOTE Ka of CH3COOH = 1.8 * 10^-5
Finding [CH3COOH]
1.8 * 10^-5 = [x][x]/[0.01864 - x]
1.8 * 10^-5 = [X^2]/[0.01864]
[X^2] = [1.8 * 10^-5][0.01864]
[X^2] = 0.033 * 10^-5 = 3.3 * 10^-7
x = 0.000579 = [CH3COOH]
Finding [H+]
[H+] = ka * [CH3COOH]/[CH3COO-]
= [1.8 * 10^-5] * [0.01864]/[0.000579]
= 3.3 * 10^-7/0.000579
= 0.00056
Finding [CH3COO-]
[CH3COO-] = ka * [CH3COOH]/[H+]
= [1.8 * 10^-5] * [0.01864]/[0.00056]
= 3.3 * 10^-7/0.00056
= 0.00058
Finding PH
PH = - log [H+]
PH = -log [0.000579]
PH = 3.23.