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 | Add Yours

sanjeetmanna's profile pic

sanjeetmanna | College Teacher | (Level 3) Assistant Educator

Posted on

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.

We’ve answered 318,980 questions. We can answer yours, too.

Ask a question