Chem Question

Question

Calculate the pH of a solution involving the addition of 35 mL of 0.1 M acetic acid, CH3COOH, to 350 mL of water. Ka for acetic acid is 1.8 x 10^(-5). My professor gave us the answer and wants us to show the work. The answer given is pH=3.39. Thank you so much.

Answer 1

First find the molarity of the new solution.

 

0.1 mol CH₃COOH/L x 35 ml x 1 L/1000 ml = 0.0035 mol CH₃COOH

0.0035 mol CH₃COOH/(350+35) ml x 1000 ml/L = .0090909 M CH₃COOH

 

K_a = [H⁺][CH₃COO^-]/[CH₃COOH]

[H⁺] = [CH₃COO^-] = x

[CH₃COOH] = .0090909 - x = .0090909  since it is a weak acid

1.8x10^-5 = x²/.0090909

x² = (1.8x10^-5)(.0090909)

x² = 1.63636x10^-7

x = 4.04475x10^-4

 

pH = -log[H⁺]

pH = -log(4.0447x10^-4)

pH = 3.393

 

Using significant digits it would be 3.39.  I hope that helps.

Answer 2

With the addition of the 350 mL of water, the concentration of acetic acid is (0.1) 0.035 /(.035 + .35) = 0.00909

 

The equilibrium is :

 

  1.8 x 10^-5  =  [ H⁺ ] [ Ac^-]/ [H Ac]

 

but [Ac^-] = [ H⁺]      call this x.    and then [ H Ac] = .00909 - x   so the equilibrium becomes

 

 1.8 x 10^-5  =  x² /( .00909 - x)

 

This  equation can be solved numerically    giving x = 3.95 x 10^-4    

 

The pH is  -log (3.95 x 10^-4 )  =  3.40

Answer 3

Concentration of CH3COOH is calculated as follows:

(35 ml)(0.1 M) = (385 ml)(x M)

x = 0.00909 M

CH3COOH <==> CH3COO- + H+

Ka = [CH3COO-][H+]/[CH3COOH] = 1.8x10^-5

1.8x10^-5 = x^2/.00909

x^2 = 1.6x10^-7

x = 4x10^-4 = [H+]

pH = -log [H+] = 3.39