Question

pH of acetic acid solutions. Remember, acetic acid is a weak acid, with a Ka = 1.8 x 10⁻⁵.

Obtain 10.0 mL of 0.10 M acetic acid and place in a clean dry 50.0 mL beaker. Predict the value of the pH. Measure the pH with the pH meter. Record the value.

Take 1.00 mL of the 0.10 M CH3COOH(aq) in the previous step, and put it in another clean beaker. Add 9.00 mL of deionized water and stir. What is the new concentration of CH3COOH(aq)? Predict the pH. Measure the pH. Record the value.

Add 90.0 mL of deionized water to the diluted CH3COOH(aq) in step 2. What is the new concentration of CH3COOH(aq)? Predict the pH. Measure the pH. Record the value.

Answer 1

acetic acid dissociates as

CH₃COOH + H₂O CH₃COO^- + H₃O⁺

K_a = [CH₃COO^- ][H₃O⁺] / [CH₃COOH]

but  [CH₃COO^- ] = [H₃O⁺] = x

K_a = [x][x] / [CH₃COOH]

Substitute the value in equation

1.810^-5 = [x]²/ 0.10

[x]² = 1.810^-5   0.10 = 1.810^-6

[x] = 1.34 10^-3

Concentration of H₃O⁺ = 1.34 10^-3 M

pH = - log[H⁺]

pH = - log (1.34 10^-3)

pH = 2.87

The pH of 0.10M acetic acid is 2.87

calculate molarity of solution after addition of 9 ml water in 1 ml solution

Use the formula C₁V₁ = C₂V₂

Where C₁ = initial concentration = 0.10 M

V₁ = initial volume = 1 ml

C₂ = final concentration = ?

V₂ = final concentration = 1 + 9 = 10 ml

Substitute the value in formula

C₁V₁ = C₂V₂

C₂ = C₁V₁/V₂

= 0.10 1/10

C₂ = 0.01 M

Final concentration is 0.01 M

K_a = [CH₃COO^- ][H₃O⁺] / [CH₃COOH]

but  [CH₃COO^- ] = [H₃O⁺] = x

K_a = [x][x] / [CH₃COOH]

Substitute the value in equation

1.810^-5 = [x]²/ 0.01

[x]² = 1.810^-5   0.01 = 1.810^-7

[x] = 4.24 10^-4

Concentration of H₃O⁺ = 4.24 10^-4

pH = - log[H⁺]

pH = - log (4.24 10^-4)

pH = 3.37

The pH of 0.01M acetic acid is 3.37

calculate molarity of solution after addition of 90 ml water in 10 ml solution of step 2

Use the formula C₁V₁ = C₂V₂

Where C₁ = initial concentration = 0.01 M

V₁ = initial volume = 10 ml

C₂ = final concentration = ?

V₂ = final concentration = 10 + 90 = 100 ml

Substitute the value in formula

C₁V₁ = C₂V₂

C₂ = C₁V₁/V₂

= 0.01 10/100

C₂ = 0.001 M

Final concentration is 0.001 M

K_a = [CH₃COO^- ][H₃O⁺] / [CH₃COOH]

but  [CH₃COO^- ] = [H₃O⁺] = x

K_a = [x][x] / [CH₃COOH]

Substitute the value in equation

1.810^-5 = [x]²/ 0.001

[x]² = 1.810^-5   0.001 = 1.810^-8

[x] = 1.34 10^-4

Concentration of H₃O⁺ = 1.34 10^-4

pH = - log[H⁺]

pH = - log (1.34 10^-4)

pH = 3.88

The pH of 0.001M acetic acid is 3.88