Question

If 20.0 ml of glacial acetic acid (pure hc2h3o2 is diluted to 1.70 l with water, what is the ph of the resulting solution? the density of glacial acetic acid is 1.05 g/ml.

Answer 1

Given:
20.0 ml of glacial acetic acid
1.70 L with water
density of the glacial acetic acid is 1.05 g/ml ; Ka = 1.8 x 10⁻⁵

1) find out how many grams of HC2H3O2 is in this glacial matter. It says here it has a density of 1.05g/mL. 
So we find the grams from that => 1.05 g/mL * (20.0 mL) = 21 grams of HC2H3O2. 
Convert this to moles => 21 grams (1 mol / 60 g) = 0.35 mol HC2H3O2 

We then get the Ka formula from that equation so it is 
Ka = [H3O+] [C2H3O2] / [HC2H3O2] (Remember H2O has an activity of 1 so we exclude it) 

We know the concentration of HC2H3O2 because we have moles of HC2H3O2 
Finding concentration of HC2H3O2 => 0.35 mol HC2H3O2/1.70 L = 0.20588 M - initial concentration

1.8 x 10⁻⁵ = (x) (x) / (0.20588 M - x)  
1.8 X 10⁻⁵ (0.20588 M) = x²
3.70584⁻ x 10⁻⁶ = x²
x = 1.9251 x 10⁻³

pH = -log (H₃O⁺) = - log (1.9251 x 10⁻³) = -(-2.7156) = 2.7156

pH of the resulting solution is 2.7156 

Answer 2

The pH of the resulting solution is 2.72

Data;

• volume of acid = 20mL
• density of acid = 1.05 g/ml
• mass = 1.05 * 20  = 21g

pH of a solution

The number of moles of acetic acid is

$n = 21/60 = 0.333 moles$

The concentration of acetic acid in 1.7L is

$c = \frac{0.333}{1.7} = 0.196M$

The equation for dissociation is

$CH_3COOH + H_2O \to CH_3COO^- +H_3O^+$

The Ka of this reaction is

$K_a = \frac{[CH_3COO^-][H_3O^+}{[CH_3COOH}$

The Ka of acetic acid = 1.8*10^-5

$1.8*10^-^5 = \frac{x*x}{0.196} \\x = 0.187*10^-^2M$

The concentration of hydrogen ion is

$[H^+] = 1.87*10^-^3M$

The pH of this solution is

$pH = -log[H^+]\\pH = -log(1.87*10^-^3)\\pH = 2.72$

The pH of the resulting solution is 2.72

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