Determine the pH of the resulting solution when the following two solutions are mixed: 20.0 mL of 0.20 M HC2H2O2 and 20.0 mL of 0.10 M NaOH. The value of Ka for HC2H2O2 is 1.8 x 10-5.

The pH of the resulting solution when 20.0 mL of 0.20 M HC₂H₂O₂and 20.0 mL of 0.10 M NaOH are mixed is 3.07.

Neutralization is a chemical reaction in which acid and base react to form salt and water. Hydrogen (H⁺) ions and hydroxide (OH⁻ ions) react with each other to form water.

The strong acid and strong base neutralization have a pH value of 7.

The balanced equation for the reaction is:

HC₂H₂O₂ + NaOH → NaC₂H₃O₂ + H₂O

Moles of HC₂H₂O₂= concentration × volume = 0.20 M × 0.020 L = 0.004 mol

Moles of NaOH = concentration × volume = 0.10 M × 0.020 L = 0.002 mol

Since HC₂H₂O₂ is a weak acid, it will partially dissociate in water according to the equation:

HC₂H₂O₂ ⇌ H⁺ + C₂H₂O₂⁻

Initial:

HC₂H₂O₂: 0.004 M

H⁺: 0 M

C₂H₂O₂⁻: 0 M

Change:

HC₂H₂O₂: -x M

H⁺: +x M

C₂H₂O₂⁻: +x M

Equilibrium:

HC₂H₂O₂: 0.004 - x M

H⁺: x M

C₂H₂O₂⁻: x M

Ka = [H⁺][ C₂H₂O₂⁻] / [HC₂H₂O₂]

1.8 x 10⁻⁵ = x × x / (0.004 - x)

Since x is small compared to 0.004, so 0.004 - x = 0.004:

1.8 x 10⁻⁵= x² / 0.004

x² = 1.8 x 10⁻⁵ × 0.004

x² = 7.2 x 10⁻⁸

x = 8.49 x 10⁻⁴ M = [H⁺]

pH = -log( 8.49 x 10⁻⁴)

pH = 3.07

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