Question

What is the ph of a solution made by mixing 30.00 ml of 0.10 m acetic acid with 50.00 ml of 0.100 m koh? assume that the volumes of the solutions are additive. k a = 1.8 × 10-5 for ch3co2h.

Answer 1

Given 30.00 ml of 0.10 m acetic acid and 50.00 ml of 0.100 m koh, the pH of the solution is 12.40

What is the pH of a substance?

This is the measure of the acidity or the basicity of an aqueous or liquidsolution.

The details that are contained in this question are:

Mole of CH3COOH = 0.1 M

Volume of CH3COOH  = 30 mL

Mole of KOH = 0.1 M

Volume of KOH = 50 mL

mol of (CH3COOH) = 0.1 Mole * 30 mL = 3 mol

mol of (KOH) = 0.1 Mole * 50 mL = 5 mol

From the solution above,

mol of CH3COOH = 3 mol

mol of KOH = 5 mol

3 mol of both substances is going to react

excess KOH remaining = 5 - 3 =  2 mmol

Volume of both Solutions

= 30 + 50 = 80 mL

OH⁻ = 2/80

= 0.025 M

use:

pOH = -log [OH-]

= -log (0.025)

pOH = 1.602

PH = 14 - pOH

= 14 - 1.60

= 12.40

The calculated pH is therefore 12.40

Read more on pH here: brainly.com/question/22390063