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
<h3>What is the pH of a substance?</h3>
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
