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
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
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