Question

calculati volumul (L) solutiei de KOH de C=0,1 M necesar neutralizarii totale a 20 mL solutiei de acid acetic de C = 1 M

Answer 1

CH3-COOH + KOH = CH3-COOK + H2O
Avem o reactie mol la mol deci vom avea nevoie de acelasi numar de moli de KOH cat avem de acid acetic.
Cm=n/V =) n= Cm x V = 1 x 0.02 = 0.02 moli acid acetic =) 0.02 moli KOH necesari
V=n/Cm = 0.02/0.1 = 0.2 litri solutie KOH necesara