Asked by Anonymous
calculate the weight of oxalic acid required to neutralise 250 ml, 0.1 normal KOH sollution?
Answers
Answered by
Graham
Find oxalic acid mass: m
Given:
(KOH hydrogen equivalence factor: 1)
KOH Concentration:
c = 0.1N = 0.1M
KOH Volume:
V = 0.250L
(COOH)2 Molar Mass:
w = 90.03517 ± 0.00006 g/mol
Stoichiometry:
(COOH)2 + 2 KOH = (COOK)2 + 2 H2O
.: m/w = cV/2
Given:
(KOH hydrogen equivalence factor: 1)
KOH Concentration:
c = 0.1N = 0.1M
KOH Volume:
V = 0.250L
(COOH)2 Molar Mass:
w = 90.03517 ± 0.00006 g/mol
Stoichiometry:
(COOH)2 + 2 KOH = (COOK)2 + 2 H2O
.: m/w = cV/2