For both nitric acid and sodium hydroxide, the acid hydrogen equivalence factor is 1. That is: 1N=1M.
Given:
22mL HNO3 neutralises 20mL, 0.5N NaOH, via the stoichiometry:
HNO3 + NaOH = NaNO3 + H2O
Molar Mass of HNO3 is 63.01296±0.00004 g/mol.
Let: c be the concentration of HNO3
Find: m, the mass of HNO3
.: (0.022L)c = (0.020L)(0.5M)
m = (0.500L)(63.01296g/mol)(0.020L)(0.5mol/L)/(0.022L)
m = 14.3211(27...) g
m ≈ 14 g
22 ml nitric acid neutralises 20 ml , 0.5 normal sodium hydroxide sollution. calculate weight of nitric acid in 500 ml?
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