I'm confused. You have titrated NaHCO3 but you give the molar mass of Na2CO3. I will assume the 0.4617 g sample did not contain Na2CO3 (or any other carbonate)
mols HCl = M x L = ?
mols HCO3^- = mols HCl since the ratio in the equation is 1:1
Convert mols HCO3^- to grams NaHCO3. g = mols x molar mass NaHCO3.
Then %NaHCO3 = (g NaHCO3/mass sample)*100 = ?
a 0.4617g sample containing sodium bicarbonate was dissolved and titrated with standard 0.106M of hydrochloric acid solution , requiring 40.72ml. the reaction is:
HCO3- + H+ -> H2O + CO2
Calculate the percent sodium bicarbonate in the sample. (Given molecular weight of Na2CO3 is 105.99 and HCL is 36.5)
Someone help me please,urgent T-T
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