O Timer Notes Evaluate Feedback Print Info 1. A sample of sodium carbonate (Na2C
ID: 581067 • Letter: O
Question
O Timer Notes Evaluate Feedback Print Info 1. A sample of sodium carbonate (Na2CO3), which is a base (assume the titration endpoint is reached when one proton has been transfered, producing NaHCO3), with a mass of 0.6514 g is dissolved in water and used to standardize a solution of hydrochloric acid. An endpoint is reached when 33.71 mL of hydrochloric acid has been added. Determine the molarity (in mol/L) of the hydrochloric acid solution. Report your answer to four significant figures. Submit AnswerTries 0/3 2. The standardized hydrochloric acid solution is then used to titrate 30.47 mL of a solution of sodium lactate (NaC2H5OCOO). An endpoint is reached when 10.97 mL of hydrochloric acid has been added. Determine the molarity (in mol/L) of the sodium lactate solution. Report your answer to four significant figures Submit Answer Tries 0/3 This discussion is closed. Send FeedbackExplanation / Answer
1. Titration of Na2CO3 with HCl
Na2CO3 + HCl ---> NaHCO3 + NaCl
moles of Na2CO3 present = grams/molar mass
= 0.6514 g/106 g/mol
= 0.00614 mol
moles of HCl reacted = moles of Na2CO3 present = 0.00614 mol
Volume of HCl used = 33.71 ml = 0.03371 L
So,
molarity of HCl solution = moles/L = 0.00614 mol/0.03371 L
= 0.1823 mol/L
2. Titration of sodium lactate with HCl
NaC2H5OCOO + HCl ---> HC2H5OCOO + NaCl
moles of HCl used = molarity x volume
= 0.1823 M x 0.01097 L
= 0.002 mol
moles of sodium lactate present = moles of HCl reacted = 0.002 mol
Volume of sodium lactate solution = 30.47 ml = 0.03047 L
So,
molarity of sodium lactate solution = moles/volume
= 0.002 mol/0.03047 L
= 0.06564 mol/L
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.