2. An unknown sample of Na2CO3 sample (mass 0.3412 g) is dissolved in distilled

Question

2. An unknown sample of Na2CO3 sample (mass 0.3412 g) is dissolved in distilled water and reacted with exactly 50.00 mL of standardized 0.09983 M HCl solution. The reaction mixture is then subjected to a back titration with standardized 0.09912 M NaOH. The volume of NaOH required to titrate the unreacted HCl in the reaction mixture is determined to be 12.50 mL. a. How many moles of HCl were added initially? b. How many moles of HCl are in excess (unreacted)? c. How many moles HCl were consumed in the reaction with the unknown sample? d. Calculate the mass of Na2CO3 in the unknown sample and the percent (%w/w) of Na2CO3 in the unknown sample. 3. If the 0.3412g Na2CO3 was 100% pure, how many moles HCl would be consumed in the reaction?

Explanation / Answer

Moles of HCl=Volume*Molarity=50.00 mL* 0.09983 M=4.9915mMoles=0.0049915moles

Moles of Excess HCl=Moles of NaOH used

=12.50mL*0.09912M=1.239mMoles=0.001239moles

Moles of HCl used with unknown=initial-excess=0.0049915-0.001239=0.003753Moles

Na2CO3 + 2HCl ==>NaCl + CO2 + H2O

2moles of HCl reacts stiochiometrically with=1mole of Na2CO3

0.003753moles of HCl reacts stiochiometrically with=1/2*0.003753moles of Na2CO3

Moles of Na2CO3 in unknown sample=0.003753/2=0.001876moles Na2CO3

Mass of Na2CO3=moles*molar mass=0.001876moles*106g/mole=0.199g

%age mass of Na2CO3=mass determined/mass taken*100=0.199/0.3412*100=58.3%

3.         If the 0.3412g Na2CO3 was 100% pure, how many moles HCl would be     consumed in the reaction?      

in case of 100% pure unknown salt of Na2CO3

Moles of Na2CO3=mass/molar mass=0.3412g/106g/mole=0.00322moles

since 1mole of Na2CO3 require=2moles of HCl

0.00322moles of Na2CO3 moles would have required=0.00644moles of HCl

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.