*PLEASE SHOW WORK FOR FULL AWARDED POINTS :)* Thank you!!!! Na2CO3(aq)+2HCl(aq)=

Question

*PLEASE SHOW WORK FOR FULL AWARDED POINTS :)* Thank you!!!!

Na2CO3(aq)+2HCl(aq)=2NaCl(aq)+H2O(aq)+CO2(aq)

Q1) A student weighed out 0.1537 grams of 100% Na2CO3 and then followed the titration procedure listed in this experiment. The volume of HCl which was used was 35.44 mL. What was the molarity of the HCl?

Q2)An unknown sample of sodium carbonate was titrated with the HCl from above.  The volume of acid used was 28.40 mL.  The weight of the sample was 0.1864 grams.  What is the percent purity of the unknown sodium carbonate sample?  

again please show work. I want to see what I am doing wrong :)

Explanation / Answer

Q1) Assuming that Na2CO3 is titrated to the second equivalence point, the procedure for solving this problem is as follows:

(0.1537 g)*(1 mole/105.988 g) = 1.45 x 10-3 moles Na2CO3

Required moles of HCl to reach second equivalence point: 2*1.45 x 10-3 = 2.90 x 10-3 moles HCl

X*(0.03544 Liters) = 2.90 x 10-3 moles

X = 0.0818 M

Q2) Moles of HCl used = (0.02840 L)*(0.0818 moles/L) = 2.32 x 10-3 moles HCl

Moles of Na2CO3 = (1/2)*(2.32 x 10-3 moles) = 1.16 x 10-3 moles Na2CO3

Grams of Na2CO3 = (1.16 x 10-3 moles)*(105.988 g/mole) = 0.1232 g Na2CO3

Purity = 0.1232/0.1864 = 0.6608 = 66.08% purity

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