Grade 12th passPhysical Chemistry
How to calculate the amount of nahco3 in a mixture of na2co3 and nahco3 via a double indicator titration
1 Answer
Vikas TU
Let the qstn. be:
A solution contains both NaHCO3 and Na2CO3. Titration of a 50.00 mL portion to a phenolphthalein end point requires 22.42 mL of 0.1170 M HCl. A second 50.00 mL aliquot requires 48.04 mL of the HCl solution when titrated to a bromocresol green end point. Calculate the molar concentration of NaHCO3 in the solution.
Solution :
here are two things that you have to know when you work with this problem.
1) when you titrate the base solution to the phenolphthalein end point you have caried out the following reaction:
Na2CO3 + HCl = NaHCO3 + NaCl
You have reacted exactly half the Na2CO3. The remaining solution then contains the unreacted NaHCO3 from this reaction plus the unreacted NaHCO3 originally in the solution.
2) When you complete the titration to the bromocresol green end point you have reacted all the original Na2CO3 and the NaHCO3 .
1st step: titration to phenolphthalein end point:
From the equation:
1mol HCl reacts with 1 mol Na2CO3 to produce 1mol NaHCO3
Mol of HCl consumed: 22.42mL of 0.1170M HCl
Moles HCl = 22.42/1000*0.1170 = 0.002623mol HCl
This reacted with half the Na2CO3, so the original solution contained 0.002623mol*2 = 0.00525 mol Na2CO3
When you titrated to the BCG end point, you will have used 22.42*2 = 44.84mL of the HCl to react completely with all the Na2CO3
You used 48.04mL HCl, therefore 48.04-44.84 = 3.2mL HCl was used to react with the original NaHCO3 in the solution:
NaHCO3 + HCl → NaCl + CO2 + H2O
1mol NaHCO3 reacts with 1mol HCl
Mol HCl in 3.2mL of 0.1170M HCl = 3.2/1000*0.1170 = 0.000374 mol HCl
Therefore in the 50mL of original solution there were 0.000374 mol NaHCO3
Molarity = 0.000374/0.050 = 0.0075M NaHCO3 solution originally.
1) when you titrate the base solution to the phenolphthalein end point you have caried out the following reaction:
Na2CO3 + HCl = NaHCO3 + NaCl
You have reacted exactly half the Na2CO3. The remaining solution then contains the unreacted NaHCO3 from this reaction plus the unreacted NaHCO3 originally in the solution.
2) When you complete the titration to the bromocresol green end point you have reacted all the original Na2CO3 and the NaHCO3 .
1st step: titration to phenolphthalein end point:
From the equation:
1mol HCl reacts with 1 mol Na2CO3 to produce 1mol NaHCO3
Mol of HCl consumed: 22.42mL of 0.1170M HCl
Moles HCl = 22.42/1000*0.1170 = 0.002623mol HCl
This reacted with half the Na2CO3, so the original solution contained 0.002623mol*2 = 0.00525 mol Na2CO3
When you titrated to the BCG end point, you will have used 22.42*2 = 44.84mL of the HCl to react completely with all the Na2CO3
You used 48.04mL HCl, therefore 48.04-44.84 = 3.2mL HCl was used to react with the original NaHCO3 in the solution:
NaHCO3 + HCl → NaCl + CO2 + H2O
1mol NaHCO3 reacts with 1mol HCl
Mol HCl in 3.2mL of 0.1170M HCl = 3.2/1000*0.1170 = 0.000374 mol HCl
Therefore in the 50mL of original solution there were 0.000374 mol NaHCO3
Molarity = 0.000374/0.050 = 0.0075M NaHCO3 solution originally.
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