wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Titration of Na3PO4 with HCl using phenolphthalein end point produces Na2HPO4 while subsequent titration using methyl orange end point produces NaH2PO4. Find the mg of Na2HPO4 present in a additional 35 mL of the same acid to reach the methyl orange end point.

A
492 mg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
170 mg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
662 mg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Insufficient data
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 170 mg
For phenolphthalein end point,
Na3PO4+HCl(25ml ,0.12 N)Na2HPO4+NaCl
Milliequivalents of HCl=25×0.12=3=Milliequivalents of Na3PO4 (in sample) = Milliequivalents of Na2HPO4 (produced)
For methyl orange end point,
Na3PO4+HCl(25 ml, 0.12 N)Na2HPO4+NaCl
Milliequivalents of HCl=35×0.12=4.2= Milliequivalents of Na2HPO4 (in sample + produced from Na3PO4)
Milliequivalents of Na2HPO4 in sample =4.23=1.2
Milliequivalents of Na2HPO4 in sample =1.2×142=170.4 170
Hence, Na2HPO4 present is 170 mg.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Preparation of Alcohols 2
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon