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20 mL of 0.1 M solution of compound NaCO...

20 mL of 0.1 M solution of compound `NaCO_(3).NaHCO_(3).2H_(2)O` is titrated against 0.05 M HCL. X mL of HCL is used when phenolphthalein is used as an indicator and y mL of HCL is used when methly orange is the indicator in two separate titrations. Hence (y-x) is:

A

80 mL

B

30 mL

C

120 mL

D

180 mL

Text Solution

Verified by Experts

The correct Answer is:
A
In presence of phenoplhthalein,
`(1)/(2)` meq. Of `Na_(2)CO_(3)` = meq of HCl
`2xx40xx0.5xx(1)/(2)=x xx0.05`
`:. X=40 mL`
with M.O.
Meq. Of `Na_(2)CO_(3)+` Meq. Of `NaHCO_(3)` = Meq. Of HCl
`2xx40xx0.05+40xx0.05=yxx0.05`
y = 120 mL
`:. (y-x)=80 mL`
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