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40 mL of 0.05 M solution of sodium sesqu...

40 mL of 0.05 M solution of sodium sesquicarbionate `(Na_(2)CO_(3). NaHCO_(3).2H_(2)O)` is titrated against 0.05 M HCl solution. X ml of HCl solution is used when phenophthalein is the indicator and y ml of HCl is used when methyl orange is the indicator in two separate titrations Hence (y-x) is

A

80 mL

B

30 mL

C

120 mL

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
In the first titration
Eq. HCl = `(1)/(2)` eq `NC_(2)CO_(3)`
`(x xx0.05)/(1000)=(1)/(2)xx(0.050xx40xx2)/(1000)impliesx=40ml`
In the second titration
Eq of HCl = Eq `NC_(2)CO_(3)+` Eq `NKHCO_(3)`
`(yxx0.05)/(1000)=(40xx0.05xx2)/(1000)+(40xx0.05xx1)/(1000)`
`y=120impliesy-x=80`
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