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In the mixture of (NaHCO3 + Na2CO3), vol...

In the mixture of `(NaHCO_3 + Na_2CO_3)`, volume of `HCI` required is x mL with phenolphthalein indicator and y mL with methly orange indicator in the same titration. Hence, volume of `HCI` for complete reaction of `Na_2CO_3` is :
(a) 2x
(b) y
(c) x/2
(d) (y-x)

A

`2x`

B

`y`

C

`(x)/(2)`

D

`(y-x)`

Text Solution

Verified by Experts

The correct Answer is:
A
(i). With phenolphthalein indicator `NaHCO_(3)` does jnot react with HCl whereas with `Na_(2)CO_(3)` is `50%` reaction `V_(HCl)=x mL` (Half titre value of `Na_(2)CO_(3)`)
(ii). With methyl orange indicator `NaHCO_(3)` reacts completely with HCl and with `Na_(2)CO_(3)` is `100%` reaction.
But y " mL of " HCl is added after `Na_(2)CO_(3)` has reacted upto `NaHCO_(3)` i.e., half titre value of `Na_(2)CO_(3)`.
`V_(HCl)=` full litre value of `NaHCO_(3)+` Half tire value of `Na_(2)CO_(3)`
`y mL=x'+x mL`
`therefore` Volume of HCl requried for complete reaction of `Na_(2)CO_(3)=x mL+x mL=2x mL`
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