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The volume of 0.1 M HCl required in mL t...

The volume of 0.1 M HCl required in mL to neutralise 20 mL of a solution containing 0.106 g of `Na_(2)CO_(3)` is

A

10

B

5

C

20

D

40

Text Solution

Verified by Experts

The correct Answer is:
C
Given,
Molarity (HCl) `[M_(1)]=0.1M`
Volume of `Na_(2)CO_(3)` used = 20mL
Mass of `Na_(2)CO_(3)` used = 0.106g
thus, molarity `(M_(2))` of `Na_(2)CO_(3) = 0.106/106xx1000/20`
`(M_(2))Na_(2)CO_(3) = 0.05M`
`because` Equation for neutralisation is
`M_(1)V_(1)Z_(1)=M_(2)V_(2)Z_(2)`
Where `Z_(1)andZ_(2)` are number of moles of `H^(+)andOH^(-)` ions, given by HCl and `Na_(2)CO_(3)` in water
`Z_(1)=1("for"HCl),Z_(2)=2("for"Na_(2)CO_(3))`
`V_(1)=(M_(2)xxV_(2)xxZ_(2))/(M_(1)xxZ_(1))=(0.05xx20xx2)/(0.1xx1)=20mL`
Hence, volume of HCl required = 20 mL
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