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30 cc of M/3 HCl, 20cc of M/2 HNO(3) and...

30 cc of `M/3` HCl, 20cc of `M/2` `HNO_(3)` and 40 cc of `M/4 NaOH` solutions are mixed and the volume was made upto `1 dm^(3)`. The pH of the resulting solution is :

A

2

B

1

C

3

D

8

Text Solution

Verified by Experts

The correct Answer is:
A
Total millimolar of `H^(+)=(30xx(1)/(3))+(20xx(1)/(2))`
`=10+10=20`
Total millimoles of `OH^(-)=40xx(1)/(4)=10`
`:. H^(+)` ions left after neutralization = 10 millimole
Volume of solution `=1 dm^(3) = 1000cc`
Hence, molarity of `H^(+)` ions `=(10)/(1000)M = 10^(-2)M`
`:. pH =- log [H^(+)]=-log 10^(-2)=2`
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