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100 ml of a solution of HCl with pH valu...

`100 ml` of a solution of `HCl` with `pH` value `3` is diluted with `400 ml` of water. The new `P^(H)` of the solution is

A

1 mL

B

10 mL

C

100 mL

D

1000 mL

Text Solution

Verified by Experts

The correct Answer is:
C
`P_("Final"^(H)=P_("Intial")^(H)-"log(V_(2))/(V_(1))`
`1=-"log"(V_(2))/(V_(1))="log"(1000)/(V_(1))V_(1)100 ml`
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