Question

100 mL of a solution of HCl with pH value 3
is diluted with 400 mL of water. The new
pH of the solution is

Answer 1

Answer:

4*log(2) =1.20

Explanation:we know that

pH= -log(conc)

from question

3= -log(conc)

=> conc = 10^-3

from this conc = no of moles /volume

no of moles =0.1

now new pH after dilution is

pH = -log(0.1/500)

=> 4 log(2)

1.20