1.

The pH of the solution obtained by mixing 100 mL of a solution of pH = 3 with 400 mL of a solution of pH = 4 is

Answer»

3 - log 2.8
7- log 2.8
4- log 2.8
5- log 2.8

Solution :pH = 3 means `10^(-3)` M and pH = 4 means `10^(-4)` M .
100 mL of `10^(-3)` M solution contain = `(10^(-3) xx 100)/(1000) = 10 xx 10^(-5)` MOL
400 mL of `10^(-4)` M solution contain= `(10^(-4) xx 400)/(1000)= 4 xx 10^(-5)` mol
Total number of moles present = `(10 xx 10^(-5))+ (4 xx 10^(-5)) = 14 xx 10^(-5)` mol
Total volume after MIXING `= (100 + 400) = 500` mL
Molarity of the mixture = `(14 xx 10^(-5))/(500) xx 1000`
`= ((14)/(5) xx 10^(-4)) M`


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