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 :

Options

A. `3-log 2.8`
B. `7-log 2.8`
C. `4-log 2.8`
D. `5-log 2.8`

Answer

C

Explanation

`{:(V_(1)=100mL,V_(2)=400mL),(pH=3,pH=4),(therefore M_(1)=10^(-3)M,M_(2)=10^(-4)M):}`
Molarity, M of resultant solution may be calculated as :
`M_(1)V_(1)+M_(2)V_(2)=MV`
`10^(-3)xx100+10^(-4)xx400=M(100+400)`
or `M=(10^(-1)+4xx10^(-2))/(500)=(10^(-1)(1+0.4))/(500)`
`=(0.14)/(500)=0.028xx10^(-2)=2.8xx10^(-4)1`
`pH=-log[H^(+)]=-log (2.8xx10^(-4))`
`=4-log2.8`.