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

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The pH of a solution obtained by mixing 100 mL of a solution pH=3 with 400 mL of a solution of pH=4 is
A. 3- log 2.8
B. 7- log 2.8
C. 4- log 2.8
D. 5- log 2.8
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Correct Answer - C
`N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))`
`10^(-3)xx100+10^(-4)xx400=N_(3)(100+400)`
or, `N_(3)=(0.1+0.04)/(500)=(0.14)/(500)=2.8xx10^(-4)M`
`pH =-log (2.8xx10^(-4))=4-"log"2.8`
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