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`10^(-6) ` M NaOH is diluted 100 times. The pH of the diluted base is
A. between 5 and 6
B. between 6 and 7
C. between 10 and 11
D. between 7 and 8

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Correct Answer - D
`[OH^(-)]` after dilution `= (10^(-6))/(100)=10^(-8)M`
`:. OH^(-)` ions from `H_(2)O` cannot be neglected.
Total `[OH^(-)]=10^(-8)+10^(-7)=10^(-8)xx11`
`:. pOH=-log(11xx10^(-8))=8-1.04=6.96`
`:. pH = 14 - 6.96 = 7.04`

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