`10^(-6) M NaOH` is diluted by `100` times. The `pH` of diluted base is

Options

A. between 5 and 6
B. between 6 and 7
C. between 10 and 11
D. between 7 and 8

Answer

D

Explanation

`[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`