A solution with `pH2.699` is diluted two time3s, then calculate the pH of the resulting sdolution. [Given antilog of `0.3010=2]`

Answer

3

Explanation

`pH=2.699`
`:. [H^(+)]=10^(-pH)=10^(-2.6999) = 10^(0.3010)xx10^(-3)`
`( :' "antilog" 0.3010=2)`
`:. [H^(+)]=2xx10^(-3)`
Now it is dilited two tomes.
Now, `[H^(+)]= (2xx10^(-3))/(2)=1xx10^(-3)`
`:. pH= -log[H^(+)]= -log[1xx10^(-3)]=3`