Question

Consider an acidic solution diluted by a factor of 10. Calculate the pH of a

0.10 M solution of H+ and a 0.010 M solution of H+. How did the pH change? Please help ASAP this is part of a lab question I need to answer! :(

Answer 1

pH of 0.1M solution : 1

pH of 0.01M solution : 2

The pH changes by 1 if the solution is diluted by a factor of 10

Step-by-step explanation:

pH =
`-log[H+]`

where ,

`[H+]` is the concentration of
`H+` ions

∴

• concentration of
  `H+` ions is :
  `10^(-1)M`

pH of this solution is :

`pH=-log[H+]\\pH=-log[10^(-1)]\\pH=1`

when the solution gets diluted by a factor of 10 , volume increases 10 times.

As Molarity is inversly proportional to volume , molarity decreases by 10 times.

Thus , molarity becomes : 0.01M

pH of this solution is :

`pH=-log[H+]\\pH=-log[10^(-2)]\\pH=2`

Δ
`pH=2-1=1`

• concentration of
  `H+` ions is :
  `10^(-2)M`

pH of this solution is :

`pH=-log[H+]\\pH=-log[10^(-2)]\\pH=2`

when the solution gets diluted by a factor of 10 , volume increases 10 times.

As Molarity is inversly proportional to volume , molarity decreases by 10 times.

Thus , molarity becomes : 0.01M

pH of this solution is :

`pH=-log[H+]\\pH=-log[10^(-3)]\\pH=3`

Δ
`pH=3-2=1`