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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! :(

Respuesta :

Answer:

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

Explanation:

pH = [tex]-log[H+][/tex]

where ,

[tex][H+][/tex] is the concentration of [tex]H+[/tex] ions

  • concentration of [tex]H+[/tex] ions is : [tex]10^{-1}M[/tex]

 pH of this solution is :

[tex]pH=-log[H+]\\pH=-log[10^{-1}]\\pH=1[/tex]

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 :

[tex]pH=-log[H+]\\pH=-log[10^{-2}]\\pH=2[/tex]

Δ[tex]pH=2-1=1[/tex]

  • concentration of [tex]H+[/tex] ions is : [tex]10^{-2}M[/tex]

 pH of this solution is :

[tex]pH=-log[H+]\\pH=-log[10^{-2}]\\pH=2[/tex]

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 :

[tex]pH=-log[H+]\\pH=-log[10^{-3}]\\pH=3[/tex]

Δ[tex]pH=3-2=1[/tex]