Question #e285b

1 Answer
Apr 8, 2017

Answer:

#"pH" = 1.30#

Explanation:

Even without doing any calculations, you should be able to say that the #"pH"# of the solution will increase as a result of the dilution.

This happens because a dilution decreases the concentration of a solute by increasing the volume of the solution.

In your case, the volume of the solution doubles

#"500 mL + 500 mL = 1,000 mL"#

which means that the concentration of hydronium cations will be halved. If you take #x# to be the number of moles of hydronium cations present in the initial solution, you can say that

#["H"_3"O"^(+)]_0 = (x color(white)(.)"moles H"_3"O"^(+)]/"500 mL solution"#

After you dilute the initial solution, you will have

#color(purple)(["H"_3"O"^(+)]) = (x color(white)(.)"moles H"_3"O"^(+))/(2 * "500 mL solution") = color(purple)(1/2 * ["H"_3"O"^(+)]_0)#

Now, the #"pH"# of the solution is defined as

#"pH" = - log(["H"_3"O"^(+)])#

The initial solution had

#color(blue)("pH"_0 = - log(["H"_3"O"^(+)]_0))#

After you dilute the solution, you will have

#"pH" = - log( color(purple)(["H"_3"O"^(+)]))#

#"pH" = - log( color(purple)(1/2 * ["H"_3"O"^(+)]_0))#

This is equivalent to

#"pH" = - [log(1/2) + log(["H"_3"O"^(+)])]#

#"pH" = - log(1/2) color(blue)(-log(["H"_3"O"^(+)]_0))#

#"pH" = color(blue)("pH"_0) - log(1/2)#

#"pH" = color(blue)("pH"_0) - [log(1) - log(2)]#

Finally, you will have

#"pH" = "pH"_0 + log(2)#

As you can see, the #"pH"# increased #(log(2) > 1)# as a result of the decrease in the concentration of hydronium cations.

#color(darkgreen)(ul(color(black)("pH" = 1 + 0.30 = 1.30)))#