Question

Question #e285b

Answer 1

`"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)))`