# Question e285b

Apr 8, 2017

$\text{pH} = 1.30$

#### Explanation:

Even without doing any calculations, you should be able to say that the $\text{pH}$ 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

$\text{500 mL + 500 mL = 1,000 mL}$

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

$\textcolor{p u r p \le}{{\left[{\text{H"_3"O"^(+)]) = (x color(white)(.)"moles H"_3"O"^(+))/(2 * "500 mL solution") = color(purple)(1/2 * ["H"_3"O}}^{+}\right]}_{0}}$

Now, the $\text{pH}$ of the solution is defined as

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

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

${\text{pH" = "pH}}_{0} + \log \left(2\right)$

As you can see, the $\text{pH}$ increased $\left(\log \left(2\right) > 1\right)$ as a result of the decrease in the concentration of hydronium cations.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 1 + 0.30 = 1.30}}}$