# Question 5f23b

Dec 23, 2017

Here's what I got.

#### Explanation:

The idea here is that you can use the $\text{pH}$ of the solution to find the initial concentration of hydroxide anions, ${\text{OH}}^{-}$.

You know that an aqueous solution at ${25}^{\circ} \text{C}$ has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH + pOH = 14}}}}$

This means that you have

$\text{pOH" = 14 - "pH}$

$\text{pOH} = 14 - 14 = 0$

Since you know that

"pOH" = - log(["OH"^(-)])

you can say that the concentration of hydroxide anions will be

$\left[\text{OH"^(-)] = 10^(-"pOH}\right)$

$\left[{\text{OH}}^{-}\right] = {10}^{- 0}$

["OH"^(-)] = "1 M"

Now, when the $\text{pH}$ of the solution is equal to $7$ at ${25}^{\circ} \text{C}$, the solution is actually neutral, meaning that you have

$\text{pOH" = "pH} = 7$

In this case, the concentration of hydroxide anions, which is equal to the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$, is equal to

["OH"^(-)] = 1 * 10^(-7)color(white)(.)"M"

So, you know that you must decrease the concentration of hydroxide anions by a factor of

"DF" = (1 color(red)(cancel(color(black)("M"))))/(1 * 10^(-7)color(red)(cancel(color(black)("M")))) = color(blue)(10^7)#

Here $\text{DF}$ is the dilution factor. Now, in order for the concentration of the solution to decrease by a factor equal to $\text{DF}$, its volume must increase by a factor of $\text{DF}$.

This means that you have

$\text{DF" = V_"diluted"/V_"stock}$

${V}_{\text{diluted" = "DF" * V_"stock}}$

In your case, the volume of the diluted solution must be equal to

${V}_{\text{diluted" = color(blue)(10^7) * "0.2 mL}}$

${V}_{\text{diluted" = 2 * 10^6color(white)(.)"mL}}$

So in order for the $\text{pH}$ of the solution to decrease from $14$ to $7$, the volume of the solution must increase from $\text{0.2 mL}$ to $2 \cdot {10}^{6}$ $\text{mL}$.

You can thus say that you can dilute this solution by adding enough water to get the total volume of the solution to $2 \cdot {10}^{6}$ $\text{mL}$, which, for all intended purposes, is equal to

${V}_{\text{water" = 2 * 10^6color(white)(.)"mL" - "0.2 mL}}$

${V}_{\text{water" = "1,999,999.8 mL}}$

However, keep in mind that you only have one significant figure for the volume of the initial solution, so you must round the answer to one significant figure.

${V}_{\text{water" = "2,000,000 mL}}$