Question #5f23b

1 Answer
Dec 23, 2017

Here's what I got.

Explanation:

The idea here is that you can use the "pH" of the solution to find the initial concentration of hydroxide anions, "OH"^(-).

You know that an aqueous solution at 25^@"C" has

color(blue)(ul(color(black)("pH + pOH = 14")))

This means that you have

"pOH" = 14 - "pH"

"pOH" = 14 - 14 = 0

Since you know that

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

you can say that the concentration of hydroxide anions will be

["OH"^(-)] = 10^(-"pOH")

["OH"^(-)] = 10^(-0)

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

Now, when the "pH" of the solution is equal to 7 at 25^@"C", the solution is actually neutral, meaning that you have

"pOH" = "pH" = 7

In this case, the concentration of hydroxide anions, which is equal to the concentration of hydronium cations, "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 "DF" is the dilution factor. Now, in order for the concentration of the solution to decrease by a factor equal to "DF", its volume must increase by a factor of "DF".

This means that you have

"DF" = V_"diluted"/V_"stock"

V_"diluted" = "DF" * V_"stock"

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

V_"diluted" = color(blue)(10^7) * "0.2 mL"

V_"diluted" = 2 * 10^6color(white)(.)"mL"

So in order for the "pH" of the solution to decrease from 14 to 7, the volume of the solution must increase from "0.2 mL" to 2 * 10^6 "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 * 10^6 "mL", which, for all intended purposes, is equal to

V_"water" = 2 * 10^6color(white)(.)"mL" - "0.2 mL"

V_"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_"water" = "2,000,000 mL"