Question

What is the pH of a 0.0605 M solution of sodium hydroxide, `NaOH`?

Answer 1

`"pH" = 12.78`

Explanation:

You're dealing with a solution that contains a strong base, so right from the start you should know that sodium hydroxide dissociates completely in aqueous solution to produce sodium cations, `"Na"^(+)`, and hydroxide anions, `"OH"^(-)`.

![http://wps.prenhall.com/wps/media/objects/476/488316/ch14.html]()

This means that for every mole of sodium hydroxide that you're dissolving in water, you will get one mole of hydroxide anions in aqueous solution.

Therefore, for a given volume of your solution, the concentration of the hydroxide anions will be equal to that of the strong base

`["OH"^(-)] = ["NaOH"] = "0.0605 M"`

Now that you know the concentration of hydroxide anions, you can calculate the pOH of the solution by using

`color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))`

Plug in your value to find

`"pOH" = - log(0.0605) = 1.22`

You know that for aqueous solutions at room temperature you have

`color(blue)(|bar(ul(color(white)(a/a)"pOH " + " pH" = 14color(white)(a/a)|)))`

Plug in the pOH of the solution to get

`"pH" = 14 - 1.22 = color(green)(|bar(ul(color(white)(a/a)color(black)(12.78)color(white)(a/a)|)))`