Question

What is the pH of a `"0.150-M"` `"NaOH"` solution?

The answer is 13.18 but I'm not sure how to get that answer? From what I know the formula is pH = -log[H+] so I just put -log(.150) which = .82 but that's not the correct answer. Can someone please correct me and show steps?

Answer 1

`13.176`

Explanation:

The `"pH"` of the solution is given by

`color(blue)(ul(color(black)("pH" = - log (["H"_3"O"^(+)]))))`

so you can't use

`color(red)(cancel(color(black)("pH" = - log(0.150))))`

because that's the concentration of the hydroxide anions, `"OH"^(-)`, not of the hydronium cations, `"H"_3"O"^(+)`. In essence, you calculated the `"pOH"` of the solution, not its `"pH"`.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a `1:1` mole ratio.

`"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)`

So your solution has

`["OH"^(-)] = ["NaOH"] = "0.150 M"`

Now, the `"pOH"` of the solution can be calculated by using

`color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))`

In your case, you have

`"pOH" = - log(0.150) = 0.824`

Now, an aqueous solution at `25^@"C"` has

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

This means that you have

`"pH" = 14 - 0.824 = 13.176`

You should leave the answer rounded to three decimal places because you have three sig figs for the concentration of the solution.