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?

1 Answer
Dec 6, 2017

Answer:

#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.