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

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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?

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

#### Answer:

#### Explanation:

The

#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, **not** of the hydronium cations,

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

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

So your solution has

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

Now, the

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

In your case, you have

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

Now, an aqueous solution at

#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.