# A particular drain cleaner contains "NaOH". What is the value of ["OH"^(-)] in a solution produced when "15.0 g" of "NaOH" dissolves in enough water to make "375 mL" of solution?

Apr 8, 2018

${\text{1.00 mol L}}^{- 1}$

#### Explanation:

Start by calculating the molarity of the sodium hydroxide solution. As you know, molarity can be calculated by dividing the number of moles of solute by the total volume of the solution in liters.

["NaOH"] = "moles NaOH"/(color(white)(underbrace(color(blue)(375 * 10^(-3) quad "L")))
$\textcolor{w h i t e}{a a a a a a a a a a a a} \uparrow$
$\textcolor{b l u e}{\text{using the fact that 1 L" = 10^3 quad "mL}}$

To find the number of moles of sodium hydroxide present in your sample, use the molar mass of the compound.

15.0 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = "0.3750 moles NaOH"

This means that the molarity of the sodium hydroxide solution is

["NaOH"] = "0.3750 moles"/(375 * 10^(-3) quad "L") = "1.00 mol L"^(-1)

Now, sodium hydroxide is a strong base, which implies that it dissociates completely in aqueous solution to produce sodium cations and hydroxide anions in $1 : 1$ mole ratios.

$\textcolor{w h i t e}{\overbrace{\textcolor{b l a c k}{\text{NaOH"_ ((aq))))^(color(blue)("1 mole dissolved"))) -> color(white)(overbrace(color(black)("Na"_ ((aq))^(+)))^(color(blue)("1 mole produced"))) + color(white)(overbrace(color(black)("OH"_ ((aq))^(-)))^(color(blue)("1 mole produced}}}}$

This means that your solution will have

$\left[\text{Na"^(+)] = ["OH"^(-)] = ["NaOH}\right]$

and so

["OH"^(-)] = color(darkgreen)(ul(color(black)("1.00 mol L"^(-1))))

The answer is rounded to three sig figs, the number of sig figs you have for your values.