# Question bc8f1

Aug 11, 2017

$1 \cdot {10}^{- 12}$ $\text{M}$

#### Explanation:

Start by calculating the number of moles of sodium hydroxide present in your sample. To do this, use the molar mass of sodium hydroxide

0.004 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.0001 moles NaOH"

Now, your first goal here is to calculate the concentration of hydroxide anions, ${\text{OH}}^{-}$, in the solution.

You can safely assume that the volume of the solution will be equal to the volume of water, and since sodium hydroxide is a strong base, you can say that your solution will contain $0.0001$ moles of hydroxide anions in $\text{10 mL}$ of solution.

This is the case because sodium hydroxide dissociates in a $1 : 1$ mole ratio to produce hydroxide anions, so

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

This means that you will have--remember, by definition, the molarity of the hydroxide anions must reflect a volume of ${10}^{3} \textcolor{w h i t e}{.} \text{mL" = "1 L}$ of solution!

["OH"^(-)] = "0.0001 moles NaOH"/(10 * 10^(-3)color(white)(.)"L") = "0.01 mol L"^(-1)

Now, an aqueous solution at room temperature has--I won't add the units for ${K}_{W} = {10}^{- 14}$

$\left[{\text{OH"^(-)] * ["H"_ 3"O}}^{+}\right] = {K}_{W} = {10}^{- 14}$

Here $\left[{\text{H"_3"O}}^{+}\right]$ is the concentration of hydronium cations, which you'll sometimes see written as hydrogen cations, ${\text{H}}^{+}$, and ${K}_{W}$ is the ionization constant of water.

This means that you have

["H"_3"O"^(+)] = 10^(-14)/(["OH"^(-)])

Plug in your value to find

["H"_3"O"^(+)] = 10^(-14)/(0.01) = color(darkgreen)(ul(color(black)(1 * 10^(-12)color(white)(.)"M")))#

The answer is rounded to one significant figure.