# Question f7407

Jan 14, 2018

$\left[{\text{H"_3"O}}^{+}\right] = 1 \cdot {10}^{- 14}$ $\text{M}$

#### Explanation:

The first thing that you need to do here is to calculate the molarity of the hydroxide anions, ${\text{OH}}^{-}$, which are delivered to the solution by the sodium hydroxide in a $1 : 1$ mole ratio.

$\text{moles of OH"^(-) = "moles of NaOH}$

To do that, use the molar mass of sodium hydroxide to convert the number of grams of solute to moles

4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.1 moles NaOH"

This means that your solution will contain $0.1$ moles of hydroxide anions, which in $\text{100 mL" = 100/1000 quad "L}$ of the solution have a molarity of

["OH"^(-)] = "0.1 moles OH"^(-)/(100 * 10^(-3) quad "L")= "1 M"

Now, you know that an aqueous solution at room temperature has

color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1 * 10^(-14)quad "M"^2)))

This means that your solution contains

["H"_3"O"^(+)] = (1 * 10^(-14)"M"^color(red)(cancel(color(black)(2))))/(1color(red)(cancel(color(black)("M")))) = 1 * 10^(-14)quad "M"#