Question 12725

Mar 24, 2017

$2.78 \cdot {10}^{- 4}$ ${\text{moles OH}}^{-}$

Explanation:

For starters, you know that the pH of the solution must increase, so you can say that you will need to add hydroxide anions, ${\text{OH}}^{-}$.

This is the case because the pH of the solution increases as the concentration of hydronium cations increases. Adding hydroxide anions will consume some of the hydronium cations present in solution, which in turn will ensure that the pH increases.

As you know, the pH of the solution is calculated by using the equation

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

You can rearrange this equation to solve for the concentration of hydronium cations

log(["H"_3"O"^(+)]) = - "pH"

This is equivalent to

10^log(["H"_3"O"^(+)]) = 10^(-"pH")

which gets you

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

${\left[{\text{H"_ 3"O}}^{+}\right]}_{0} = {10}^{-} 3.39$

["H"_ 3"O"^(+)]_ 0 = 4.07 * 10^(-4)"M"

After you add the hydroxide anions, you have

${\left[{\text{H"_ 3"O}}^{+}\right]}_{1} = {10}^{- 3.89}$

["H"_ 3"O"^(+)]_ 1 = 1.29 * 10^(-4)"M"#

As you can see, the pH of the solution increased because the concentration of hydronium cations decreased.

Since you're dealing with $\text{1 L}$ of solution, you can treat molarity and number of moles interchangeably.

This means that the number of moles of hydronium cations decreases by

$4.07 \cdot {10}^{- 4} \textcolor{w h i t e}{.} \text{moles" - 1.29 * 10^(-4)color(white)(.)"moles" = 2.78 * 10^(-4)color(white)(.)"moles}$

Since hydronium cations and hydroxide anions react in a $1 : 1$ mole ratio

${\text{H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O}}_{\left(l\right)}$

you can say that in order to consume $2.78 \cdot {10}^{- 4}$ moles of hydronium cations you must add $2.78 \cdot {10}^{- 4}$ moles of hydroxide anions.

I'll leave the answer rounded to three sig figs, but you should round it to two sig figs because you have two decimal places for the two pH values, and that if we don't take into account the fact that you have one significant figure for the volume.