Calculate the "pH" of a solution when "1 g" of acetic acid and "4 g" of sodium hydroxide are dissolved in "2 L" of water?

Apr 3, 2018

$\text{pH} = 12.6$

Explanation:

Start by using the molar masses of the two compounds to determine how many moles of each you are adding.

1 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05 color(red)(cancel(color(black)("g")))) = "0.0167 moles CH"_3"COOH"

4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.10 moles NaOH"

Now, acetic acid and sodium hydroxide react in a $1 : 1$ mole ratio to produce aqueous sodium acetate and water.

${\text{CH"_ 3"COOH"_ ((aq)) + "NaOH"_ ((aq)) -> "CH"_ 3"COONa"_ ( (aq)) + "H"_ 2"O}}_{\left(l\right)}$

As you can see, you have more moles of sodium hydroxide than of acetic acid, so right from the start, you can say that sodium hydroxide is in excess, which implies that acetic acid is the limiting reagent.

This means that acetic acid will be completely consumed by the reaction. After the reaction is complete, the solution will contain $0$ moles of acetic acid and

$\text{0.10 moles " - " 0.0167 moles" = "0.0833 moles NaOH}$

Sodium hydroxide is a strong base, so it dissociates completely in aqueous solution to produce hydroxide anions in a $1 : 1$ mole ratio. This means that the resulting solution will contain $0.0833$ moles of hydroxide anions.

You can thus say that the molarity of the hydroxide anions in the resulting solution will be

["OH"^(-)] = "0.0833 moles"/"2 L" = "0.04165 M"

Finally, you know that an aqueous solution at ${25}^{\circ} \text{C}$ has

$\text{pH + pOH = 14}$

Since

"pOH" = - log(["OH"^(-)])

you can say that the $\text{pH}$ of the solution is given by the equation

"pH" = 14 + log(["OH"^(-)])

Plug in your value to find

$\text{pH} = 14 + \log \left(0.04165\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{12.6}}}$

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