# Question 70830

Nov 4, 2015

$\text{pH} = 0.641$

#### Explanation:

Your starting point here is the balanced chemical equation for this neutralization reaction - I'll use the net ionic equation

${\text{H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-) -> 2"H"_2"O}}_{\textrm{\left(l\right]}}$

Hydrochloric acid, $\text{HCl}$, is a strong acid, which means that it dissociates completely in aqueous solution to form hydronium ions, ${\text{H"_3"O}}^{+}$, in a $1 : 1$ mole ratio.

Sodium hydroxide, $\text{NaOH}$, is a strong base, which means that it too dissociates completely, but it forms hydroxide ions, ${\text{OH}}^{-}$, in a $1 : 1$ mole ratio.

Before doing any calculations, try to predict what will happen. Notice that you're dealing with two solutions that have the same molarity, which means that the number of moles of solute depends exclusively on the volume.

Since you're adding a smaller volume of sodium hydroxide, you can expect the hydroxide ions to be consumed completely by the reaction. Automatically, you can expect the pH of the solution to be quite acidic, since you're left with excess hydronium ions coming from the hydrochloric acid solution.

Use the molarity and the volume of the hydrochloric acid solution to determine how many moles of hydronium ions you had in solution before adding the sodium hydroxide

$c = \frac{n}{V} \implies n = c \cdot V$

${n}_{\text{HCl" = "0.500 M" * 20.0 * 10^(-3)"L" = "0.0100 moles H"_3"O}}^{+}$

Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide ions were added

${n}_{\text{NaOH" = "0.500 M" * 7.45 * 10^(-3)"L" = "0.003725 moles OH}}^{-}$

So what happened when the two solutions were mixed? The hydronium and hydroxide ions neutralize each other to form water, but notice that you don't have enough moles of the latter to have a complete neutralization.

More specifically, the reaction will consume all the moles of hydroxide ions and leave thesolution with excess hydronium ions.

The number of moles hydronium ions in excess will be

${n}_{{H}_{3} {O}^{+}} = {\text{0.0100 moles" - "0.003725 moles" = "0.006275 moles H"_3"O}}^{+}$

Do not forget that the volume of the solution changes! The total volume will now be

${V}_{\text{total" = V_"HCl" + V_"NaOH}}$

${V}_{\text{total" = "20.0 mL" + "7.45 mL" = "27.45 mL}}$

The cocnentration of hydronium ions will now be

["H"_3"O"^(+)] = "0.006275 moles"/(27.45 * 10^(-3)"L") = "0.22860 M"

The pH of the solution will thus be

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

$\text{pH} = - \log \left(0.22860\right) = \textcolor{g r e e n}{0.641}$