# Question a3b8b

Nov 30, 2015

$11.48$

#### Explanation:

Right from the start, you can say for a fact that the pH of the resulting solution will be greater than $7$.

You're dealing with a neutralization reaction between sodium hydroxide, a strong base, and hydrochloric acid, a strong acid. The balanced chemical equation for this reaction looks like this

${\text{HCl"_text((aq]) + "NaOH"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Now, when equal number of moles of hydrochloric acid and sodium hydroxide are added, a complete neutralization takes place. That is, both the acid and the base will be completely consumed and the resulting solution will become neutral.

If you have more moles of acid than you have of base, the resulting solution will be acidic. Likewise, if you have more moles of base than of acid, the resulting solution will be basic.

In your case, you have $3.60 \cdot {10}^{- 3}$ moles of sodium hydroxide and $5.95 \cdot {10}^{- 4}$ moles of hydrochloric acid. This means that the hydrochloric acid will be completely consumed in the reaction, and you'll be left with excess sodium hydroxide.

${n}_{N a O H} = 3.60 \cdot {10}^{- 3} - 5.95 \cdot {10}^{- 4} = \text{0.003005 moles NaOH}$

Now, when you're dealing with a basic solution, you can find its $\text{pH}$ by calculating its $\text{pOH}$.

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

As you know, sodium hydroxide will dissociate completely to form sodium cations and hydroxide anions

${\text{NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

This means that every mole of sodium hydroxide will produce one mole of hydroxide ions in solution. The concentration of the hydroxide ions will thus be

["OH"^(-)] = "0.003005 moles"/"1.00 L" = "0.003005 M"#

The $\text{pOH}$ of the solution will be

$\text{pOH} = - \log \left(0.003005\right) = 2.52$

The pH of the solution will be

$\text{pH" = 14 - "pOH}$

$\text{pH} = 14 - 2.52 = \textcolor{g r e e n}{11.48}$