# Question 9b388

Apr 14, 2016

$\text{pH} = 10.83$

#### Explanation:

You're titrating sodium hydroxide, $\text{NaOH}$, a strong base, with hydrochloric acid, $\text{HCl}$, a strong acid, so right from the start you know that a complete neutralization will produce a neutral solution of pH equal to $7$.

The balanced chemical equation for this neutralization reaction looks like this

${\text{NaOH"_ ((aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

The two reactants react in a $1 : 1$ mole ratio, which means that the reaction consumes one mole of strong acid for every mole of strong base present in solution.

Use the molarity and volume of the sodium hydroxide solution to determine how many moles of strong base it contains

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

n_(NaOH) = "0.0500 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(50.00 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))

$= \text{0.00250 moles NaOH}$

Now, if you have fewer moles of strong acid in that $\text{24.50-mL}$ sample than needed for a complete neutralization, the pH of the solution will be higher than $7$.

That happens because the moles of hydrochloric acid will be completely consumed before all the moles of sodium hydroxide get the chance to react.

On the other hand, if you more moles of strong acid than needed for a complete neutralization, the pH of the solution will be lower than $7$.

The number of moles of hydrochloric acid added is equal to

n_(HCl) = "0.1000 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(24.50 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))

$= \text{0.00245 moles HCl}$

As you can see, you have fewer moles of hydrochloric acid than of sodium hydroxide, which means that the resulting solution will be basic.

The two reactants are consumed in a $1 : 1$ mole ratio, which means that after the reaction is complete you'll be left with

${n}_{H C l} = \text{0 moles } \to$ completely consumed

${n}_{N a O H} = \text{0.00250 moles" - "0.00245 moles" = "0.0000500 moles NaOH}$

Sodium hydroxide dissociates in a $1 : 1$ mole ratio to form sodium cations, ${\text{Na}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$

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

This means that the resulting solution will contain

${n}_{O {H}^{-}} = {n}_{N a O H} = {\text{0.0000500 moles OH}}^{-}$

The total volume of the solution will be

${V}_{\text{total}} = {V}_{N a O H} + {V}_{H C l}$

${V}_{\text{total" = "50.00 mL" + "24.50 mL" = "74.50 mL}}$

The concentration of hydroxide anions will be

["OH"^(-)] = "0.0000500 moles"/(74.50 * 10^(-3)"L") = "0.000671114 mol L"^(-1)

Now, the pOH of the solution can be found by using

color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))#

In your case, you will have have

$\text{pOH} = - \log \left(0.00067114\right) = 3.17$

Now, aqueous solution at room temperature have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the pH of the solution will be

$\text{pH} = 14 - 3.17 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 10.83 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As predicted, the pH is indeed higher than $7$ because the acid was completely consumed before all the moles of sodium hydroxide were neutralized.