# Question 70577

Jan 30, 2016

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation for this neutralization reaction. Since sodium hydroxide is a strong base that dissociates completely in aqueous solution, you can represent it by using the hydroxide anions, ${\text{OH}}^{-}$

${\text{HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((aq])^(-) + "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

Now, you have $1 : 1$ mole ratios across the board, you can say that for every mole of strong base added to the weak acid solution, one mole of acid will be consumed and one mole of conjugate base, ${\text{A}}^{-}$, will be produced.

Use the molarities of the two solutions to figure out how many moles of each you're mixing

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{H A} = \text{0.45 M" * 25 * 10^(-3)"L" = "0.01125 moles HA}$

${n}_{O {H}^{-}} = {\text{0.34 M" * 15 * 10^(-3)"L" = "0.005100 moles OH}}^{-}$

So, you have more moles of weak acid than of strong base. This means that the number of moles of hydroxide anions will be completely consumed by the reaction.

Don't forget that the reaction will also produce ${\text{A}}^{-}$, the conjugate base of the weak acid, in the aforementioned $1 : 1$ mole ratio with the strong base and the weak acid.

The resulting solution will thus contain

${n}_{O {H}^{-}} = \text{0 moles}$

${n}_{H A} = 0.01125 - 0.005100 = \textcolor{g r e e n}{\text{0.0062 moles HA}}$

${n}_{{A}^{-}} = \textcolor{g r e e n}{{\text{0.0051 moles A}}^{-}}$

The answers are rounded to two sig figs.

Now, in order to calculate the concentrations of the two species in the resulting solution, you first need to calculate the total volume of the solution.

${V}_{\text{total}} = {V}_{H A} + {V}_{O {H}^{-}}$

${V}_{\text{total" = "15 mL" + "25 mL" = "40 mL}}$

SIDE NOTE As you know, one milliliter is equivalent to one cubic centimeter, which is why I used milliliters instead of the given cubic centimeters.

The concentrations of the weak acid and conjugate base will thus be

$\left[\text{HA"] = "0.0062 moles"/(40 * 10^(-3)"L") = color(green)("0.16 M}\right)$

$\left[\text{A"^(-)] = "0.0051 moles"/(40 * 10^(-3)"L") = color(green)("0.13 M}\right)$

To find the pH of the solution, write a balanced chemical equation for the equilibrium that gets established when the weak acid partially ionizes in aqueous solution

${\text{HA"_text((aq]) + "H"_2"O"_text((aq]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "A}}_{\textrm{\left(a q\right]}}^{-}$

You know that the acid dissociation constant, ${K}_{a}$, can be expressed using the equilibrium concentrations of the weak acid, conjugate base, and hydronium ions

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["A"^(-)])/(["HA}\right]\right)$

Rearrange to solve for the concentration of hydronium ions

["H"_3"O"^(+)] = (["HA"])/(["A"^(-)]) * K_a

Plug in the values you have for the concentrations of the weak acid and conjugate base

["H"_3"O"^(+)] = (0.16color(red)(cancel(color(black)("M"))))/(0.13color(red)(cancel(color(black)("M")))) * K_a = 1.231 * K_a

The pH of the solution will thus be

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

$\text{pH} = - \log \left(1.231 \cdot {K}_{a}\right)$

$\text{pH} = - \log \left(1.231\right) {\overbrace{- \log \left({K}_{a}\right)}}^{\textcolor{p u r p \le}{p {K}_{a}}}$

$\text{pH} = p {K}_{a} - 0.0902$

As you know, the $P {K}_{a}$ of the acid is defined as

$\textcolor{b l u e}{p {K}_{a} = - \log \left({K}_{a}\right)}$

So, does this result make sense?

Take a look at the concentrations of the weak acid and conjugate base. Notice that you have more acid than conjugate base, which means that you can expect the pH of the solution to be lower than the $p {K}_{a}$ of the acid.

That happens because the pH of the buffer solution is equal to the $p {K}_{a}$ of the acid when you have equal concentrations of weak acid and conjugate base present.

This is why more acid implies $\text{pH} < p {K}_{a}$. By comparison, more conjugate base would imply $\text{pH} > p {K}_{a}$.