A buffer solution is prepared by mixing 1 mole of HA and 1 mole of NaA into 1 L distilled water. Calculate the change in pH when 2t mL of 0.20 M NaOH is added into 500mL of the buffer?

Jan 16, 2016

Here's what I got.

Explanation:

SIDE NOTE Since you mistyped the volume of strong base added to the buffer, I'll pick a sample of sodium hydroxide solution and use it as an example.

So, you're dealing with a buffer solution that contains a weak acid, $\text{HA}$, and its conjugate base, ${\text{A}}^{-}$, delivered to the solution by its salt, $\text{NaA}$.

As you know, the pH of a buffer solution that contains a weak acid and its conjugate base can be calculated using the Henderson - Hasselbalch equation

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))" "

Here you have

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

${K}_{a}$ - the acid dissociation constant for the weak acid

Since the problem doesn't provide you with a value for ${K}_{a}$, you will have to find a way to express the change in pH, ${\Delta}_{p H}$, without using the $p {K}_{a}$ of the acid.

So, you prepare your buffer solution by mixing $1$ mole of weak acid and $1$ mole of conjugate base.

SIDE NOTE The salt of the conjugate base dissociates in a $1 : 1$ mole ratio with the conjugate base, so any number of moles of salt you have will be equal to the number of moles of conjugate base produced in solution.

The molarity of the two chemical species in the stock solution will be

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

["HA"] = ["A"^(-)] = "1 mole"/"1 L" = "1 M"

Now, you're going to take a $\text{500-mL}$ sample of this stock solution, This sample will contain

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

${n}_{H A} = {n}_{{A}^{-}} = \text{1 M" * 500 * 10^(-3)"L" = "0.50 moles}$

of weak acid and of conjugate base.

Before adding the strong base, the pH of the buffer will be equal to the acid's $p {K}_{a}$, since you're dealing with equal concentrations of weak acid and conjugate base

"pH"_1 = pK_a + overbrace(log( (1 color(red)(cancel(color(black)("M"))))/(1color(red)(cancel(color(black)("M"))))))^(color(purple)(=0))

Let's assume that you're adding $\text{25 mL}$ of sodium hydroxide solution. The hydroxide anions will react in a $1 : 1$ mole ratio with the weak acid (neutralization reaction) and produce conjugate base

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

The number of moles of hydroxide anions added to the $\text{500-mL}$ buffer will be

${n}_{O {H}^{-}} = {\text{0.20 M" * 25 * 10^(-3)"L" = "0.0050 moles OH}}^{-}$

After the neutralization reaction takes place, you'll be left with

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

${n}_{H A} = \text{0.50 moles" - "0.0050 moles" = "0.495 moles}$

${n}_{{A}^{-}} = \text{0.50 moles" + "0.0050 moles" = "0.505 moles}$

The total volume of the buffer will now be

${V}_{\text{total" = "500 mL" + "25 mL" = "525 mL}}$

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

["HA"] = "0.495 moles"/(525 * 10^(-3)"L") = "0.9429 M"

["A"^(-)] = "0.505 moles"/(525 * 10^(-3)"L") = "0.9619 M"

The pH of the solution will now be

"pH"_2 = pK_a + log( (0.9619 color(red)(cancel(color(black)("M"))))/(0.9429color(red)(cancel(color(black)("M")))))

${\text{pH}}_{2} = p {K}_{a} + 0.00866$

Therefore, the change in pH will be

${\Delta}_{p H} = {\text{pH"_2 - "pH}}_{1}$

${\Delta}_{p H} = \textcolor{red}{\cancel{\textcolor{b l a c k}{p {K}_{a}}}} + 0.00866 - \textcolor{red}{\cancel{\textcolor{b l a c k}{p {K}_{a}}}}$

${\Delta}_{p H} = \textcolor{g r e e n}{0.00866}$

Now do the same with the actual volume of sodium hydroxide given to you in the problem.