# A buffer solution is prepared by mixing 1.0 mole of HA and 1.0 mole of NaA into 1.0 L distilled water. Calculate the change in pH when 25.0 mL of 0.20 M NaOH is added into 500 mL of the buffer? Given Ka(HA) is 1.75×10^(-5) M

Jan 24, 2016

Here's what I got.

#### Explanation:

I'll show you hot to solve this one without using the Henderson - Hasselbalch equation. This method is a bit long, but I think that it's a great help to understanding the general idea behind how buffers work.

So, your solution is said to contain $\text{HA}$, a weak acid, and $\text{NaA}$, the salt of its conjugate base, the ${\text{A}}^{-}$ anion.

You know that this buffer is prepared by dissolving $\text{1 mole}$ of each chemical species in $\text{1.0 L}$ of distilled water. Use this information to find the molarity of the weak acid and of the conjugate base

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

["HA"] = ["A"^(-)] = "1.0 moles"/"1.0 L" = "1.0 M"

Keep in mind that the salt dissociates in a $1 : 1$ mole ratio to form sodium cations, ${\text{Na}}^{+}$, and the conjugate base of the acid, ${\text{A}}^{-}$.

Now, you take $\text{500 mL}$ of this buffer solution. Calculate how many moles of weak acid and conjugate base will be present in this sample

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

${n}_{H A} = {n}_{{A}^{-}} = \text{1.0 M" * 500 * 10^(-3)"L" = "0.50 moles}$

To find the pH of this buffer solution, you need to use an ICE table - remember to use the molarities of the weak acid and conjugate base!

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

color(purple)("I")" " " "1.0" " " " " " " " " " " " " " " " " " "0" " " " " " " " " " "1.0
color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)
color(purple)("E")" "1.0-x" " " " " " " " " " " " " " " "x" " " " " " " " "1.0+x

By definition, the acid dissociation constant, ${K}_{a}$, will be equal to

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["A"^(-)])/(["HA}\right]\right) = \frac{x \cdot \left(1 + x\right)}{1 - x} = 1.75 \cdot {10}^{- 5}$

Because the value of ${K}_{a}$ is so small, you can use the approximation

$1.0 \pm x \approx 1.0$

This will get you

$1.75 \cdot {10}^{- 5} = \frac{x \cdot 1}{1} = x$

Here $x$ represents the equilibrium concentration of the hydronium ions, so

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

${\text{pH}}_{0} = - \log \left(1.75 \cdot {10}^{- 5}\right) = 4.757$

SIDE NOTE This is why the pH of a buffer solution is said to be equal to the acid's $p {K}_{a}$ when equal concentrations of weak acid and conjugate base are present.

Now, sodium hydroxide is a strong base, which means that it will neutralize the weak acid to produce water and the conjugate base of the acid.

This means that you can expect the pH of the solution to increase a little after the addition of the strong base.

The balanced chemical equation for this reaction looks like this - I'll use the hydroxide anion to represent the strong base

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

Notice that the hydroxide anions react in a $1 : 1$ mole ratio with the weak acid and produce conjugate base in the same $1 : 1$ mole ratio.

Use the molarity and volume of the sodium hydroxide solution to find how many moles of strong base are added to the buffer

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

${n}_{O {H}^{-}} = {\text{0.20 M" * 25.0 * 10^(-3)"L" = "0.0050 moles OH}}^{-}$

This means that after the neutralization reaction takes place, you will be left with

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

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

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

The new volume of the buffer will be

${V}_{\text{total" = V_"initial" + V_"base}}$

${V}_{\text{total" = "500 mL" + "25.0 mL" = "525 mL}}$

Now calculate the new molarities of the weak acid and conjugate base

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

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

Now it's back to the ICE table. To keep the answer at a reasonable length, I won't add it again. This time, the initial concentrations of the weak acid and conjugate base will no longer be $\text{1.0 M}$.

Once again, you will have

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

In this case, this will be equal to

${K}_{a} = \frac{x \cdot \left(0.9619 + x\right)}{0.9429 + x} = 1.75 \cdot {10}^{- 5}$

Using the same approximation, you can say that

$1.75 \cdot {10}^{- 5} = x \cdot \frac{0.9619}{0.9429}$

Therefore,

$x = \frac{1.75 \cdot 0.9429}{0.9619} \cdot {10}^{- 5} = 1.7154 \cdot {10}^{- 5}$

The pH of the solution will now be

${\text{pH}}_{1} = - \log \left(1.7154 \cdot {10}^{- 5}\right) = 4.766$

The change in pH can be calculated as

${\Delta}_{\text{pH" = "pH"_1 - "pH}} _ 0$

${\Delta}_{\text{pH}} = 4.766 - 4.757 = \textcolor{g r e e n}{0.009}$

Jan 24, 2016

Δ"pH" = +0.01

#### Explanation:

1. Write the chemical equation for the buffer.
2. Calculate the $\text{pH}$ of the original buffer.
3. Calculate the moles of base added.
4. Calculate the new moles of $\text{HA}$ and ${\text{A}}^{-}$
5. Calculate the $\text{pH}$ of the new solution.
6. Calculate the change in $\text{pH}$.

1. Chemical Equation

${\text{HA" + "H"_2"O" ⇌ H_3"O"^(+)+ "A}}^{-}$; K_"a" = 1.75 × 10^-5

2. $\text{pH}$ of Buffer

$\textcolor{w h i t e}{X X X X X X l} {\text{HA" + "H"_2"O" ⇌ H_3"O"^+ + "A}}^{-}$; K_"a" = 1.75 × 10^-5
${\text{E/mol·L}}^{-} 1 : 1.0 \textcolor{w h i t e}{X X X X X X X X X X l l} 1.0$

["HA"] = ["A"^-] = "1.0 mol"/"1.0 L" = "1.0 mol/L"

We can use the Henderson-Hasselbalch equation to calculate the $\text{pH}$.

"pH" = "p"K_"a" + log(("[A"^(-)"]")/"[HA]") = -log(1.75× 10^-5) + log(1.0/1.0) = 4.757 + 0 = 4.757

$\text{Moles of OH"^(-) = 0.0250 color(red)(cancel(color(black)("L"))) × "0.20 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.0050 mol}$

4. New moles of $\text{HA}$ and ${\text{A}}^{-}$

$\textcolor{w h i t e}{X X X X X X} {\text{HA" + "H"_2"O" ⇌ H_3"O"^+ + "A}}^{-}$
$\text{I/mol:} \textcolor{w h i t e}{X X l} 0.50 \textcolor{w h i t e}{X X X X X X X X X X} 0.50$
$\text{C/mol:} \textcolor{w h i t e}{l} - 0.0050 \textcolor{w h i t e}{X X X X X X X l} + 0.0050$
$\text{E/mol:} \textcolor{w h i t e}{X X} 0.495 \textcolor{w h i t e}{X X X X X X X X X l} 0.505$

You are using only half of the original buffer, so you are starting with 0.50 mol each of $\text{HA}$ and ${\text{A}}^{-}$.

The added ${\text{OH}}^{-}$ will decrease the amount of $\text{HA}$ and increase the amount of ${\text{A}}^{-}$ by 0.0050 mol.

5. $\text{pH}$ of new solution

Since $\text{HA}$ and ${\text{A}}^{-}$ are in the same solution, the ratio of their concentrations is the same as the ratio of the moles.

"pH" = "p"K_"a" + log(("[A"^(-)"]")/"[HA]") = 4.757 +log(0.505/0.495) = 4.757 + 0.009 = 4.766

6. Change in pH

$\text{ΔpH = 4.766 – 4.757 = 0.009}$

We usually express $\text{pH}$ to only two decimal places, so

$\text{ΔpH = 0.01}$