Experiment 8: Acids, Bases, Salts, and Buffers

GOAL AND OVERVIEWHydrolysis of salts will be used to study the acid-base properties of dissolved

ions in aqueous solutions. The approximate pH of these solutions will be determined using acid-base indicators. A buffer solution will be prepared, and its ability to moderate pH will be investigated alongside solutions that cannot function as buffers.

Objectives of the data analysis understand conjugate acid-base pairs and equilibria of weak acids and bases perform calculations involving ionic equilibria understand the components of buffer solutions and how they work to resist

changes in pH

SUGGESTED REVIEW AND EXTERNAL READING reference section on data analysis; relevant textbook information on acids,

bases, salts, and buffers

BACKGROUNDAqueous solutions of substances such as HCl (hydrochloric acid) or HC2H3O2

(acetic acid) are expected to be acidic ([H3O+] > [OH-], pH < 7), while aqueous solutions of substances such as NaOH (sodium hydroxide) or NH3(ammonia) are expected to be basic ([H3O+] < [OH-], pH > 7).

Recall that Brønsted-Lowry acids are proton donors and bases are proton acceptors. In water, an acid can donate a proton to water to form H3O+ and the conjugate base; a base can accept a proton from water to form OH- and the conjugate acid. Also, when an acid and base undergo a neutralization reaction, the products usually include water and a “salt”. For example,

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

It may be surprising to discover that the dissolution of some salts into water can affect the pH. For example, aqueous solutions of NaNO2 and KC2H3O2 are basic, whereas those of NH4Cl and FeCl3 are acidic. Many salts are strong electrolytes and exist as ions in aqueous solutions. It is the dissolved ions that have the potential to undergo proton transfers with water to generate H3O+ or OH-. These reactions are sometimes called hydrolysis reactions (or proton transfer reactions).

Anions such as C2H3O2- or OCl- that are the conjugate bases of weak acids

(HC2H3O2 or HOCl) react with water to form OH-. Cations such as Cr3+ or NH4

+ come from weak bases and react to form H3O+.

A weak acid is one which does not completely dissociate in or react with the water solution. Instead, this equilibrium is established. The equilibrium constant is called the acid dissociation constant and is given the symbol, Ka.

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HA(aq) + H2O(l) A-(aq) + H3O+

(aq) or HA(aq) A-(aq) + H+

(aq) (1)

or (2)

HA represents a weak acid; A- stands for its conjugate base.

A weak base is one which does not completely dissociate in or react with the water solution. Instead, this equilibrium is established. The equilibrium constant is called the base dissociation constant and is given the symbol, Kb.

A-(aq) + H2O(l) HA(aq) + OH-

(aq) (3)

(4)Looking at the behavior of the conjugate acid-base pair

HA and A- and their behavior in water:HA(aq) A-

(aq) + H+(aq) (1)

A-(aq) + H2O(l) HA(aq) + OH-

(aq) (3)the net ionic equation is: H+

(aq) + OH-(l) H2O(l)

and:

Kw at 25oC is 1.010-14. This relationship allows you to find Kb for the conjugate base of a weak acid or Ka for the conjugate acid of a weak base.

The stronger an acid is, the larger its Ka and the weaker its conjugate base (with a smaller Kb). Likewise, the weaker an acid is, the smaller its Ka and the stronger its conjugate base (with a larger Kb). Similar statements apply to bases and their conjugate acids.

Hydrolysis – Anions of Weak AcidsAnion of weak acids can react with proton sources. In water, the anions react

with water to some extent to form OH- (and the conjugate acid). The OH- causes the solution pH to be greater than 7.

For example: NO2-(aq) + H2O(l) HNO2(aq) + OH-

(aq) Note the similarity to: NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq)

The Ka of HNO2 is 4.510-4, so Kb for NO2- is:

Compare this to the value of Kb for NH3: 1.810-5. Which is the stronger base, NO2

- or NH3?

To calculate the pH of a solution containing one of these ions, you will need to know the concentration of the H3O+ at equilibrium. The standard approach is write the balanced chemical equation with a grid below it where rows give initial amounts, changes in amounts, and final equilibrium amounts and concentrations. This is an “ICE” table:

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X + Y XY

I_nitial amount (mole/L) [X]initial [Y]initial 0

C_hange in amount (mole/L)

−x −x +x

E_quilibrium amount (mole/L)

[X]initial −x [Y]initial −x +x

initial amounts of reactants (amounts before the reaction begins): find via the dilution equation

initial amount of product (the amount before the reaction begins): here, assume zero

equilibrium concentration of product (after equilibrium is attained): find from pH

As an example, find the pH of 0.10 M NaOCl. Ka of HOCl is 3.010-8.OCl-

(aq) + H2O(l) HOCl(aq) + OH-(aq)

OCl- + H2O HOCl + OH-

0.10 --- 0 ~0-x --- +x +x

0.10 - x --- x xKb is small, so x can be neglected relative to 0.10, so 0.10 – x 0.10

Anions derived from strong acids, such as Cl- from HCl or NO3- from HNO3, do

not react with water to affect pH. Br-, I-, SO42-, ClO4

- are also nonbases in water (spectator ions). The parent acids are so strong in water that the conjugate bases are exceedingly weak.

Anions with ionizable protons such as HCO3-, H2PO4

-, and HPO42- are

amphiprotic – they may be acidic or basic, depending on the values of Ka and Kb for the ion. These types of species will not be considered in this lab.

Hydrolysis – Cations of Weak BasesCations derived from weak bases react with water to increase the H3O+

concentration (acidic). Consider NH4+ in water: NH4

+(aq) + H2O(l)

NH3(aq) + H3O+(aq)

or: NH4+

(aq) NH3(aq) + H+ (aq)

Ka for NH4+, the conjugate acid of NH3, can be determined using the Kb of NH3

and Kw:

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Cations of the group 1A metals (Li+, Na+, K+, Rb+, Cs+) and the group 2A metals (Ca2+, Sr2+, Ba2+) do not react with water and are nonacids. They do not affect the pH of the solution.

Hydrated cations of many other metals do hydrolyze to produce acidic solutions. For example:

Fe(H2O)63+

(aq) + H2O(l) Fe(H2O)5(OH)2+(aq) + H3O+

(aq) The waters of hydration are sometimes omitted: Fe(H2O)6

3+(aq) ≡ Fe3+

(aq); Fe(H2O)5(OH)2+

(aq) ≡ Fe (OH)2+(aq).

Additional hydrolysis reactions can occur to produce species such as Fe(H2O)4(OH)2

+(aq), even leading to the precipitation of Fe(OH)3. The equilibria of

these types of ions can be complex. The examples illustrate the acidic nature of dipositive and tripositive ions, and most of the H+

(aq) in such a solution come from these reactions.

Summary of Hydrolysis of SaltsThe acidity, basicity, or neutrality of an aqueous salt solution can be predicted

based on the strengths of the acid and base from which the salt was derived.

1. Cation from strong base; anion from strong acid Ex. NaCl, KNO3

Solution has pH = 7 (neutral)

2. Cation from weak base; anion from strong acid Ex. NH4Cl, Zn(NO3)2

Solution has pH < 7 (acidic) due to the hydrolysis of the cation.

3. Cation from strong base; anion from weak acid Ex. NaF, KNO2

Solution has pH > 7 (basic) due to the hydrolysis of the anion.

4. Cation from weak base; anion from weak acid Ex. NH4F, NH4C2H3O2

Solution pH is determined by the relative Ka and Kb of the cation and anion.

In part 1 of this experiment, the pH of water and several salt solutions will be tested. The pH and initial concentration of each solution (and using the approximation that the extent of dissociation is small relative to that concentration) allows an approximate value of Ka or Kb to be calculated. A set of acid-base indicators will be used to determine pH (see figure).

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color changespH

13

12

11

10

9

8

7

5

4

2

6

3

Methyl

orange

Methyl

red

Bromothymolblue

Phenol red

Phenol-phthale

in

Alizarin

yellow-R

4.4Yellow

3.1Red

4.8Red

6.0 Yellow

6.0Yellow

7.6 Blue

6.6Yellow

8.0Red

8.2Colorl

ess

10.0Red

10.1Yellow

12.0Red

As an example, a solution of 0.10 M NaOBr has a pH of 10.85.This solution would appear: Yellow in methyl orange (pH > 4.4); yellow in methyl red (pH > 6.0); blue in bromothymol blue (pH > 7.6); red in phenol red (pH > 8.0); red in phenolphthalein (pH > 10.0), and yellow-red in alizarin yellow-R (pH between 10.1 and 12.0).

Determine which ion hydrolyzes in water and calculate its Ka or Kb. Na+ is a spectator ion and does not affect the pH of the solution. OBr- is the conjugate base of a weak acid, HOBr. The hydrolysis equation is:

OBr-(aq) + H2O(l) HOBr(aq) + OH-

(aq) The pOH of the solution is 14.00 – 10.85 = 3.15, so the [OH-] is 10-3.15 =

7.110-4.Because the [OH-] concentration derived from the OBr- hydrolysis is much

greater than that from the auto-ionization of water, the [OH-] is essentially equal to [HOBr] formed. The [OBr-] concentration is not significantly changed by the hydrolysis.

OBr- + H2O HOBr + OH-

0.10 --- 0 ~07.110-4 --- +7.110-4 +7.110-4

~0.10 --- 7.110-4 7.110-4

The solutions will generally be more acidic than predicted primarily due to the presence of dissolved CO2. CO2 reacts with water to generate H3O+ (aqueous protons, H+

(aq)):CO2(g) + H2O(l) H2CO3(aq) HCO3

-(aq) + H+

(aq) The solubility of CO2 is greatest in basic solutions; intermediate in neutral; and least in acidic. Boiling can help remove the CO2.

Buffer Solutions (“Buffers”)The pH of a solution frequently needs to be controlled. The effect of the

addition of even a small amount of strong acid or strong base to water is large – for example, the addition of 0.001 mol HCl to 1 L water causes the pH to drop from 7.0 to 3.0; the addition of 0.001 mol of NaOH to 1 L water causes the pH to rise to 11.0. These types of changes can be disastrous, particularly in biological systems.

A buffer solution resists large changes in pH upon the addition of small amounts of strong acid or strong base. A buffer has two components: one that will react with added H+ and one that will react with added OH-. Usually these two parts are an acid and its conjugate base. Buffers are often prepared by mixing a weak acid (or weak base) with a salt of that acid (or base). For example,

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a buffer could be made by adding NaC2H3O2 solution to an HC2H3O2 solution. Buffers of almost any pH can be made by proper choice of components and concentrations.

If significant concentrations of both a weak acid and its conjugate base are present in the buffer:

added OH- reacts with the weak acid: HA + OH- H2O + A-

K1 = 1/Kb, A-

added H+ reacts with the conjugate base: A- + H+ HAK2 = 1/Ka, HA

K1 and K2 are large, so the above reactions essentially go to completion. Once equilibrium is again reached, HA and A- (and H2O) are the dominant species in the solution:

HA H+ + A-

The expression for [H+] indicates that the pH depends on two factors: the value of Ka for the acid component of the buffer and the ratio of the weak acid to its conjugate base ([HA]/[A-]).

Looking again at the equilibrium expressions above: small amounts of added OH- would slightly increase the A- concentration; small amounts of added H+ would slightly increase the HA concentration. As long as the ratio of [HA]/[A-] is relatively constant, the pH change is small.

Buffers work most effectively when [HA]/[A-] is roughly equal to 1 ([HA] ≈ [A-]). When [HA] equals [A-], [H+] equals Ka:

Since pH is usually of interest, the Ka expression is rearranged to give the Henderson-Hasselbach equation:

(5)

Eq. 5 is particularly useful for calculating the pH of a buffer, particularly since the small amounts of acid or base that ionize can normally be ignored. Thus, the initial concentrations of HA and A- can be used directly in Eq. 5.

As an example, find the pH of a buffer that is 0.150 M lactic acid (HC3H5O3, “HLc”) and 0.130 M sodium lactate (NaC3H5O3, “NaLc”). Ka for HLc = 1.410-4.

HLc-

(aq) H+(aq) + Lc-

(aq

0.150 ~0 0.130-x +x +x

~0.150 x ~0.130

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Addition of strong acid or base to buffersWhen strong acid is added to a buffer solution containing HA and A-, the H+

consumes A- to form HA. So, [HA] increases while [A-] decreases. When strong base is added to the buffer, HA is consumed by the added OH- to form A-. Here, [HA] decreases while [A-] increases. Two steps are required to calculate the resulting pH – first, a stoichiometry problem for the neutralization; second, an equilibrium problem for the new concentrations.

For example, calculate the pH of 1 L of the HLc-Lc- buffer (see above) if 0.001 mol NaOH is added.

Stoichimetry: HLc(aq) + OH-(aq) Lc-

(aq) + H2OHLc + OH- Lc- + H2O

Before rxn

0.150 0.001 0.130 ---

Change -0.001 -0.001 +0.001 ---After rxn ~0.149 ~0 0.131 ---

Equilibrium: HLc(aq) H+(aq) + Lc-

(aq)

PRELAB HOMEWORK (to be filled out in your bound lab notebook before you perform the experiment)Title and date Define: (1) pH, (2) weak acid, (3) weak base, (4) buffer, (5) Henderson-Hasselbach equation, Answer:1. When salts dissolve in water, certain ions may be present. Label each of the

following aqueous ions as an acid, base, or neither. Examples: Na+, neither (non-acid); NO2

-, basea. K+ b. NH4

+ c. C2H3O2- d. Cl- e. Zn(H2O)6

2+ e. Al(H2O)63+

f. CO32-

note: SO42- (there is a stronger acid in solution we use here; so this will be

considered non-base in this lab)2. Write the chemical equilibrium that describes the acid-base behavior of each of

the following species in aqueous solution. Indicate whether the solution should be acidic, basic, or neutral. For example, for NO2

-: NO2- + H2O HNO2 + OH- basic

a. ClO- b. (CH3)3NH+ c. F- d. Zn(H2O)62+

3. If 50.0 mL of 1.00 M formic acid (HCOOH) is mixed with 50.0 mL of a 1.00 M sodium formate (COOH-) solution, what is the pH of the resulting solution?

HLc(aq) H+(aq) + Lc-

(aq)

0.149 ~0 0.131-x +x +x

~0.149 x ~0.131

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Could this solution act as a buffer? Ka of HCOOH is 1.810-4

4. Write the equations showing how a buffer solution containing HF and F- would react with added acid and with added base (2 reactions). Based on the products of these two reactions, explain why the pH is held fairly constant when a small amount of acid or base is added.

5. Calculate the molarity of an acetic acid solution (HC2H3O2 = HAc) prepared by diluting 5 mL 1M HAc to a final volume of 100 mL.

6. Calculate the molarity of an acetate ion solution (C2H3O2- = Ac-) prepared by

diluting 5 mL 1M NaAc to a final volume of 100 mL.7. Calculate the molarities of HAc and Ac- in a solution prepared by mixing 5 mL

1M HAc with 5 mL 1M NaAc and diluting to a final volume of 100 mL.Procedure (Experimental Plan)Data Tables

PROCEDURE

CAUTION: You are working with acids and bases. Handle them with caution and clean up spills.

Part 1. Hydrolysis of Salts and Solution pH (proton transfer reactions in water)1) Boil approximately 250 mL deionized water and allow it to cool to room

temperature (be careful when heating/handling).2) Determine the approximate pH of two water samples and six 0.1M salt

solutions (by observing the colors of six different indicators; see the figure). Test the solutions in two batches, four at a time.

3) Place 1 drop of an indicator into a well (see below). Add a small volume (about 1 ‘squirt’ from the pipet) of the solution to be tested. Record the color. Do not fill each well (about ½ way).

4) In addition to unboiled deionized water and boiled deionized water, the six 0.1 M salt solutions to test are: NaCl, NH4Cl, NaC2H3O2, ZnCl2, KAl(SO4)2, and Na2CO3. The indicators to use are: methyl orange, methyl red, bromothymol blue, phenol red, phenolphthalein, and alizarin yellow-R.

5) Use the schematics below as a guide. Rinse the well plate with a small amount of boiled deionized water and dry it before and after each set of observations.

6) Record your estimate of the pH based on the indicator colors. 7) Wash the well plate thoroughly when finished.

Waste disposal: Neutralize all liquid waste with NaHCO3 and dispose down the drain with a lot of water unless instructed otherwise by your TA.

Note – your TA may instruct you to split up the procedure for part 1 and share with another lab pair.

Make sure you record all the required data for both sets of solutions.

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Note – your TA may instruct you to start with part 2 and do part 1 after that. Please follow instructions.

Set 1:Methyl orange

Methyl red

Bromo-thymol blue

Phenol red

Phenol-phthalein

Alizarinyellow-R

deionizedH2O

boileddeionized

H2O

NaCl

NaC2H3O2

Set 2:Methyl orange

Methyl red

Bromo-thymol blue

Phenol red

Phenol-phthalein

Alizarinyellow-R

NH4Cl

ZnCl2

KAl(SO4)2

Na2CO3

Part 2a. pH of Buffer Solutions1) Obtain about 10 mL 1M acetic acid (HC2H3O2, “HAc”) and about 10 mL 1M

sodium acetate (NaC2H3O2, “NaAc”) solutions in small, labeled beakers.

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2) Pipet 5 mL 1M HAc into a labeled 100mL volumetric flask and dilute to the mark with water.

3) Pipet 5 mL 1M NaAc into a labeled 100mL volumetric flask and dilute to the mark with water.

4) Pipet 5 mL 1M HAc into a labeled 100mL volumetric flask; pipet into the same flask 5 mL 1M NaAc(aq). Dilute to the mark with water.

5) The above steps should provide you with 100 mL of each of the following: acetic acid solution (HAc), acetate (NaAc) solution, and acetic acid/acetate solution (HAc/Ac-). The calculation of each solution’s molarity was performed in your prelab.

6) Measure the pH of the solutions using a calibrated pH meter (see instructions in lab). The pH meters are delicate and you need to read the printed instructions carefully (and follow the TA’s instructions, too).Compare the measured values to those predicted based on literature values.

7) Save the solutions for part 2b.

Part 2b. Effect of Acid and Base on Buffer pH9) Pipet 1 mL 1.0 M NaOH into the HAc solution and record the pH. Note – there is a weak acid present to react with the added OH -. Do you expect a huge jump in pH here?10) Pipet 1 mL 1.0 M NaOH into the Ac- solution and record the pH. Note – there is only a weak base present in the initial solution, so what do you think added OH -

will do to the pH here?11) Pipet 1 mL 1.0 M NaOH into the HAc/Ac- and record the pH. Note – the original solution is a buffer, so do you expect the pH change to be relatively large or small here?12) Please clean the pH meter and leave the electrode immersed in the pH = 7 buffer (yellow solution).

Waste disposal: Neutralize all liquid waste with NaHCO3 and dispose down the drain with a lot of water unless instructed otherwise by your TA.DATA ANALYSIS Part 1. Hydrolysis of Salts and Solution pH (proton transfer reactions in water)1) For each liquid or solution tested, write the equilibrium expression for the

hydrolysis reaction that would explain your observations.2) Calculate the Ka or Kb for each cation or anion that hydrolyzes. Compare your

calculated value to the literature value and give possible explanations for the deviation (if any).

Literature values:

Species Ka at 25oC

HC2H3O2, acetic acid 1.810-5

Al(H2O)63+, hydrated aluminum

cation110-5

Zn(H2O)62+, hydrated zinc

cation110-9

H2CO3, carbonic acid 4.310-7

HCO3-, hydrogen carbonate ion 5.610-11

NH4+, ammonium ion 5.610-1092

Part 2a. pH of Buffer Solutions3) Calculate the expected pH of the 0.05M HAc, 0.05M Ac-, and 0.05M

HAc/0.05M Ac-. Compare your measured value to the theoretical value and give possible explanations for the deviation (if any).

Part 2b. Effect of Acid and Base on Buffer pH 4) Calculate the expected pH of 0.05M HAc, 0.05M Ac-, and 0.05M HAc/0.05M

Ac- after the addition of 1mL 0.1 M NaOH to each. Compare your measured pH value to the theoretical value and give possible explanations for the deviation (if any).

REPORTING RESULTS – Complete your lab summaryIf a report is required in place of a lab summaryAbstractResults:

Part 1. Hydrolysis equation for each solutionKa or Kb of each liquid or solution; %error

Part 2a. measured and expected pH of solutions before OH- additionPart 2b. measured and expected pH of solutions after OH- addition

Sample Calculations: Part 1. Ka or Kb of each hydrolyzed species; %errorPart 2a. expected pH of solutions before OH- additionPart 2b. expected pH of solutions after OH- addition

Discussion/ConclusionPart 1. Did the pH of the solutions match what might be predicted based on

the cation and anion provided by the salt? How good were your measured Ka or Kb values?

Part 2a. Did the measured pH match the expected pH?Part 2b. Did the measured pH match the expected pH? Which solution showed

the greatest change in pH? Which showed the least? Why? Review Questions

REVIEW QUESTIONS1. Indicate whether the following could act as buffers:

a. 100 mL 0.01 M HNO2 mixed with 100 mL 0.02 M NaNO2.b. 100 mL 0.01 M HNO2.c. 100 mL 0.01 M HNO2 mixed with 25 mL 0.02 M NaOH.

2. The Ka of lactic acid, HC3H5O3, is 1.410-4. Over what pH range would a buffer prepared using equal volumes of 0.1M HC3H5O3 and 0.1M C3H5O3

- be most effective?

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3. Find the pH of a 210-5 M HCl solution. If 1 mL of 0.1 M NaOH is added to 1 L of the acid, what is the resulting pH? How do these pH values compare to the buffer you made in part 2b?

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