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PURPOSE

To calculate an approximate K, or K6 value for weak acids and weak bases based on the pH ofthe solution determined by acid-base indicators.

MATERTALS AND EQUIPMENT

. Spot plates . Hot plate and small beaker

CHEMICALS

o Q. l0 M solutions of NaC[, NaC2H3O2, NH4CL, ZnCl2, KAI(SO4)2, and Na2CO3.

Solutions of the indicators methyl orange, methyl red, bromothymol blue, phenol red,

phenolphthalein, and alizarin yellow-R

INTRODUCTION

A salt is an ionic compound formed by the reaction between an acid and a base. Salts are

strong electrolytes. They dissociate completely into their constituent ions in aqueous solution.

Generally, the reaction of a cation or an anion or both of a salt with water is cdled salt hydro-

lysis (hydro =w&ter, lysis = splitting). But in fact for weak acids (NH4-), only the dissociation

of H. ion occurs. However, for simpliciry the reactions of the cations and anions of salts withwater are generally cdled hydrolysis reactions. Hydrolysis generates hydrogen or hydronium ions

and hydroxide ions. Hence the pH of the solution is affected.

\When a salt is dissolved in watet the resulting solution can be either acidic, or basic or neutral

depending on the ions contained in the salt. Usually, when a salt is derived from a strong acid

and a sffong base, its aqueous solution is neutral.

l4l2Experiment 5 . Hydrolysis Reactions of Anions and Cations of Salts

This is because the cation of a strong base and the anion of a strong acid have little tendency to

hydrolyse. Th.y are classified as neutral ions (middle tab[e). For example, aqueous solutions ofNaCl, KNO3, CaBrr, CsClOn etc. are neutral (pH = 7).

\Urhen weak acids and bases react, the relative strength of the coniugated acid-base pair in the salt

determines the pH of its solutions. 'W'[ren a salt formed from a strong acid and a weak base is

dissolved in water, the solution is acidic (pH . 7). For example, NH4CI andZnClr. The cations

of such salts wifl hydrolyse (metal cations alone) or undergo proton transfer (conjugate acids ofweak bases, NH4.) generating hydronium ions (or hydrogen ions). The anions of these salts will

not hydrolyse.

lZn(HrO)u1'. (rq) -

lZn(HrO),OHl. (aq) + H. (aq)

NHn. (aq) +HrO (l) NH, (aq) +HrO'("q)

Or simply,

NH.. (aq) -

NH, (aq) + H. (aq)

The aqueous solution of a salt formed from a weak acid and a strong base is basic (pH , 7). For

example, NaCrHrOr. ft. anion of such a salt will hydrolyse generating hydroxide ions. The

cations of these salts will not hydrolyse.

CrHrOr- (aq) + HrO (l) -

HC2H3O, (aq) + OH- (aq)

If the salt contains both the conjugate acid of a weak base and the coniugate base of a weak acid,

then the pH of the resulting solution will depend on the relative strengths of each.

In summary determining whether a salt solution will be acidic, basic, or neutrd, depends on the

narure of the cation and the anion. If both the cation and anion are neutral ions, the salt will not

affect the pH and the salt solution will be neutral. If the cation is acidic and the anion is neutral,

the salt wifl dissolve in water to produce an acidic solution. If the cation is neutral and the anion

is basic, the salt will dissolve in water to produce a basic solution. If the cation is acidic and the

anion is basic, the resulting salt solution may be acidic or basic depending on the reladve acid and

base strengths of the ions. Some general trends for acidic and basic ions are given below:

-.-l--rr_

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-{2:{

ffiffiSkh}'*ffi+#tri+ffi

CHEM l4I2Experiment 5. Hydrolysis Reactions ofAnions and Cations of Salts

General Rules for Acidic and Basic Ions in Salts

It is the conjugate acid of a weak base, or It is the conjugate base of a weak acid, or

Group IA Metal cations and Group ILAcations larger than calcium are neutral

Conjugate bases of strong acids are neutral

In this experiment, you will examine hydrolysis reactions of various salts. A hydrolysis reaction

wifl have occurred if additional H. or OH- ion is formed when the salt is dissolved in water. The

trick is to be able to figure out which ion in the salt is producing the hydrogen or hydroxide ions

which are making the solution acidic or basic.

Negatiae ions

In terms of conjugate acid-base relationships, remember that the stronger the acid, the weaker

its conjugate base, and the sffonger a base, the weaker its conjugate acid. For example, consider

the chloride ion, C[- which is the conjugate base of HCl, a strong acid. fu the conjugate base ofa strong acid, chloride ion is too weah a base n hydrolyze in aqueous solution, and a solution of achloride salt such as NaCl will be neutralwith a pH of 7. the group IA and ILA metal cadons(such as Na* ion) do not hydrolyze significantly in aqueous solution. Similarly, a solution ofKNO3 will have a pH of 7, since the nitrate ion is the conjugate base of nitric acid (a strong acid)

and the potassium ion wi[ not hydrolyze.

Now consider a solution of sodium acetate, NaC2H3O2. This salt consists of a sodium ion,

which does not hydrolyze, and an acetate ion. Acetate ion is the conjugate base of acetic acid,

HC2H3O2, which is a weak acid. Therefore, we predict that the acetate ion will hydrolyze inaqueous solution. In any base hydrolysis reaction of a base B-, the general equilibrium is:

B- ("q) + HzO (l) -

BH (aq) + OH- (aq)

In the above example, for the base hydrolysis of acetate ion, we would write

C2H3O2-(rq) + HzO (l) -

HC2H3oz(aq) + oH-(aq)

The equilibrium constant for this reaction is termed K6, since it describes a base reaction, and is

given by

Kb = [HC2H3O2][OH] l[CzHtOzl

From the measured pH of this solution, we can calculate [H-]. From which we can calculate the

equilibrium [OH-], since [H-][OH-]= 1.0 X 10-14. And if we know the hydroxide concentration

at equilibrium, we automatically know the acetic acid concentration [HC2H3O2], since

HC2H3O2and OH- are formed in 1:1 amounts in the reacdon. Finally, if we know the initialconcentration of CzH3Oz-, we can calculate K6.

CHEM l4l2 Experiment 5 . Hydrolysis Reactions ofAnions and Cations of Salts 59

For example, the pH of a 0.10 M solution of sodium acerare was determined to be approximately8.5. From this, calculate K6 for the acetare ion, C2H tOz-.

SincepH = -log[H.], tHl = 10-PH = l0-t'5 = 3.16X t0-eM.

Andsince tHItOHl = 1.0X l0-", tOHl = l.0X 10-'4 t3.l6X lO-e = 3.16X t0-6M.

Kb is calculated by plugging the equilibrium concentrations into the equilibrium consranr expression:

CzH3oz- (aQ + H2O (l) - - HC2H3O, ("q) + OH- (aq) (omit the specator Na. ion)

Initially: 0.10 MChange - x

Equilibrium: 0.10 - x

Kb = [HC2H3Oz][OH-] llCzF{3Oz-) = (")(*) / (0.10-*) = x2 l(0.10-x)

where x = tOH-l = 3.16 X 10-6 M (which is also the HC2H3O2 concentration)

Therefore, Kb = O.16 X 10-6)2 / (0.10 - 3.16x l0-1 = 1.0 x 10-t'.

Positiae fons

Tiansition metal ions and many other metal ions act as weak acids ion aqueous solution. These

metal ions bind \Mater molecules (usually about six) more strongly than, for example, the GrouplA metal ions do, and it turns out that a water molecule strongly associated with the metal iondissociates H* ion more easily. In the case of zinc(Il) ion, we have

lZn(H2O)ol'. ("q) -

[Zn(HzO):OH]. (aq) * H. (aq)

If a salt solution of something other than a transirion metal salt gives an acidic pH, then the

hydrolysis reaction is generally given by the dissociadon of a weak acid HA*:

HA ("q) + H2O (l) -

H3O+ (aq) + A-(aq)

Or more simply, we can say,

HA (aq) =- H- ("q) + A (aq)

For exarnple, in the case of ammonium chloride, the ammonium ion dissociates as follows:

NHr. (aq) -

H- (aq) + NH, (ag)

The weak acid dissociation constanr K, is

Iq = HIINH;I /[NH+.]

00+x +x+x +x

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CHEM l4l2 Experiment 5 . Hydrolysis Reactions ofAnions and Cations of Salts

If we have a 0.10 M solution of NHaCl, the calculation of K" is ser up as follows:

NHa. (aq)

- H. (aq) + NH3 (aQ (omitting the spectator CI ion)

0.10M 0 0Initially:Change - x

Equilibrium: 0.10 - x+x +x

xx

K = xz I (0.10 - x) where x is the hydrogen ion concentration, calculated from the pH.

Conjugate Pairs

A conjugate pair differs by one (and only one) H. in the formulas.

H2SOa and HSO 4 is a conjugate pair, but H2SOa and SO42- is not.

Remember:

The conjugate acid of a compound is obtained by adding one If n its forrnuk, andThe coniugate base of a compound is obtained by remouing one Il fro* itsformuk.

The hydrolysis reactions of weak acids and bases in water thus foflow the parrerns:

\TeakAcid + Hzo -- H3o. + Conjugate Base of therUfleakAcid

Or, \TeakAcid =- H- (aq) + Conjugate Base of the\TeakAcid

rveak Base + Hzo -

conjugate Acid of the rureak Base + oFr

Th. I(" of a weak acid is related to the K5 of its conjugate base by

I(" X Kb = Kw = 1.0X 10-t4

For example, if we know the vdue of Iq for a weak acid, we can calculate K6 of the conjugate

base of the weak acid: Kb = 1.0 X l0-t4 / Iq

From the equation we can see that the larger K" ir, the smaller Kb musr be, in order for I(. X Kb

to equal 1.0 X 10-t4. Or, the larger Kb ir, the smaller K" must be. The larger I(" ir, the srongerthe acid, and the larger Kb ir, the stronger the conjugate base of the acid is. So, the stronger andacid is, the weaher its conjugate base will be, or the other way around; the stronger a bdse is, the

weaker its conjugate acid will be.

CHEM l4l2Experiment 5 . Hydrolysis Reactions ofAnions and Cations of Salts 6r

Salts of Weak Acids and Weak Bases

Here, neither ion in the salt is a spectator ion. A salt formed between a weak acid and a weak

base can be neutral, acidic, or basic depending on the relative sffengths of the acid and base:

J

>t>l!-)>l!>t!,>t,>t>I>I>I-)>t>I>I!>I>Iv!>t>I>r2>t>t>I:{>{:lvvv>l:Tvv

If K^of the cation

If K^of the cation

If K"of the cation

K1o of the anion, the solution of the salt will be acidic.

K6 of the anion, the solution of the salt will be neutral.

K1o of the anion, the solution of the salt will be basic.

Deterntination of pH

The semi-quantitative procedure in this experiment uses only approximate concentration and pH

values. pH indicators gives pH values which are sufficiently precise for this purpose. For exact

work, very accurate pH meters are used to obtain the pH values of the salt solutions. In today's

experiment, you wifl estimate the acidic and basic properties of various salts by measuring the

effect of 0.10 M solutions of those salts on a series of six pH indicator solutions. A pH indicator

is a substance that changes color over a particular range of pH. Indicator solutions can be used

ro esrimate the pH of a solution. The pH interval over which an indicator changes its color is a

properry of the particular indicator.

The pH intervals for all of these indicators are also diagrammed in your textbook (in color) You

can estimate the pH of a solution by determining the color of each of these indicators when

added to the solution.

The colors and pH ranges of the indicators used in this experiment are given below:

The idea is ro narrow the estimate of the pH to within a certain small range. For example, if a

solution is yellow when methyl orange is added, the pH must be greater than 4.4. If the color

is yellow when methyl red is added, the pH must be greater than 6.0. If the color is blue when

bromothymol blue is added, the pH must be greater than7.6. If the color is red when phenol

red is added, the pH must be greater than 8.0. If the solution is colorless when phenolphthalein

is added, the pH must be less than 8.2. If the color is yellow when alizarin-R is added, we know

the pH of the solution is less than 10.1. From these results, we would conclude that the pH of

the solution is between 8.0 and 8.2, or about 8.1.

62 CHEM l4l2Experiment 5 . Hydrolysis Reactions ofAnions and Cations of Salts

t-

trEErb]t}tltbbltl-Ei;EEEEEEEEEEEEEEEEEEtsEbI'b

Wtren measuring the pH of solutions exposed to air, you will generally find that the solutions are

slighdy more acidic than you would calculate them to be. This is due to the absorption of carbon

dioxide from the atmosphere. COz dissolves in water to produce H* according to the reacdon:

CO2Q) + IHzO(| - - HzCOtkd =- H*(aq) +HCO3-fuq)

Even distilled water will be somewhat acidic unless it is boiled to remove the dissolved CO2. You

will measure this difference in the lab.

Satnple koblem.s

Fxample l. Calculate the pH of a 0.050 M solution of ammonium chloride, NHaCl. K" fotammonium ion = 5.5 X 10-ro.

Since the ammonium ion, NH4*, is the conjugate acid of theweakbase ammonia, NH3, we

know that it will be a weak acid. Set up the problem by writing the acid dissociation reaction

and the initial and equilibrium concentrations underneath:

NH<. (aq)

- H. (aq) + NHa (aq)

Initially: 0.050M 0 0

Change -x +x +xEquilibrium: 0.050-x x x

Next, plug the equilibrium terms into the equilibrium constant expression and solve for x:

Ka = 5.6X 10-'0 = tHltNHal / [NH+.] = (*)(*) / (0.0500 -x) = x2 l(0.050 -x)

'We will try the approximation that 0.050 - x is 0.050 if x is very small. If x is less than about

5o/o of the initial concentration of 0.050 M, this approximation is OK to use, rounding the

results to one or two places after the decimal. If x turns out to be more then 5o/o of the initialconcentradon, we must solve for x from the original equation using the quadratic formula.

5.6x 10-to = x2 10.050, x = 5.2915 X 10-6 M.

Check Is 5.3 X 10-6less than 5o/o of0.050 M? 5.3X 10-6 M / 0.050 M X 100 = 0.0llo/o,so the approximation was OK!

Since x is the H. ion concenffation, the pH of the solution is -log(5.2915 X 101 = 5.28.

Note on rounding: In multistep calculations such as these, it is best to not round offnumbersuntil the very end. A pH value calculated from numbers with two significant figures would be

rounded to two places after the decimal point according to the rounding rule for logarithms.

F*ample 2. 'lU7'hat is the pH of a 0.10 M solution of potassium hypochlorite, KCIO? Ka forhypochlorous acid, HCIO, is 3.0 X l0-8.

L4l2 Experiment 5 . Hydrolysis Reactions ofAnions and Cations of Salts

Since hypochlorous acid is a weak acid (with K" of 3.0 x l0-8) we know that its conjugate base,the hypochlorite ion, clo-, is a weak base. K6 for clo- is

&Ku = 1.0X10-'n, Kb = l.0Xt0-'4/Iq = l.0Xl0-,4 l3.0Xl0-8 = 3.3333X10-7.

'Writing the base hydrolysis reacrion of the hypochlorite ion, and omitting the specrator K. ion,

we have

IIttIITIIIIIIIIIIIIITITITTT,TTTIIIIIIIII,

Initially:

Change

Equilibrium:

Initiallp

Clo- (aq)

0.10 M-x0.10 - x

+ HzO (aq) HCIO (aq)

0

+xx

+ OH- (aQ

0

+xx

Iq = 3-3333x10-7 = [Hclo]toH-l /tctol = k)k)/(0.10-x) = *t(0.10-x)

Agri, using the approximation that 0.10 - x is about 0.10, we have

3.3333X10-7 = *10.10, x - 1.82574X104M - tOHl(Check that 1.8 X l0-4 M is less than 5o/oof 0.10 M)

POH = -log[OH-] = -log(1.82574x t0r) = 3.74 (rounded ro rwo digits after the decimal)

pH = 14 - pOH = 14 - 3.74 = 10.26.

Example 3. Calculate [H3O.], [OHl, and K5 for the BrO- ion if a 0.10 M solution of NaBrOhas a pH of 10.85.

The setup is just as in the previous problem:

BrO- (aq) + HrO (l) -

HBrO (aq) + OH- (aq)0.10M 0 0

AtEquilibrium: 0.10-x x x

'We can obtain the hydroxide ion concentration at equilibrium from the pH:

tHl = IO-PH = l0-r0'85 - t.4lz5x l0-rr MtoHl = l.0x 10-'4/ Hl = 1.0x l0-r4 tt.4t25x l0-'r = 7.07g46X 10-4M

Therefore, at equilibrium,

tOH-l = [HBrO] = 7.07946 X t0-4 MlBrO-l = 0.10M -7.07946X l0-4M - 0.a99292M

Kb = tHBrOltOH-l t tBrol = (7.U7946X rc4 M)0.Otg46Xl0+ M) t 0.099292M =5.0X t0-6(answer rounded ro rwo significant figures)

64 CHEM l4l2Experiment 5 ' Hydrolysis Reactions ofAnions and Cations of Salts

EXPERTMENTAL PROCEDURE

The pH of the following will be determined using acid-base indicators:

1) Unboiled deionized water 5) 0.10 M NH4CI solution

2) Freshly boiled deionized water 5) 0.10 M ZnCI2 solution

3) 0.10 M NaCl solution 7) 0.10 M KAI(SO4)2 solution

4) 0.10 M NaC2H3O2 solution 8) 0.10 M Na2CO3 solution

The solutions will be tested with the following indicators:

1) methyl orange

2) methyl red

3) bromothymol blue

4) phenol red

5) phenolphthalein

5) alizarinyellow-R

Obtain a hot plate and boil approximately 100 mL of deionized water for about 10 minutes to

remove dissolved COz from the water.

Place about four drops of solutions 1-8 into the wells of a spot plate and add about rwo drops ofmethyl orange indicator to each well. \7rite the observed colors in Thble I of the report form.

E*p.y the spot test plate and carefully rinse it with deionized water. Then repeat the procedure

using the next indicator, methyl red, and so forth, until the solutions have been tested with each

indicator. This can be divided among the members of your group if desired.

Note: It is more beneficial to tesr each indicator individually as above. This allows you to see the

range of color of the sarne indicator at the various pHs of the solutions, which in turn allows you

to make a better comparison of the colors of the indicator. ![hat looks like an "orange" color in

an isolated test may actually be the "red" color of that particular indicator instead.

Next, estimate the pH of each solution using the colors recorded in Thble I of the report form.

From the pH values, calculate the hydrogen ion concentration, [H.], and the hydroxide ion

concentration, tOHl in each of the solutions. Complete Thbles 2 and 3 on the report form and

calculate the Il or K6 values of the salt solutions in Thble 3.

CHEM l4l2Experiment 5 . Hydrolysis Reactions of Anions and Cations of Salts

EXPERIMENT-5

REPORTFORM NAME,

INSTRUCTOR

DATE,

TABLE I - Indicator Colors

Hzo(unboiled)

0.r0 MNaCl

ffi ffiffiffi ffi ffiffi *iiiffi*tflff#rfffi *tf-H++{iii.iffi

0.10 MNH4CI

fi#ffi ffi +i#iffittffi

f##ffi0.r0 Mr(Al(sor,

ffi ffi ffi ",'Wffiffi ffi

CHEM l4l2Experiment 5 'Hydrolysis Reactions ofAnions and Cations of Salts 67

:-!r-_

tr--

tl--

tl__

tl__

tL-

tr_

!a--

tl__

ra_

!-_

!-_

!-_

tl_

rI_-

tr__

r-_

!-_

tl--

tr_

tl,_

1_

tl_

rt_

rL-

tl-

!a_

r,_

:__

tr_

!-J{I!!'-J*J-J-J*J

_ sg;r*rli

ThBLE 2 - Ions From Salts

TABLE 3 - Reactions and Calculations

CHEM L4l2Experiment 5 . Hydrolysis Reactions of Anions and Cations of Salts

EXPERIMENT-5 NAME

DAIE

Pre-Laboratory Review Questions and ExercisesDue before lab begins. Answer on separate sheet of paper.

1. Define the following terms:

") Bronsted acid and base

b) Arrhenius acid and base

2. 'Write net ionic equation for the following ions which undergo hydrolysis.

NO; CtO3- Zn'*

3. Identify the fo[owing salts as acidic, basic, or neutral:

NaClO, KCN NarSO,

4. Aniline is a weak organic base with the formula C6HrNHr. The anilinium ion has the formula

C6HsNHr* and its IQ = 2.3 X 10r.

a) 'Write the chemical reaction showing the hydrolysis of the anilinium ion.

b) Calculate the pH of a 0.0400 M solution of anilinium chloride.

5. Find the pH of a 0.20 M solution of sodium propionate, NaCrHrOr.

IQ of propionic acid, HC3H,O, is I .34X t0-5. (Hint: K* = I("Kb)

Mffiq'w

*e*wrwes#

.M. @&V

CHEM l4l2Experiment 5 . Hydrolysis Reactions ofAnions and Cations of Salts

EXPERIMENT-5 NAME

DATE

Post-Laboratory Questions and ExercisesDue after completing lab. Answer in space provided or on separare sheets if needed.

1. Determine the I(" of the weak acid HX knowing that a 0. 10 M solution of LX has pH of8.90.

2. Pyridine, c5H5N, is a weak base and reacts with HCI as follows:

C,H,N (aq) + HCI (aq) ) C,H,NH. ("q) + Cl- (aq)

\7trat is the pH of a 0.015 M solution of the pyridinium ion (CrHrNH)? Th. I(U for ppidine is

1.6 X l0-e. (Hint: Calculate the K" for the pyridinium ion and use it in the calculation.)

3- Find the pH of a 0.30 M solution of sodium benzoate, NaCrHrOr. I$ for the benzoateion, CrHrO, is 1.55 X 10-ro.

4. I(" fot nitrous acid, HNO2, is 4.5 X l0-4. Wlrat is the value of Kb for the nitrite ion,NO;?

5. Identifr the acids and bases which reacr to form the following salts:

a) ZnSOa

b) KBr

c) GCo,

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tu****#ffiffis,E

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=_*.$CHEM l4l2 F-xperiment 5 ' Hydrolysis Reactions ofAnions and Cations of Salts 7t