Acids and Bases Chapter 8. Polyprotic acids However, the most ionization occurs in the first step. ...

Acids and Bases

Chapter 8

Polyprotic acids

However, the most ionization occurs in the first step.Ka1>> Ka2 > Ka3 .. . .

Consequently, the [H+] is predominantly established in the first ionization with the Ka1 value. Subsequent ionizations (Ka2 & Ka3) only add minimal amounts of [H+].

Use Ka1 to determine the pH of the solution at equilibrium.

Acid-Base properties of Salt Solutions

A salt is a solid that when dissolved in water dissociated into ions.

Some salts may contain ions that alter the [H+] or pH of a solution.

The acidic or basic properties of a salt solution arise from the reaction of the dissociated ions of the salt with water.

Salts that form neutral solutions

The salt of a strong acid/strong base dissolves in water to form neutral solutions. Strong acid supplies the anion of the salt

HCl H+ + Cl-

Strong base supplies the cation of the salt NaOH Na+ + OH-

When the water is removed (evaporation) a salt remains

HCl + NaOH H+ + OH- + Cl- + Na+ NaCl

Includes groups 1 & 2, but not Be2+

The solution has a pH of 7

Salts that form acidic solutions

The salt of a weak base (cation) and strong acid (anion) dissolves in water to form acidic solutions.

The cation reacts with water to liberate H+

The solution has a pH less than 7

NH4Cl dissociates NH4+ & Cl-

NH4+ + H2O H3O+

(aq) + NH3(aq)

Salts that form basic solutions

The salt of a strong base (cation) and weak acid (anion) dissolves in water to form basic solutions.

The anion reacts with water to liberate OH-

The solution has a pH greater than 7

NaC2H3O2 dissociates Na+ & C2H3O2-

C2H3O2- (aq)+ H2O HC2H3O2(aq) + OH-

(aq)

Weak base/weak acid salts

The salt of a weak base (cation) and weak acid (anion) dissolves in water and both ions react with water.

If Ka>Kb, the solution is acidic.If Kb>Ka, the solution is basic.

When an ion reacts with water to produce an acidic or basic solution (break water into its ions) it is called hydrolysis.

Weak base/weak acid salts NH4HCO3 is the salt of a weak base (NH3) and a

weak acid (H2CO3). Both ions react with water.

NH4+ + H2O NH3 + H3O+

HCO3 + H2O H2CO3 + OH-

Ka for NH4+ = 5.6 x 10-10

Kb for HCO3- = 2.2 x 10-8

Therefore, the solution would have more OH- in solution (Kb>Ka) and have a large pH. It is basic!

Anion from Strong acid(Large Ka)

Anion from Weak acid (Small Ka)

Cation from Strong base(Large Kb)

Neither ions reacts with water so the pH is neutral.

Anion reacts with water to produce OH- so the pH is basic.Very large Kb.

Cation from Weak base (Small Kb)

Cation reacts with water to produce H+ so the pH is acidic.Very large Ka.

Both ions react with water.Ka = Kb neutral

Ka > Kb acidic

Kb > Ka basic

Acid-Base Reactions

Neutralization reactionDouble displacement reaction between an

acid and a base to produce a salt and water (solvent).

The acidic and basic properties are destroyed – neutralized.

Acid-Base Reactions

TitrationAnalytical lab technique used to determine the

concentration of a solution.Quantitative neutralization reactionTitrant – the solution of known concentration in

buret.Sample – the solution of unknown

concentration, but known volume (in flask).

Acid-Base Reactions

TitrationStandard solution – a stock solution that is

of known concentration that is used to create the titrant.

Primary standard – a chemical that is available in a pure and stable form that can be used to produce an accurate concentration.

Standardization – a titration used to find the concentration of the titrant using a a primary standard.

Acid-Base Reactions

Titration Indicator – an acid base indicator that will

change colour at a known pH to signify a specific pH in the neutralization reaction.

End point – the point in the titration when the indicator changes colour.

Equivalence point – the point in the titration when chemically equivalent amounts of reactants have reacted. Usually an equilibrium has been established.

Titration of a strong acid with a strong base At the equivalence point [H+]=[OH-] or

pH=7. Phenolphthalein is a popular indicator

because it is colourless in acidic solutions and pink in basic.

Remember to consider the reaction at the molar level. (convert to moles!)C=n/V and C1V1 = C2V2

Titration of a strong acid with a strong base Titration curve – a plot of the pH vs. Volume of

titrant added.

pH

Vol. of titrant

7

0

14

Equivalence point (pH=7.0)

Titration of a strong base with a strong acid

pH

Vol. of titrant

7

0

14

Equivalence point (pH=7.0)

Titration of a weak acid with a strong base In the titration of a weak acid with a strong

base the neutralization occurs equilibrating the number of H+ and OH- ions.

In doing so, a salt ion is produced that adjusts the hydrolysis of water due to the ions “desire” to form a weak acid.

The shift in the ionization of water produces a pH other than 7 at the equivalence point.

Titration of a weak acid with a strong base

pH

Vol. of titrant

7

0

14

Equivalence point (pH> 7.0)

Buffer region

½ way to Equivalence point

Titration of a weak base with a strong acid

pH

Vol. of titrant

7

0

14

Equivalence point (pH<7.0)

Buffer region

Titration of a polyprotic acid with a strong base

pH

Vol. of titrant

7

0

14

Second Equivalence point

First Equivalence point

Buffer

During the “buffer region” one has a combination of weak acid and its conjugate base.

A buffer solution can “absorb” either H+ or OH –

ions added to the solution. A buffer is an equal mixture of a weak acid

and its conjugate base.

Buffer solution of H2CO3 & HCO3-

Consumes H+

H +(aq) + HCO3

- (aq) H2CO3 (aq) + H2O(l)

Conjugate Weakbase acid

Consumes OH-

OH -(aq) + H2CO3 (aq) HCO3

-(aq) + H2O

(l)

Weak Conjugateacid base

Acid-base calculations

Determine [H+] and pH Percent ionization Use ka to determine [H+] and pH Neutralization reactions

Determine [H+] and pH

602.5

)10 x 50.2log(

]log[

10 x 50.2][

6-

Lmol-6

H - pH

H

10 x 37.4

10

10][

36.8

Lmol9-

36.8

pHH

pH

Determine [H+] and pH

The pH and pOH add up to 14 for a conjugate acid-base pair.

Determine pKa and pKb

The pKa and pKb add up to 14 for a conjugate acid-base pair.

Percent Ionization

%100HA][

]H[

%100solution] [acid

acid] ionized[ionization %

Use ka to determine [H+] and pH

Given the concentration and equilibrium constant for an acid one can determine the hydronium ion concentration and the pH of the solution.

Calculate the hydrogen ion concentration and the pH of a 0.15 mol/L solution of hydrofluoric acid (HF(aq)) with a ka of 6.6 x 10-4.

1. Produce a balanced chemical equation for the hydrolysis equilibrium.

HF (aq) + H2O (l) H3O+

(aq) + F- (aq)

2. Create an equilibrium expression.

4-

3

10 x 6.6

][

]][[

HF

FOHka


3. Generate and ICE table to account for concentration changes.

HF (aq) + H2O (l) H3O+

(aq) + F- (aq)

Initial 0.15 mol L-1 0 0

Change -x +x +x

Equilibrium 0.15 -x 0 + x = x 0 + x = x


4. Use the equilibrium expression to generate a mathematical model for the equilibrium.

5-4-2

4-4-2

24-

4-

3

10 x 9.910 x 6.60

)15.010 x 6.6()10 x 6.6(0

)15.0(10 x 6.6

)15.0(

))((10 x 6.6

][

]][[

xx

xx

xx

x

xx

HF

FOHka


5. Resolve for “x” and relate to [H+].

Lmol4-

Lmol4-

2

-5-42

10 x 3.3or 10 x 3.3

2

4

10 x 9.910 x 6.60

x

a

acbbx

xx

x = [H+] = 3.3 x10-4 mol/L


6. Calculate the pH.

48.3

)10 x 3.3log(

]log[

10 x 3.3][

4-

Lmol-4

H - pH

H

[H+] = 3.3 x10-4 mol/L has two significant digits (3.3) while the exponent is a

certain value. The pH value must reflect the certainty of

the exponent and significance of the

measured coefficient. (3 – indicates the exponent,

.48 – the two SD of the coefficient)3.48 2SD

The [H+] = 3.3 x10-4 mol/L and the pH= 3.48.

Neutralization of a weak/strong combination Stoichiometric calculation of quantities

during neutralization. Determine adjusted specie concentration

due to modified volumes Look at the common ion affect on the

hydrolysis of water equilibrium.

Calculate the pH of a sample when 25.00 mL of 0.200 mol/l HClO (aq) is titrated with 10.00 mL of 0.200 mol/L KOH (aq) .

There are two reactions to consider in this situation: The neutralization reaction which is a quantitative

reaction that is governed by stoichiometry.

HClO (aq) + KOH (aq) → H2O (l) + KClO (aq) The neutralization produces a salt that adjusts an acid-

base equilibrium system for the conjugate acid-base pair.

HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq) This might be considered as a common ion effect where

the neutralization generates and ion that affects an equilibrium.


1. Use the neutralization reaction to determine the quantity of acid and base that react.

HClO (aq) + KOH (aq) → H2O (l) + KClO (aq)HClO (aq) KOH (aq)

Concentration (C) 0.200 mol/L 0.200 mol/L

Volume (V) 0.02500 L 0.01000L

Moles (n)

mol 10 x 00.5

L 02500.0L 1mol 0.200

3-

mol 10 x 00.2

L 01000.0L 1mol 0.200

3-

2. Use the neutralization reaction to determine the quantity of “salt” produced in the reaction.

HClO (aq) + KOH (aq) → H2O (l) + KClO (aq)2.00 x 10-3 mol of KOH (aq) is used up completely and produces 2.00 x 10-3 mol of KClO (aq) or ClO- (aq).

In doing so there is a quantity of HClO (aq) that is left unreacted in the neutralization reaction.

The unreacted HClO (aq) is 5.00 x 10-3 mol – 2.00 x 10-3 mol = 3.00 x 10-3 mol


2. Alternate method to use the neutralization reaction to determine the quantity of “salt” produced in the reaction.

HClO (aq) + KOH (aq) → H2O (l) + KClO (aq)

The unreacted HClO (aq) is . . . 5.00 x 10-3 mol – 2.00 x 10-3 mol = 3.00 x 10-3 mol HClO


3. Using the quantities of weak acid and conjugate base determine the affects on the hydrolysis of water.

HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)


HClO (aq) H3O+ (aq) ClO- (aq)

[Initial]= 3.00 x 10-3 mol/

0.035 L= 0.0857 mol/L

0= 2.00 x 10-3

mol/ 0.035 L= 0.0571 mol/L

[Change in] -x +x +x

[Equilibrium] 0.0857 - x x 0.0571 + x

4. Use the ka value and equilibrium expression to solve for [H+] and pH



x)-(0.0857

x) (0.0571 (x)

[HClO]

][ClO ]O[H -3

ak

Lmol4-

8-

8-

10 x 2.09

0571.0

)0857.0)(10 x 9.2( x

10 x 9.20.0857

(0.0571) (x)

4. Use the ka value and equilibrium expression to solve for [H+] and pH



3.68

)10 x 2.08log(

][H log- pH

10 x 2.08 ]H[

4-

Lmol-4

Acids and Bases Chapter 8. Polyprotic acids However, the most ionization occurs in the first step. ...

Documents

Transcript of Acids and Bases Chapter 8. Polyprotic acids However, the most ionization occurs in the first step. ...