Ionic Eq.docx

CHAPTER 9 IONIC EQUILIBRIUM 9.1 Theories of Acids & Bases 9.2 Acid-Base Equilibria 9.3 Hydrolysis of Salt 9.4 The Common Ion Effect: Buffer Solutions 9.5 Acid-Base Titrations 9.6 Solubility Equilibria 9.1 Theories of Acids & Bases 9.1.1 Arrhenius Definition Acids and bases are electrolytes (form ions when dissolved in water).

Svante Arrhenius, came up with the concept or idea that: +



 acids dissociated in water to give H and + HCl(aq)  H (aq) + Cl−(aq)  bases dissociated in water to give hydroxide ions. −

NaOH(aq)  Na+(aq) + OH (aq) Acids and bases neutralize one another because the hydrogen ion and the hydroxide ion combine with one another to form water. +

−

H (aq) + OH (aq)  H2O(l) In general, this reaction will also produce a salt as a by-product. The full equation would be a salt HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) 1

9.1.2 Brønsted-Lowry Definition +

With the Brønsted-Lowry concept, we usually refer to a H as a proton. That is because a proton is all that is left when… + − HH +e ACID  An acid donates a proton to another substance.   When an acid reacts, the proton is transferred from one substance to another. +

−

HCl(aq) + H2O(l)  H3O (aq) + Cl (aq) Brønsted Acid (donate proton)

 When an acid dissolves and dissociates in water it gives a proton to the water.  The following equation shows HCl giving a proton (H+) to a hydroxide ion (OH−). HCl(aq) + OH−(aq)  H2O(l) + Cl−(aq) Brønsted Acid (donate proton)

BASE  Bases are the opposite of acids. Bases accept protons.   OH−, for example can accept a proton to form water. +

−

H (aq) + OH (aq)  H2O(l)   Brønsted and Lowry realized that not all bases had to have OH−.   As long as something can accept a proton it is a base.

H2O(l) + HCl(aq)  H3O+(aq) + Cl−(aq) Brønsted Base (accept proton)

NH3(aq) + H2O(aq)  NH4+(aq) + OH−(aq) Brønsted Base (accept proton)

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Let’s take a closer look at the acid-base equation again…

Notice that H2O is a proton acceptor. Therefore, it is a base.  Likewise, HCl is an acid because it is a proton donour. 

 Because H3O+ is directly derived from water, it is its conjugate pair. Since H3O+ is also capable of donating a  proton, it is a conjugate acid.  Likewise, Cl− is the directly derived from HCl, it is its conjugate pair. Since Cl− is also capable of accepting a proton,  it is a conjugate base.  Therefore, H2O is a base and H3O+ is it’s conjugate acid.   And, HCl is an acid and Cl− it it’s a conjugate base.

NOTICE THAT…

 A weak base gives a strong conjugate acid.   A strong acid gives a weak conjugate base.

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And likewise…  A strong base gives a weak conjugate acid.   A weak acid gives a strong conjugate base.

Exercise 1: Identify the two conjugate acid-base pair for the reactions below: a) HSO4−(aq) + PO43−(aq) SO42−(aq) + HPO42−(aq) b) HCO3−(aq) + H2PO4−(aq) H2CO3(aq) + HPO42−(aq) c) N2H4(aq) + H2O(l) N2H5+(aq) + OH−(aq)

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9.1.3 Lewis Definition Lewis Acid-base definition is based on the electron pairtransfer concept. Lewis focused on what happens when a base accepts a proton. So, Lewis' definition of acid and base is… A BASE is an electron pair donour An ACID is an electron pair acceptor For example, When a hydroxide ion accepts a proton, H+, it attaches itself to a pair of electrons from the oxygen. H

+ H+

O

Lewis Base (electron pair donour)

H

O H

Lewis Acid (electron pair acceptor)

Likewise, when an ammonia molecule accepts a proton, H+, it attaches itself to a pair of electrons from the nitrogen. H H

N

H

+ H+

H

H

+

N H H

Lewis Base Lewis Acid (electron pair (electron pair donour) acceptor)

What happens in each case is that the base donates a pair of electrons to the H+ in order to allow it to form a coordinate covalent bond.

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Neutralization, the reaction between acid and base is defined as: the formation of a coordinate covalent bond between the electron pair donour (base) and the electron pair acceptor (acid) H

O

Lewis Base

+ H+

H

Lewis Acid

O

(formation of dative bond)

Exercise 1: Identify the Lewis acids and bases for the reactions below: a) b) c) d)

OH− + HCO3−  CO3− + H2O HSO4− + PO43−  SO42− + HPO42− 2AlCl3  Al2Cl6 Cu2+ + 4H2O  Cu(H2O)42+

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H

COMPARING THE CONCEPTS a) The Arrhenius view: HCl dissociates to form H+ and Cl−. HCl(aq)  H+(aq) + Cl−(aq) This concept is only limited to acid-base reactions in aqueous conditions involving H+ and OH−. b) The Brønsted-Lowry view: shows a closer approximation to what is really going on. The HCl doesn't just dissociate, but reacts with water to H3O+.

HCl(aq) + OH−(aq)  H2O(l) + Cl−(aq) The idea here is that a proton is transferred from one chemical to another in an acid-base reaction. c) The Lewis view: A H+ from HCl combines with a water molecule to form H3O+. Cl

H + O

H

Cl

-

+

+ H O H

H

H

The reaction occurs because H2O provides an electron pair for the hydrogen ion to accept. The electron dots emphasize the Lewis point of view. +

H

O

H

-

+

O

H

H

O

+ H O

H

H

H

Likewise, when a hydroxide ion accepts a proton, H+, it attaches itself to a pair of electrons from the oxygen. We use different acid-base theories to explain different reactions. This is because each of the acid-base theory has its limitations…

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Normally if the acid-base reaction is between strong mineral acids and bases in aqueous solutions, the Arrhenius theory is adequate: HCl(aq)  H+(aq) + Cl−(aq) But in a acid-base reaction that does not involve H+ or/and OH− ions, the Brønsted-Lowry Definition is required. NH3(aq) + H2O(aq) NH4+(aq) + OH−(aq) This theory is also useful to describe reactions of weak acids or bases that do not completely ionize. When we are dealing with acid-base reactions in a non- aqueous environment, whereby ionization may not occur, we use the Lewis acid-base theory. H Cl H Cl H

N +

B

H

Cl

Cl

H

N

B

H

Cl

Cl

9.2 Acid-Base Equilibria 9.2.1 The Auto-ionization of Water Water, the most common solvent for acids & bases is a very weak acid. Although water is a molecule, its high polarity enables it to dissociate and form ions. One water molecule will donate a proton to another to form a hydronium ion and a hydroxide ion. H3O+(aq) + OH−(aq)

H2O(l) + H2O(l)

= 1.0 x 10-14 (at 25oC) The water equilibrium constant for this reaction is better known as the auto-ionization constant of water, Kw.

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The auto-ionization constant of water, Kw, is also known as the ion-product constant for water. In pure water, the concentration of H3O+ and OH− produced by the auto-ionization are equal. Therefore… Kw = [H3O+(aq)] [OH−(aq)] = [H3O+] [H3O+] 1.0 x 10-14 = [H3O+]2 [H3O+] =

1.0 10 14 = 1.0 x 10

-7

M

Molar concentrations of H3O+ and OH− in many solutions; weak acids, bases and salts will always be very small. To simplify analysis and discussions, a scale based on the negative logarithm of molar concentrations is used. This is known as the pH scale + pH = − log [ H3O ] + Or more simply… pH = − log [ H ] From this relationship, we can determine the concentration of

H3O+. [ H+ ] = antilog (− pH) Therefore the pH of water (at 25oC) is… pH of water =

Another example: calculate the pH of 0.1M of HCl. Since HCl ionizes completely in water, HCl(aq)  H+(aq) + Cl−(aq)

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9.2.2 The pH Scale Definitions of acidic, basic and neutral according to pH.  pH is normally measured using…  pH meter (most  accurate)  litmus paper  (least precise)  pH paper   Acid-base indicators

9.2.2 The pH Scale The concept of pOH. pOH = − log [OH−] Because the auto-ionization of water produces OH− as well as H3O+, the pOH of water is… [ OH− ]water = 1.0 x 10-7 M pOH of water = Just the same… pKw = − log Kw Therefore…

pKw = 10

Notice that… Kw = [H3O+] [OH−] = 1.0 x 10-14 −log Kw = −log ([H3O+] [OH−]) Exercise 2: Calculate the [H3O+], [OH−], pH & pOH of:

a) b)

2.0 M HNO3 0.45 M NaOH

c)

0.002 M H2SO4

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9.2.3 Acid & Base Ionization Constants Based on the Bønsted-Lowry Acid-Base theory, reactions between acids & bases are seen as reversible. H2PO4−(aq) + NH3(aq) NH4+(aq) + HPO42−(aq) As such, the net direction of the reaction is based on the relative strengths of the acids & bases involved. Strong Acids ionizes completely (stronger tendency to donate proton)

HCl(aq) + H2O(l)  H3O+(aq) + Cl−(aq) Weak Acids ionizes partially (weaker tendency to donate proton) H3O+(aq) + HS−(aq)

H2S(aq) + H2O(l)

General guide on predicting strength of acidity: 

 Mineral acids are generally stronger than organic acids.

 The higher difference in electronegativity between H and  the other atom, more acidic.  The larger the size of the anion, more acidic.   The more “O” atom in the acid, more acidic.   Strengths of binary acids increase from top to bottom in a group of the periodic table & from left to right.

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General guide on predicting strength of acidity: 

 Mineral bases are generally stronger than organic bases.

 Bases decreases in strength across the period from left to  right (Group 1 being strongest, followed by group 2).  The larger the (metal) cation, the stronger the base.    A) Weak Acids Since a weak acid do not ionize completely, H3O+(aq) + A−(aq)

HA(aq) + H2O(l) K =

[H 3O + ][A - ]

But since [H2O] is a constant, [H 3O + ][A - ]

K=

[HA] (Ka : acid dissociation constant) a

Strength of an acid is BEST judged from it’s Ka value (NOT pH!)

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Exercise 3: HNO2 is a weak acid with a Ka value of 7.1 x 10-4. Write a chemical equation to show its dissociation, and then calculate the pH of 0.010 M HNO2

Simplified calculations…

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Exercise 4: Acetic acid, CH3COOH, is a very weak acid with a Ka value of 1.8 x 10-5. Calculate the pH of 0.010 M acetic acid.

Simplified calculations…

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When to use “simplified method”?  When difference between Ka and concentration is more than 1 x 10-3 (e.g. when: [HA]0 = 0.0100M, and Ka = 1.00 x 10-5 safe [HA]0 = 0.0010M, and Ka = 1.00 x 10-5 not safe!  For these equilibrium calculations to work properly, the acid  must not dissociate more than 5% in water.  Double check by dividing the final (ionized) [H+] with the initial (unionized) [HA]. Using the last example… Acetic acid, CH3COOH, is a very weak acid with a Ka value of 1.8 x 10-5. Calculate the pH of 0.010 M acetic acid.

Ostwald Dilution Law If conditions are right, calculations for weak acids can be summarized as follows… H+(aq) + A−(aq)

HA(aq)

[H + ][A - ] Ka = [HA] [H + ]2 = [HA]

but at equilibrium [H+] = [A−] and [HA] = [HA]0 − [H+]

But in weak acids, Ka value is very small. Thus, [H+] << [HA]0 As such, [HA]0 − x ≈ [HA]0 [H + ]2 Therefore, Ka = [HA] 0

Giving us… + [H ] = Ka [HA]o

This is called the Ostwald Dilution Law.

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That means, the pH (or [H+]) of weak acids, can be determined if we know: the Ka & the initial concentration. Sometimes strength of acids (or bases) is given in the form: percent ionization or percent dissociasion, α.

α

=

[H ]

100% [HA]0

A high α indicates a strong acid or a base. Acid

HCl

α

99.99%

CH3COOH 1.3%

HSO4

−

29.0%

HNO2 6.5%

Exercise 5: Calculate the pH of an unknown acid, HX, if it 0.50 M of this acid dissociates 9.5% when dissolved in water.

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B) Weak Bases Since a weak acid do not ionize completely, BH+(aq) + OH−(aq)

B(aq) + H2O(l)

[BH + ][OH - ]

Kc =

[B][H 2O]

But since [H2O] is a constant,

K=

[BH + ][OH - ]

[B] (Kb: base dissociation constant) Strength of a weak base is BEST judged from it’s Kb value (NOT pH!) Exercise 6: b

NH3 is a weak base with a Kb value of 1.8 x 10-5. Calculate the pH of 0.010 M NH3.

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If conditions are right, similar to weak acids, simplified calculations for weak bases can be summarized as follows… BH+(aq) + OH−(aq)

B(aq) + H2O(l)

[BH + ][OH - ] Kb = but at equilibrium [BH+] = [OH−] [B] [OH - ]2 = [HA]

and [B] = [B]0 − [OH−]

But in weak bases, Kb value is very small. Thus, [BH+] << [B]0 As such, [B]0 − x ≈ [B]0 [OH - ]2

Therefore,

K

b

=

[B]0

Giving us… − [OH ] = Kb [B]o

This is called the Ostwald Dilution Law. That means, the pOH (or [OH−]) of weak bases, can be determined if we know: the Kb & the initial concentration. Exercise 7: Calculate the pH, pOH and percent ionization of the following solutions: a) 0.03 M HF (Ka = 9.8 x 10-4) b) 0.05 M N2H4 (Kb = 1.7 x 10-6) c) 0.25 M HOCl (Ka = 3.0 x 10-8)

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Exercise 8: 0.100 M of an unknown monoprotic weak acid is found to have a pH of 2.38 at 25oC. Determine its pKa at this temperature.

C) Relationship between Ka & Kb According to the Brønsted-Lowry definition…

The conjugate base can react with water…

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Since both equations are closely related (they have same species of anions & molecules)… Notice that when… Ka x Kb =

Exercise 9: CH3COOH, a weak acid, has a Ka value of 1.8 x 10-5. Determine the strength of its conjugate base in pKb.

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Acid Dissociation for Polyprotic Acids A polyprotic acid is an acid that can donate more than one H+ ion per acid molecule. Dissociation of polyprotic acids differs at different levels, regardless if the polyprotic acid is strong or weak. Examples:

K

H2SO4: H2SO4 HSO4 − + H+ HSO4 − H2SO3: H2SO3 HSO3 − H3PO4: H3PO4 H2PO4−

HSO4 2− + HSO3 − +

H2PO4 −

+

K

a1

+H

K

a1

= 1.3 x 10-2 M

a2

= 6.50 x 10-8 M

= 7.90 x 10-3 M

K

HPO4 2− + H+

a2

= 1.40 x 10-2 M

K

H+ +

= very large

K

H+

H

HSO3 2− +

a1

a2

= 6.31 x 10-8 M

K

a3 = 3.98 x 10-13 M HPO4 2− PO4 3− + H+ Overall Ka is determined by multiplying the dissociation constants for all levels of dissociation. Example:

Ka for H2SO4

= very large (due to very large Ka1)

Ka for H3PO4

= Ka1 x Ka2 x Ka3 = (7.90 x 10-3) x (6.31 x 10-8) x (3.98 x 10-13) =

Exercise 10 Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0 x 10-5 and Ka2 = 5 x 10-12) found in citrus fruit. Calculate [H2Asc], [HAsc−], [Asc2-], and the pH of 0.050M H2Asc.

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9.3 Hydrolysis of Salt Weak acids & weak bases are not limited to molecular substances. SALTS (ionic compounds) can also function as a Brønsted acid or base. This is especially evident in salts of: STRONG ACID – WEAK BASE  (ACIDIC SALTS) WEAK ACID – STRONG BASE  (BASIC SALTS) Salts ionize completely in water, for example… NH4Cl  NH4+(aq) + Cl−(aq)

Its ions, however, can function as either as a Brønsted acid or base when reacting with water… NH4+(aq) + H2O(l) Cl−(aq) + H2O(l)

NH3(aq) + H3O+(aq) HCl(aq) + OH−(aq)

a) Acidic Salts Acidic salts are produced when a strong acid reacts with a weak base.

Once the Kh (or Ka) of the salt is known, it’s pH can be determined. 23

For example, the pH of 0.50 M of NH4Cl… NH4Cl  NH4+(aq) + Cl−(aq)

NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) Acidic salts are extremely weak acids, and have very small dissociation (e.g. Kh NH4Cl = 5.56 x 10-10)

b) Basic Salts Basic salts are produced when a weak acid reacts with a strong base.

The salt that is formed ionizes completely (as with all ionic compounds)… CH3COONa  CH3COO−(aq) + Na+(aq) Its ions, however, can function as either as a Brønsted acid or base when reacting with water… CH3COO−(aq) + H2O(l) Na+(aq) + H2O(l)

CH3COOH(aq) + OH−(aq) NaOH(aq) + H+(aq)

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For example, the pH of 0.50 M of CH3COONa… CH3COONa  Na+(aq) + CH3COO−(aq) CH3COO−(aq) + H2O(l) CH3COOH(aq) + OH−(aq) Basic salts are extremely weak bases, and have very small dissociation (e.g. Kh CH3COONa = 5.56 x 10-10)

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Summary Remember that salts ionize 100%. Salts can be either acidic or basic.

Acidity or basicity of salts are due to one of their ions. Acidity or basicity of salts are very weak (even weaker than weak acids or bases). The hydrolysis constant for salts are very small, therefore it is normally safe to assume that: [ salt ion ]0 = [ salt ion ]eq Therefore, short cut method is usually safe to use…

Exercise 11 Calculate the pH of the following solutions: a) 0.3 M sodium benzoate (Ka benzoic acid = 9.5 x 10-5) b) 1.5 M N2H5Cl (Kb N2H4 = 1.7 x 10-7)

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9.4 The Common Ion Effect: Buffer Solution What happens when we mix a weak acid with its salt? The molecules of the weak acid would ionize partially…

H+(aq) + CH3COO−(aq)

CH3COOH

CH3COONa  Na+(aq) + CH3COO−(aq) Based on equilibrium concepts, we know that the addition of the weak acid’s salt would cause a change in the ionic equilibrium. Therefore, it is reasonable to say that the acidity of the weak acid would have been effected. We call the change the common ion effect. Let’s look at the equations again… CH3COONa  Na+(aq) + CH3COO−(aq) CH3COOH H+(aq) + CH3COO−(aq) Since the salt ionizes completely, equilibrium law does not apply for that equation. As for the weak acid… [H + ]soln [CH 3 COO - ]soln

Ka =

[CH 3 COOH]soln

Let’s analyze a few facts… at equilibrium; [CH3COO−]soln = [CH3COO−]acid + [CH3COO−]salt  But since salt ionizes 100%, and acid ionizes < 5%; [CH3COO−]acid << [CH3COO−]salt  and [CH3COOH]0 ≈ [CH3COOH]soln due very little ionisation of acid molecules. So we can safely assume that… [CH3COO−]salt ≈ [CH3COO−]soln (value of [CH3COO−]acid is negligible) 27

But remember, the salt is a weak acid, so… [CH3COO−]soln = [H+]acid Applying these statements onto the ionic equilibrium equation… [H + ]soln [CH 3 COO - ]salt

Ka =

[CH 3 COOH]0

Rearranging the equation would give…

Likewise for base… BHY  Y−(aq) + BH+(aq) B(aq) + H2O(l)

OH−(aq) + BH+(aq)

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Exercise 12 Calculate the pH of the following solutions: a) 50 mL 0.3 M sodium formate is mixed with 50 mL 0.1 M formic acid, HCOOH (Ka HCOOH = 1.7 x 10-4) b) 100 mL 0.2 M NH3 is mixed with 200 mL 0.1 M NH4Cl (Kb NH3 = 1.8 x 10-5)

Buffer solution is… 

 a solution of a weak acid and its salt.  a solution of a weak base and its salt.

Example of an Acidic Buffer: Mixture of acetic acid & sodium acetate (its salt) CH3COONa  Na+(aq) + CH3COO−(aq) CH3COOH

H+(aq) + CH3COO−(aq) 29

Example of a Basic Buffer: Mixture of ammonia & ammonium chloride (its salt) NH4Cl(aq)  Cl−(aq) + NH4+(aq) NH3(aq) + H2O(l) OH−(aq) + NH4+ (aq) Buffer solution resists changes in pH of small amounts in acids or bases. Let’s compare what happens when 1.0 mL 0.1 M HCl is added to a 100 mL pure water compared to 100 mL buffer solution.

In the buffer solution…

When H+ is added to an acidic buffer, it reacts with the conjugate base in the buffer solution. HCOO−(aq) + H+(aq)  HCOOH(aq) Therefore, the quantity of [HCOO−] is reduced. and the quantity of [HCOOH] is increased. 30

n

HCOOH

n

HCOO−

Before + H+ 0.01 mol

+ 0.0001 mol

After + H+ 0.0101 mol

0.01 mol

− 0.0001 mol

0.0099 mol

H+

Using the Henderson-Hasselbalch Equation

If Ka HCOOH = 1.7 x 10

-4

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Exercise 13 Using the same sample buffer solution: 100 mL 0.1M HCOOH, 0.1M HCOONa, determine what is the pH change when 1.0 mL 1.0 M NaOH is added.

Exercise 14 A buffer solution is prepared by mixing 300 mL of 2.0 M ammonium chloride, NH4Cl, solution 200 mL to 0.5 M aqueous ammonia. a) Calculate the pH of the buffer b) Determine the pH of the buffer after addition of i. 5.0 mL 0.5 M HNO3 ii. 10.0 mL 0.2 M KOH (Kb NH3 = 1.8 x 10-5)

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Exercise 15 Calculate the mole ratio between NaF and HF if a 3.00 pH buffer solution is needed to be prepared. (Ka HF = 7.1 x 10-4)

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9.5 Titration Acid-base titration normally gives an “S” curve when a pH vs. volume of titrant is plot.

a)

Strong Acid vs Strong Base

For example, titration between 25.00 mL 0.10 M HCl and 0.10 M NaOH

EQUIVALENCE POINT is the point at which equimolar amounts of acid and base have reacted. At this point, all acids have completely reacted with all bases. Thus, the solution is 100% salt. pH of Equivalence point is the pH of the salt. Before Eq. point, pH of the solution is the pH of the solution in the conical flask. (in this case, the pH of HCl that have not been neutralized yet) Near equivalence point, pH rises steeply. After Eq. point, pH of the solution is the pH of the excess solution added into the conical flask. (in this case, the pH of excess NaOH) 34

b)

Strong Acid vs Weak Base

For example, titration between 25.00 mL 0.10 M HCl and 0.10 M NH3

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c)

Weak Base vs Strong Acid

For example, titration between 0.10 M NH3 and 25.00 mL 0.10 M HCl.

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c)

Strong Base vs Weak Acid

For example, titration between 25.00 mL 0.10 M CH3COOH and 0.10 M NaOH

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Summary of Shape of Titration Curves: Strong Acid vs Strong Base

Strong Base vs Strong Acid

Strong Acid vs Weak Base

Weak Base vs Strong Acid

Weak Acid vs Strong Base

Strong Base vs Weak Acid

Weak Acid vs Weak Base

Weak Base vs Weak Acid

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Exercise 16 Calculate the pH for the following steps within the titration of 20.00 mL 0.50 M KOH with 0.60 M benzoic acid, C6H5COOH; a) Before titration begins b) After addition of 10.00 mL benzoic acid c) At equivalence point d) After addition of 30.00 mL benzoic acid e) Sketch the pH vs. volume of titrant graph for this titration. (Ka benzoic acid = 9.5 x 10-5)

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Indicators During an acid-base titration process, the equivalence point is determined using acid-base indicators. This is usually done by dropping a few drops of indicators before the titration process. The end point occurs when the indicator changes colour. Not all indicators change colour at the same pH or pH range,

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Therefore, a good & precise titration can only be achieved if a suitable indicator is used. That is one with its end point overlapping the equivalence point of the titration. Let’s look at the use of three indicators for the titration of a strong acid vs. strong base. For example, the titration between 25.00 mL 0.10 M HCl and 0.10 M NaOH (titrant).

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Although phenolphthalein may be suitable when the titrant is NaOH, it may not be the case for the same titration in which HCl is the titrant!

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Let’s look at the use of three indicators for the titration of a strong acid vs. weak base. For example, titration between 25.00 mL 0.10 M HCl and 0.10 M NH3 (titrant).

43

Exercise 17 For the titration between 20.00 mL 0.5 M CH3COOH and 0.5 M NaOH, determine the pH of the solution in the conical flask… a) before titration begins

b) after addition of 10.00 mL NaOH c) after addition of 19.00 mL NaOH d) at equivalence point e) after addition of 21.00 mL NaOH Sketch the pH vs. volume of titrant graph for this titration. Following this, determine if phenolphthalein (pH range: 8.3 – 10.0) or methyl orange (pH range: 3.1 – 4.4) is more suitable as the indicator for this titration. Explain. (Ka CH3COOH = 1.8 x 10-5)

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9.6 Solubility Equilibria When salt, an ionic compound, is added to water, it will dissolve. When dissolved, the salt ionizes completely… CuCl2(s)  Cu2+(aq) + 2Cl−(aq) Salt will continue to dissolve until a point whereby the solution is SATURATED. When this happens, addition of more salt will not cause to dissolve. At this point, an equilibrium exist in the system. The quantity of undissolved (solid) salt that ionizes equals the amount of salt ions that precipitates. This equilibrium is called the solubility equilibrium. CuCl2(s) Cu2+(aq) + 2Cl−(aq) As in all equilibria systems, the equilibrium law can be written as...

Since CuCl2(s) is a constant… Ksp = [Cu2+(aq)] [Cl−(aq)]2 Ksp is the solubility product for CuCl2 Exercise 17 Write the solubility product expressions for the following compounds: a) Silver chloride, AgCl b) Sliver carbonate, Ag2CO3 c) Magnesium phosphate, Mg3(PO4)2 d) Iron(III) hydroxide, Fe(OH)3

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Exercise 18 If the solubility product, Ksp, of a compound is known, its solubility can be determined (and vice versa). For example, the solubility product, Ksp, of silver chloride, AgCl, is 1.6 x 10−10. Determine its solubility.

Ksp = [Ag+(aq)] [Cl−(aq)]

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Exercise 19 Write the solubility product expression for lead(II) fluoride and determine its molar solubility. Ksp PbF2 = 4.1 x 10−8.

Exercise 20 Determine the solubility product, Ksp, for copper(II) hydroxide if its molar solubility is 1.8 x 10−7mol L-1.

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Based on the solubility product, Ksp, we can predict…  Whether addition of certain quantity of salt would yield a dissolved solution, a saturated solution or a super-saturated solution.   A mixture of two aqueous solution would cause a precipitation to occur. Similar to the concepts in chemical equilibrium, we can compare the solubility product quotient, Qsp, with Ksp. CuCl2(s) Cu2+(aq) + 2Cl−(aq) Note that… Ksp = [Cu2+(aq)]equilibrium [Cl−(aq)]2equilibrium Wheas… Qsp = [Cu2+]0 [Cl−]02 Example What happens when 1.00 g of PbI2 is mixed with 200 mL of water? Will the compound dissolve completely? Explain. (Ksp, of PbI2, is 7.9 x 10−9)

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Exercise 21 Predict if a precipitate occur when 10.0 mL 0.15 M of calcium nitrate, Ca(NO3)2, is added to 100.0 mL of 0.50 M sodium sulphate, Na2SO4, solution. (Ksp, of CaSO4, is 2.4 x 10−5)

Exercise 22 At 25oC, the molar solubility of CoCO3 in a 0.10 M Na2CO3 solution is 1.0 x 10−9 mol L−1. What is the Ksp for CoCO3?

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Exercise 23 Would a precipitate of silver acetate, CH3COOAg, form if 22.0 mL of 0.100 M AgNO3 were added to 45.0 mL of 0.0260 M CH3COONa? Ksp for CH3COOAg is 4.0 x 10−3.

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EXERCISE 1. Identify Lewis acid and Lewis base in each of the following reactions: (a)

SnCl4(s) + 2Cl-(aq)

(b)

Hg2+(aq) + 4CN-(aq)

SnCl62-(aq) Hg(CN)42-(aq)

2. Classify each of the following species as a Bronsted acid, base or both: H2O, OH-, H3O+, NH3, NH4+, NH2-, CO32-, HBr, HCN 3. Identify the acid-base conjugate pairs in each of the following reactions: (a)

CH3COO- + HCN

CH3COOH + CN-

(b)

HCO3- + HCO3-

H2CO3 + CO32-

(c)

H2PO4- + NH3

(d)

HClO + CH3NH2

HPO42- + NH4+ CH3NH3+ + ClO-

4. The concentration of OH- in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of H+ ions. 5. Indicate whether the following solutions are acidic, basic or neutral: (a)

0.62 M NaOH

(b)

1.4 x 10-3 M HCl

(c)

2.5 x 10-11 M H+

(d)

3.3 x 10-10 M OH-

6. The concentration of H+ ions in a bottle of vinegar was 3.2 x 10-4 M right after the cork was removed. Only half of the vinegar was consumed. The other half, after it had been open in the air for a month, was found to have a hydrogen ion concentration of 1.0 x 103 M. Calculate the pH of the vinegar on these two occasions. 7. Calculate the pH of a 0.027 mol L-1 solution of propionic acid at 250C. (Ka = 1.3 x 10-5)

8. How much NaOH (in grams) is needed to prepare 500 mL of solution with a pH of 10.00? 9. Calculate the concentration of H+ ions if the pOH of the solution is 9.40. 10. The pH of a 0.100 M solution of a weak acid HA is 2.85. What is the Ka of the acid?

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11. Calculate the percentage of ionization for 0.60 M hydroflouric acid solution? (Ka HF = 7.1 x 10-4) 12. What is the pH of a 0.40 M ammonia solution? (Kb NH3 = 1.8 x 10-5) 13. Calculate the pH and percentage of hydrolysis (a)

0.15 M solution sodium acetate

(b)

0.10 M NH4Cl solution ( Kh = 5.6 x 10-10 )

14. Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa. What would be the pH of the 0.20 M CH3COOH solution if there was no salt present? ( Ka CH3COOH = 1.8 x 10-5) 15. Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What would the pH be after the addition of 20.00 mL of 0.050 M NaOH to 80 mL of the buffer solution? 16. In the titration of 25.0 mL of 0.100 M acetic acid with sodium hydroxide, calculate the pH after the addition of: (Ka = 1.8 x 10-5) (a) 10.00 mL of 0.100 M NaOH (b)

25.00 mL of 0.100 M NaOH

(c)

35.00 mL of 0.100 M NaOH

(d)

to the acid solution. ( Kh = 5.6 x 10-10 )

17. In the titration of 25.0 mL of 0.100 M HCl with ammonia, calculate the pH after adding: (a)

10.0 mL of 0.100 M NH3

(b)

25.0 mL of 0.100 M NH3

(c)

35.0 mL of 0.100 M NH3

to the acid solution. ( Kh = 5.6 x 10-10 ) 18. The solubility of silver sulphate is 1.5 x 10-2 mol/L. Calculate the solubility product of the salt. 19. It was found experimentally that the solubility of calcium sulphate is 0.67 g/L. Calculate the Ksp for calcium sulphate. 20. Will precipitate form if 200 mL of 0.0040 M BaCl2 are added to 600 mL of 0.0080 M K2SO4? ( Ksp BaSO4 = 1.1 x 10-10 )

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STPM Past Year Questions Q6-P2-2003 a) Ethanoic acid is an example of a weak acid. i)

Describe the properties of ethanoic acid as a Bronsted-Lowry acid ii) Describe the use of ethanoic acid in the preparation of a buffer solution. b) Phenolphthalein, HPh, is an acid-base indicator with a pH range between 8.2 and 9.8. The equilibrium equation of phenolphthalein in aqueous state is as follows: HPh(aq) + H2O(l) H3O+(aq) + Ph−(aq) Explain how phenolphthalein acts in acid-base titration. c) A solution is prepared by mixing 250.0 cm3 0.05 mol dm-3 nitric acid with 250.0 cm3 0.05 mol dm-3 calcium hydroxide solution. Calculate the pH of the solution.

Q5-P2-2002 a) Boric acid, B(OH)3, is an antiseptic found in some eyewash. In aqueous state, boric acid acts as a Lewis acid. i) Define Lewis acid. ii) Explain the acidic property of aqueous boric acid. b) Phenol, C6H5OH, is a weak organic acid. A solution containing 0.385 g of phenol in 2.00 dm3 of water has a pH of 6.29 at 25oC. i) Explain why phenol is a weak organic acid.

ii) Calculate the acid dissociation constant, Ka, of phenol at 25oC.

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