ACIDS & BASES

PropertiesAcids Bases

TheoriesArrhenius Bronsted-Lowry

HCl (g) + H2O (l) à Cl- (aq) + H3O+ (aq)

HCl (g) + NH3 (g) à Cl- (g) + NH4+ (g)

NH3 (aq) + H2O (l) à NH4+(aq) + OH- (aq)

Hydrogen ion =

Hydronium ion =

Hydroxide ion =

Amphoteric =

Conjugate Acids and Bases

1. In the following reaction identify the acid, base, conjugate acid and conjugate baseHBr + NH3 NH4

+ + Br -

2. What is the conjugate base of H2S?

3. What is the conjugate acid of NO3-?

Polyprotic Acids donate only 1 proton at a time!Monoprotic –

Diprotic –

Triprotic –

4. Write the ionization/dissociation reactions for each of the following acids (omit water)a) HCl

b) HC2H3O2

c) The ammonium ion NH4+

d) H2SO4

Relative Acid and Base Strength

Strength is determined by the position of the dissociation equilibrium DO NOT CONFUSE CONCENTRATION WITH STRENGTH!

Strong Acids

Weak Acids

Factors Affecting Acid Strength

Polarity Proton donated

Nonpolar – no effect on pH

Strength of bondH-X bond Bond Strength

(kJ/mol)Acid Strength

H-F 565 WeakH-Cl 427 StrongH-Br 363 StrongH-I 295 Strong

Oxyacids

C

Cl

OH is bonded to a central atom, Y.

The higher the electronegativity of the central atom, the ______________ the acid

The more oxygen that is attached to Y, the ______________ the acid.For a series of oxyacids, acidity increases with the number of oxygens.

Strong Bases

Weak Bases

The stronger the acid, the weaker its conjugate base. The converse is also true.

In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.

HCl(aq) + H2O(l) ¾¾® H3O+(aq) + Cl−(aq)

H2O is a much stronger base than Cl−, so the equilibrium lies very far to the right (K>>1).

Weak Acids and BasesThe vast majority of Acids & Bases are weakWeak acids and bases (weak ≠ dilute) are only partially ionized

ACIDSFor a generalized acid dissociation, HA (aq) + H2O (l) A-(aq) + H3O+ (aq) the equilibrium expression would be:

Ka =

This equilibrium constant is called the acid-dissociation constant, Ka.

Polyprotic Acids• Have more than one acidic proton.• If the difference between the Ka for the first dissociation and subsequent Ka values is 103

or more, the pH generally depends only on the first dissociation.

BASESBases react with water to produce hydroxide ion.B (aq) + H2O (l) HB+(aq) + OH- (aq) The equilibrium constant expression for this reaction is

Kb =

See Textbook Appendix for Ka and Kb values

Ways to describe Acid strengthProperty Strong Acid Weak AcidKa value

Position of dissociation equilibrium

Equilibrium concentration of H+ compared to original concentration HAStrength of conjugate base compared to that of water

The Autoionization of WaterH2O (l) + H2O (l) OH- (aq) + H3O+ (aq)

The equilibrium expression for this process is Kw=This special equilibrium constant is referred to as the ion-product constant for water, Kw.

At 25⁰C, Kw = 1.0 ´ 10−14 So if you know either [OH-] or [H3O+] you can calculate the other!

5. At 60⁰C, Kw = 1.0 ´ 10−13 a) Calculate [OH-] and [H3O+] in a neutral solution (where [OH-] = [H3O+] ) at 60⁰C.

b) Using Le Chatelier’s principle, predict whether the reaction below is endothermic or exothermic2H2O (l) OH- (aq) + H3O+ (aq)

Relationship between Ka and Kb

Ka and Kb are related in this way: Ka ´ Kb = Kw Therefore, if you know one of them, you can calculate the other.

THE pH SCALE pH is defined as the negative base-10 logarithm of the hydronium ion concentration.

pH = −log [H3O+]

Knowing that at 25⁰C, Kw = [H3O+] [OH−] = 1.0 ´ 10−14 and in pure water [H3O+] = [OH−], [H3O+] =

Therefore, in pure water, pH =

An acid has a ____________ [H3O+] than pure water, so its pH is <7

pH _____________ as acidity & [H+] increases

A base has a ___________ [H3O+] than pure water, so its pH is >7.

pH ____________ as basicity & [OH-] increases

How Do We Measure pH?For less accurate measurements, one can use

Litmus paper• “Red” paper turns blue above ~pH = 8• “Blue” paper turns red below ~pH = 5

An indicatorFor more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

Solution type [H+] [OH-] pH ValueAcidicNeutralBasic

[OH-] = [H3O+] _____________

[OH-] > [H3O+] _____________

[OH-] < [H3O+] _____________

Other “p” scalesThe “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).Some similar examples are

pOH = −log [OH−] pKw = −log Kw

In a neutral solution at 25⁰C we know that [H3O+] = [OH−] = Kw = 1.0 ´ 10−14,

Taking the -log of this….. −log [H3O+] + −log [OH−] = −log Kw = ________

or, in other words, pH + pOH = pKw = ________

6. The pH of a sample of blood was measured to be 7.41 at 25⁰C. Calculate the [H+], pOH and [OH-] for the sample.

CALCULATING pH OF A STRONG ACID OR BASERemember:• For the monoprotic strong acids, [H3O+] = [acid]• For the monobasic strong acids, [OH-] = [base]

(Autoionization of water is a negligible H+ contributor unless acid/base is very dilute <10-6M)

7. Calculate the pH of 0.10 M HNO3

8. Calculate the pH of 5.0x10-2 M NaOH

CALCULATING pH OF WEAK ACIDS

Remember: weak ≠dilute!

These acids are in equilibrium HA (aq) + H2O (l) A-(aq) + H3O+ (aq)

Calculating pH from K a You will be given the concentration of the weak acids (initial concentrations) and a Ka value. We can use BRICE TABLES to solve for the [H+] and then solve for pH.

9. Calculate the pH of a 0.10 M solution of acetic acid, HC2H3O2, at 25⁰C.HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2

−(aq) Ka = 1.8 ´ 10−5

Often the –x term in a Ka expression can be neglected. (This simplifies the math tremendously as we now do not have to use the quadratic formula!)

How do you know when to neglect x? One method tells us to neglect x when the percent ionization is less than 5%.

Percent Ionization [ H 3O + ] ´ 100 In the above example: [HA] [H3O+]eq = 1.34 ´ 10−3 M

[HC2H3O2]initial = 0.10 M

Go ahead and try the above example using the quadratic and again by neglecting x. The answers should be the same.

Good news – The AP exam will not require you to do the quadratic equation. No promises about homework problems though Calculating K a from the pH

To solve for Ka we need to know the [H+] (and the anion and the acid concentrations too).

[H+] we can get this from the pH. (pH = -log H+ so 10-pH = H+)

[A-] assuming it is a 1:1 ratio (we are dealing with a monoprotic acid) we can assume [A-] = [H+]

[HA] this will be given to us in the problem. We can probably use the initial concentration of the acid given to us in the K expression. Really the concentration will be a bit lower (because some of the acid has dissociated), but it won’t be much. If we want to be accurate, we can use an ICE table.

Let’s try both methods!

10. The pH of a 0.10 M solution of formic acid, HCOOH, at 25⁰C is 2.38. Calculate Ka for formic acid at this temperature.

HCOOH (aq) H+ (aq) + COOH- (aq)

Kb can be used similarly to find [OH−] and, through it, pH.

11. What is the pH of a 0.15 M solution of NH3 where Kb = 1.8´ 10−5? NH3 (aq) + H2O (l) NH4

+ (aq) OH- (aq)

ACID-BASE PROPERTIES OF SALT SOLUTIONS

Salts are ionic compounds produced when acids and bases react. Soluble salts will dissociate into their respective cations and anions. Those ions may affect the pH of the resulting solution.

Acid + Base Water + Salt

Write the equation for the formation of the salt NaNO2(aq).

______ + ______ ______ + ________ Acid + Base Water + Salt

Where does the anion of the salt come from? _________________

Where does the cation of the salt come from? _________________

Use this to determine the acid/base strength of each of these ions.Na+ =

NO2- =

pH of a salt solution• Salts derived from strong base & strong acid: ___________

Neither anion nor cation will hydrolyze water Example:

• Salts derived from strong base & weak acid: ___________anion is a conjugate base of a weak acid, ph>7 Example:

• Salts derived from weak base & strong acid: ___________cation is a conjugate acid of a weak base, pH<7 Example:

• Salts derived from weak base & weak acid: ___________both anion & cation hydrolyzes. pH depends on extent Example: For each, tell whether the substance is:

strong acid (SA) strong base (SB) weak acid (WA) weak base (WB) acidic salt (AS) basic salt (BS) neutral salt (NS)

1. HCl 7. HC2H3O2

2. NaBr 8. KOH

3. NH4Cl 9. NH4I

4. NH3 10. HF

5. KF 11. H3PO4

6. H2SO4 12. MgCl2

15. Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2 x 10-4.

16. Calculate the pH of a 0.1M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5

Common Ion effect and Ksp Solubility ProductConsider the equilibrium that exists in a saturated solution of BaSO4 in water:

First figure out which ion will react with water . Write out that reaction.

Since this is a reaction of a base, we need to find Kb. Remember Ka Kb = Kw

Using the value of K b solve for [OH-]

Now you can calculate the pOH and then the pH!

BaSO4 (s) Ba+2 (aq) + SO4-2 (aq)

Write the K expression for the above reaction. Ksp =

Common Ion Effect says that if one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.

AP Free Response 2004 Answer the following questions relating to the solubilities of two silver compounds, Ag2CrO4 and Ag3PO4.Silver chromate dissociates in water according to the equation shown below:

Ag 2CrO4 (s) 2Ag+ (aq) + CrO42- (aq) Ksp = 2.6 x 10-12 @ 25°C

a) Write the equilibrium constant expression for the dissolving of Ag2CrO4 (s)

b) Calculate the concentration, in M, of Ag+(aq) in a saturated solution of Ag2CrO4 @ 25°C.

c) Calculate the maximum mass, in grams, of Ag2CrO4 that can dissolve in 100mL of water at 25°C.

d) A 0.100 mol sample of solid AgNO3 is added to a 1.00L saturated solution of Ag2CrO4. Assuming no volume change, does [CrO4

2-] increase, decrease or remain the same? Justify your answer.

In a saturated solution of Ag3PO4 at 25°C, the concentration of Ag= (aq) is 5.3 x 10-5M. The equilibrium constant expression for the dissolving of Ag3PO4 in water is shown below.

Ksp = [Ag+]3[PO43-]

e) Write the balanced equation for the dissolving of Ag3PO4 in water.

f) Calculate the value of Ksp for Ag3PO4 at 25°C.

g) A 1.00 L sample of saturated Ag3PO4 solution is allowed to evaporate at 25°C to a final volume of 500mL. What is [Ag+] in the solution? Justify your answer.

What does this have to do with acids and bases??

Adding a common ion prevents a weak acid or base from ionizing as much as it would without the common ion.

Look at the following equation for the reaction of acetic acid in water:HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2

−(aq)

If we added some sodium acetate (NaC2H3O2) to this solution, we will effectively be adding more C2H3O2

−. Which way would that shift the equilibrium? (Think Le Chatelier……...think about Q)

As a result of that shift, what would the consequence be to the [H3O+]?

So what would happen to the pH?

17. a) Calculate the pH of a 2.0M HF solution. Ka = 7.2 x 10-4.

b) Calculate the pH of a 2.0M HF solution that is mixed with 1.0M NaF.

BuffersThe most important application of acid base solutions containing a common ion is for buffering. A buffered solution ______________________________

So they must contain an acid (that can deal with the addition of any base) and a base (that can deal with the addition of any acid). But, this acid and base must not react with each other. We need a weak conjugate acid-base pair!

Calculate the pH of a buffer18. a) Calculate the pH of a solution containing 0.5M acetic acid (HC2H3O2, Ka=1.8x10-5) and 0.5M sodium acetate (NaC2H3O2).

b) Calculate the change in pH that occurs when 0.010 mole of NaOH is added to 1.0 L of the previous solution.

Buffering CapacityThe amount of acid/base that can be neutralized before the pH changes significantly.The pH of a buffered solution depends on the ratio of [HA]

[A-]But the capacity of a buffered solution depends on the magnitude of [A-] & [HA]

Compare the following two solutions:1.0M HC2H3O2 & 1.0M NaC2H3O2

0.1M HC2H3O2 & 0.1M NaC2H3O2

[H+] = Ka [Acid] so it would be the same in both of the above solutions. [A-]

However, the 1M solution has more acid and anion so it would have a larger capacity to neutralize any added acid or base

Write the balanced reaction showing which species will react with the added base.

How many moles of acetic acid are in solution? How many moles of base are added?Use stoichiometry to determine how much acetic acid remains after neutralizing the base.

Now we can calculate [H+] and then pH

Henderson Hasselbach Manipulating the Ka expression, we get

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

Taking the –log of each side, we get

pH = pKa – log [acid] à [base]

The pH of a buffered solution depends on the ratio of [HA] [A-]

Large changes is in this ratio would produce large changes in pH, so an optimal buffer occurs when [HA] = [A-]. This would make the ratio = 1.

Plugging this into Henderson-Hasselbach…..

When choosing a buffer, the most effective will have pH = pKa

19. What is the pH of a buffer that is 0.12M in lactic acid, HC3H5O3 and 0.1M in sodium lactate? Ka for lactic acid = 1.4 x 10-4

20. What is the pH of a buffer solution that is 0.45M acetic acid, HC2H3O2, and 0.85M sodium acetate? Ka=1.8x10-5

pH = pKa + log [base] [acid]

21. A buffer is made by adding 0.30 mol HC2H3O2, and 0.30 mol NaC2H3O2 to enough eater to make 1.00 L of solution. The pH of the buffer solution is 4.74.a) Calculate the pH after 0.020 mol of NaOH is added

b) Calculate the pH after 0.020 mol of HCl is added

TitrationsA sample of known concentration acid or base (titrant) is slowly added to a sample of unknown concentration of base or acid (analyte). Formula MH+VH+ = MOH-VOH- can be used to calculate the molarity of the analyte.

A pH meter, or indicator, is used to determine when the solution has reached the end point.End point =

Equivalence point =

Only titrations of strong acids with strong bases will result in an equivalence point at pH=7.

Titration curves

Graph of the titrant volume vs. pH

4 distinct regions of the curve:Initial

Before the equivalence point

Half equivalence point

At the equivalence point

After the equivalence point

Strong Base added to a Strong AcidLet’s calculate the pH changes as 0.10M NaOH is added to 50.0 mL of 0.20 M HNO3

A. Initial pH – 0 mL of NaOH has been added

B. 10 mL of 0.1M NaOH has been added

Before any reaction occurs, we have:H+ .2M x .05L = NO3

- .2M x .05L = Na+ .1M x .01L =OH- .1M x .01L =

Use stoichiometry to determine what remains in solution.

Calculate the new molarity of the H+ ion (using new volume of 50mL + 10mL). Then calculate the pH

C. 20 mL of 0.1M NaOH has been addedAt the point where 20 mL of total base has been added…..Before any reaction occurs, we have:

H+ .2M x .05L = NO3

- .2M x .05L = Na+ .1M x .02L =OH- .1M x .02L =

Use stoichiometry to determine what remains in solution.

Calculate the new molarity of the H+ ion (using new volume of 50mL + 10mL). Then calculate the pH

D. 50 mL of 0.1M NaOH has been added

E. 100 mL of 0.1M NaOH has been added

F. 150 mL of 0.1M NaOH has been added Now OH- is in excess and will determine the pH. Calculate the concentration of hydroxide remaining in solution after neutralization. Then calculate the [H+] using the formula [H+][OH-] = Kw. Then solve for pH

G. 200 mL of 0.1M NaOH has been added

Indicator choiceNotice how steep the slope is surrounding the equivalence point in this titration of a strong acid and strong base. Adding a very small amount of titrant caused the pH to change dramatically. It would not matter much if we chose to use an indicator such as methyl red (which changes pH between 4-6) or phenolphthalein (which changes pH between 8-10). The results would have been almost the same.

However, if this vertical part of the graph was not as large, it would mean we would have to be more selective when choosing an indicator. A useful indicator has a pH range that covers the pH at the equivalence point.

Titration of Weak Acid with Strong Base

Let’s calculate the pH changes as 0.10M NaOH is added to 50.0 mL of 0.10 M HC2H3O2

(HC2H3O2, Ka=1.8x10-5)

A. Initial pH – 0 mL of NaOH has been added

B. 10 mL of 0.1M NaOH has been added The added strong base will react with the acid. Calculate how many moles of acid (and acetate ion) are left over from this neutralization. Then calculate the molarity of the remaining ions.Next use those values, and the K value, to calculate the hydrogen ion concentration (and then the pH)

C. 25 mL of 0.1M NaOH has been added

Notice that for part C, we have used up half of the acid. Initially we had (.05mL)(.1M) = .005moles acid. After adding .025moles of base we saw that half of the original acid (HC2H3O2) was converted to conjugate base (C2H3O2

-).

Neglecting the dissociation (ignoring the –x), we can say that after neutralization, [HC2H3O2] = [C2H3O2

-]Plugging this into the Ka expression,

Ka = [H + ] [C 2H3O2- ] Ka = [H + ] [C 2H3O2

- ] Ka = [H+] so, pH = pKa

[H C2H3O2] [H C2H3O2]

D. 40 mL of 0.1M NaOH has been added

E. 50 mL of 0.1M NaOH has been added This is the equivalence point. After .005 moles of base are added to .005 moles of acid, the only species left in solution after the neutralization are Na+, C2H3O2

-, and H2O.As the conjugate of a strong base, Na+ will not affect the pH. However, as a weak base C2H3O2

- will remove a proton from water.Write this reaction.

Calculate Kb (knowing KaKb=Kw)

Then solve for x (which will be the [OH-])

Next solve for [H+] and then pH

Why is the ph at the equivalence point of this weak acid and strong base equal not to 7? F. 60 mL of 0.1M NaOH has been added

G. 75 mL of 0.1M NaOH has been added

For this titration, would methyl red (which changes pH between 4-6) or phenolphthalein (which changes pH between 8-10) be equally as effective?

Titrations with Polyprotic acidsThere is an equivalence point for each dissociation.