Topic 12: Acid-Base Equilibria

Question 1

What is the Bronstead-Lowry definition for acids and bases?

Answer 1

Acids are proton donors (H+) and bases are proton acceptors.

To accept a proton, a base must have a lone pair of e to form a dative cv bond w the proton. So a base must have an atom on the right side of the periodic table.

Question 2

What are conjugate base pairs?

Use HCl + H20 –> H3O+ + Cl- in your answer

Answer 2

In the forward reaction, HCl is an acid bc it’s donating a proton to H20. H20 is a base bc it’s accepting a proton from HCl.

In the reverse reaction H3O+ behaves as an acid because It’s donating a proton to Cl-. Cl- is the base because it’s accepting a proton. Therefore HCl and Cl-, H3O+ and H20. Both of these are conjugate base pairs

Question 3

Identify the acid and base in each of the following reactions:

Answer 3

Question 4

State definitions for monoprotic and diprotic acids

Answer 4

Monoprotic acids: can only donate one proton, eg HCl

Diprotic acids: can donate 2 protons, eg H2SO4

Question 5

State definitions for monoprotic and diprotic/diacidic bases

Answer 5

monoprotic base: accepts one proton

diprotic/diacidic base: accepts 2 protons, eg C03 2-

Question 6

What is the difference between strong and weak acids?

Answer 6

Strong acids almost completely disassociate in aqueous solution.

Weak acids only partially dissociate in aqueous solution.

H+ conc is directly related to acid concentration. E.g. a HCl solution of 0.1moldm^3 will produce a H+ ion conc of 0.1moldm3

Question 7

What is pH?

Answer 7

pH= -log10 [H+]

This can be rearranged to:

[H+]= 10^-pH

Question 8

Calculate the pH of a solution formed when 100 cm³ of of water is added to 50 cm³ of 0.1moldm3 HN03

Answer 8

[H+] in original solution= 0.1

[H+] in dilute solution= 0.1 x old volume/new volume

= 0.1 x 50/150= 0.333..

pH= -log10[0.333..]

pH= 1.48

Question 9

Calculate the pH of the solution formed when 250 cm³ of 0.3 moldm3 H2SO4 is made up to 1000 cm³

solution with water

Answer 9

[H+] in original solution equals 0.3 x 2= 0.6

You multiply by two because H2SO4 is diprotic!

[H+] in dilute solution= 0.6× 250/1000=0.15

pH= -log10 [0.15]= 0.82

Question 10

What do you assume when finding the pH of strong acids?

Answer 10

Assume that [H+] = [HA]

Question 11

How do you find [H+] of a weak acid using the acid dissociation constant?

Answer 11

If HA represents a weak acid the eqn for its disassociation into aq solution is: HA —> H+ + A-.

Using eqm law to find Ka: Ka= [H+(aq)] [A-(aq)] / [HA(aq)]

Assume the conc of H+ and A- is the same. This simplifies the expression to: Ka = [H+]^2 / [HA]

Therefore √Ka [HA] = H+

Question 12

Calculate pH of an aqueous solution of ethanoic acid of 0.05moldm3. The value of Ka for ethanoic acid is 1.74 x 10^-5 moldm3 at 298K.

CH3COOH(aq)—-> CH3COO- + H+

Answer 12

Question 13

What is Kw?

Answer 13

Pure water self ionises: H20(l) —-> H+(aq) + OH-(aq)

Applying the law of eqm: [H+] [OH-] / [H20] =Kw

As water is constant at a given temperature, the expression may be simplified to: [H+] [OH-] =Kw

298K, Kw= 1x 10^-14 mol2dm-6

Question 14

What is the relationship between Kw and pKw?

Answer 14

pKw = -log10 Kw

Question 15

Describe what happens to the pH of acids when diluted

Answer 15

Strong acids: PH increases by one for each tenfold decrease in concentration

Weak acids: PH increases by 0.5 for each tenfold decrease in concentration

Question 16

What is the difference between the endpoints and the equivalence point of a titration?

Answer 16

Endpoint: When the indicator changes colour

Equivalence point: when acid + base react juntos in exact propotions.

pH at the equivalence point depends on the acid and base, e.g. strong acid +weak base will have pH<7 at the equivalence point.

Question 17

What are indicators?

Answer 17

Most are weak acids which dissociate to form an ion of a DIFF colour to the acid molecule: Hind(aq) —> H+ + ln-(aq).

HInd and In- have diff colours in solution. For methyl orange these are red and yellow. When H+ is large eqm shifts left so solution turns red in acids. Sera a stage where [Hind]= [Ind-] and the indicator looks orange. Find the exact pH of this using the eqm constant for methyl orange:

KInd= [H+] [in-] / [Hind] = 2 x 10^-4 moldm3

When HInd= Ind-, expression simplifies: Kind= [H+]= 2 x 10^-4. This gives a pH 3.7 at the halfway stage and is the same value as pKind

The best indicator to choose for a titration has a pKind value closest to the pH at the equivalence point

Question 18

Describe and explain the titration curve for:

Strong base + Strong acid

Answer 18

Any indicators within the straight line can be used in the titration. this graph has a longer straight line

Question 19

Describe and explain the titration curve for:

Titration of a strong acid with a weak base

Answer 19

The straight line happens at neutralisation. It doesn’t reach such a high pH bc it is a weak base. Straight line is shorter.

Question 20

Titration curve to show the titration of a weak acid with a strong base?

Answer 20

The dashed line is the same as the end point.

Question 21

What does a titration curve of a weak acid with a weak base look like?

Answer 21

There is no steep section in this graph. Instead hay a point of inflection. The lack of a steep section means that its difficult to do a titration using an indicator.

Question 22

Calculate the pH of 0.1moldm3 H2SO4

Answer 22

2 x 0.1= 0.2 (DIPROTIC!)

-log10 [0.2]= 0.698 = 0.70

Question 23

Calculate the pH of 0.25moldm-3 methanoic acid. Ka= 1.7 x 10^-4

Answer 23

[H+]= Square root of Ka[HA]

= Square root of (1.7 x 10^-4 x 0.25)= 6.5 x 10^-3

pH= -log10 [6.5 x 10^-3]

= 2.19

Question 24

What are buffers?

Answer 24

A solution which resists any change in pH when a small amount of acid or base is added, so pH remains almost constant. Hay acid and alkalai buffer solutions:

Acid buffer solutions: mixture of a weak acid and its conjugate base of similar conc, eg enthanoic acid and Na ethanoate.

Alkaline buffer solutions: a mixture of a weak base and its conjugate acid of similar conc, eg ammonia and ammonium chloride.

Question 25

Which acid is stronger: lactic acid or ethanoic acid?

Answer 25

Ethanoic acid has pKa of 4.8 and lactic acid has pKa of 3.86

3.86 is less than 4.8 therefore lactic acid is a stronger acid.

Question 26

Complete the following equation for the reaction of methanoic acid with propanoic acid. Use your data booklet to help you.

HCOOH + C2H5COOH —>

Answer 26

According to the data booklet, methanolic acid is stronger because the K a value is bigger hence the equation should look like this:

HCOOH + C2H5COOH –> HCOO- + C2H5COOH2+

Methanoic acid is stronger so it becomes the H+ donator!!!

Question 27

A buffer solution with a pH of four is made using lactic acid and sodium lactate. A small volume of hydrochloric acid is added to the buffer solution. Explain why the PH does not change significantly.

Answer 27

There is a large reservoir of lactate ions. Therefore the addition of hydrogen ions from hydrochloric acid will not completely change the ratio of undissociated lactic acid to lactate. (HA: A-)

Question 28

Calculate the pH of 0.15moldm3 solution of lactic acid. State to assumptions you have made.

Ka of lactic acid is 1.38×10^ -4

Answer 28

H+ = √Ka x HA

=√1.38×10^ -4 x 0.15 = 4.55x10^-3

pH= -log10 (4.55x10^-3) = 2.34

Assumption 1: All H+ ions came from the acid and not from the dissociation of water

Assumption 2: [H+] = [A-]

Question 29

Sulfurous acid, H2SO3, is also a diprotic acid. Diprotic acids require two OH– ions per molecule for complete neutralization.

Sketch the likely shape of the titration curve for sulfurous acid, H2SO3, during the neutralization process.

Answer 29

Because its a diprotic acid, it will have 2 equivalance points.

Question 30

Explain why the calculation of the pH of a solution of
sodium hydrogenethanedioate gives a more accurate value than a similar calculation for ethanedioic acid.

Answer 30

Ethanedioic acid is a stronger acid than sodium hydrogenethanedioate. So [H2C2O4]equilibrium is not equal to [H2C2O4]initial