F325: Module 1: Acids, Bases and Buffers

Question 1

C) What is a Bronsted Lowry acid and base?

Answer 1

• Acid= proton donor

- Base= proton acceptor

Question 2

C) What is a conjugate acid base pair?

Answer 2

Species differing only in a H

Question 3

C) What is a strong acid? What is a weak acid? (lol easy question soz)

Answer 3

A strong acid is one that completely dissociates.

A weak acid is one that only partially dissociates in solution.

Question 4

C) What is pH?
What is Ka?
What is pKa?
What is a buffer?

Answer 4

• pH is the negative log to the base ten of the hydrogen ion concentration.
• Ka is the acid dissociation constant of a weak acid. Ka = [H+][A-]/[HA]
• pKa is the negative log to the base ten of the Ka of a weak acid
• A buffer is a solution that resists changes in pH when small quantities of acid or alkali are added

Question 5

C) How does pure water conduct (some) electricity?

Answer 5

Water contains ions that can carry charge. Indeed, water can be electrolysed by a direct current but its conductivity is very low.

Question 6

C) What is the ionic equation when a metal carbonate reacts with an acid
When a metal (2+) neutralises an acid
When a base neutralises an acid

Answer 6

Co3^2- + 2H+ –> CO2 + H2O

M + 2H –> H2 + M^2+

H+ + OH- –> H2O

Question 7

C) What is the enthalpy change of neutralisation?

Answer 7

When a sufficient amount of acid or alkali is neutralised to produce one mole of water.

Question 8

C) What is the expected value for enthalpy of neutralisation for strong acids and bases? Why?

Answer 8

57kJ per mole of water produced. This is because the neutralisation equation is always the same for acid base reactions: H+ + OH- –> H2O

Question 9

C) State the equation for the ionic product of water.

What is this value at 298K?

Answer 9

• Kw= [OH-][H+] (no H2O on the bottom of the equation because we assume the conc of H2O to be constant as it is so great).
• At 298K it is 1x10^-14 mol^2 dm^-6

Question 10

C) If an acid is a strong acid, will its conjugate base be a strong or weak base?

Answer 10

It will be a weak base.

Question 11

C) How do you calculate the pH of a strong acid?

Answer 11

Do negative log of the concentration of the acid (which is the same as the conc of the H+ ions)

Question 12

C) How do you calculate the pH of a strong base

Answer 12

You will be given Kw. Sub into the equation and solve for [H+] then do the negative log of this value.

Question 13

C) What does a large Ka value indicate?

Answer 13

A stronger weak acid.

Question 14

C) How do you calculate the pH of a weak acid?

Answer 14

Substitute into the equilibrium equation for Ka and solve for [H+] and then do the negative log of this value.

Question 15

C) Unless we wish to determine the pH to more than two decimal places, what assumptions must we make to simplify the calculation?

Answer 15

• We assume that [H+]=[A-] (an approximation becasue some water molecules will have dissociated to form H+ ions however this is only a very small number of water molecules so this does not affect the pH calculated to 2 decimal places).
• We also assume that [HA] is approx equal to the concentration of the acid, assuming that only a very small number of the HA molecules have dissociated (does not significantly affect 2dp pH value because as it is a weak acid, only a very small proportion of molecules will dissociate

Question 16

C) How do you calculate [H+] from the pH value?

Answer 16

10^-pH

Question 17

C) Describe the neutralisation curves of

• Strong acid against strong base
• Strong acid against weak base
• Weak acid against strong base
• Weak acid against weak base

Answer 17

• Starting at very high pH, large steep change in pH over a small range, ending at very low pH
• Starting at a high (ish) pH, less large steep change in pH over a small range, ending at a very low pH
• Starting at a very high pH, less large steep change in pH, ending at a low (ish) pH
• Starting at a high (ish) pH, small less steep pH change, ending at a low (ish) pH

Question 18

C) How does the body maintain a constant blood pH?

Answer 18

• pH of blood maintained at around a pH of 7.35 to 7.45
• Carbonic acid buffer- dissociates into H+ and HCO3- ions (equilibrium) or H2O + CO2 H+ + HCO3-
• Increase in H+ ions reacts with HCO3- and forms more H2CO3 to minimise the change in pH
• Increase in OH- ions reacts with the H+ ions and reforms the carbonic acid again, maintaining a constant pH

Question 19

C) How does a buffer solution work? Use the example of ethanoic acid and sodium ethanoate. Why must the buffer be based on a weak acid conjugate pair?

Answer 19

• An equilibrium is set up and an increase in H+ or OH- ions reacts with one side of the equation to form more of the other side, thereby minimising the change in pH.
• Ethanoic acid provides the CH3COOH (as few molecules dissociate as it is a weak acid) and sodium ethanoate provides CH3COO- ions (as we assume all of the molecules dissociate from the sodium in solution). This provides a conjugate acid base pair
• The equilibrium shifts when H+ or OH- ions are added to either reform more base to counteract the H+ ion increase or more acid to counteract the OH- ion increase.
• It must be a conjugate pair so that both the un-dissociated and dissociated forms of the weak acid are present so that an equilibrium can be set up.

Question 20

C) Why does pH of a buffer change with temperature? Why does pure water remain neutral even though the pH decreases at higher temperatures?

Answer 20

Le Chatelier’s principle- there is an equilibrium and depending on if the forward/ backward reaction is endothermic/exothermic the equilibrium will be shifted to minimise the change when the temperature increases/ decreases.

Pure water remains neutral because even though the equilibrium is shifted at higher temperatures to produce more H+ ions, the same number of OH- ions are formed (the concentration of both is equal) so as there is no excess of either, the solution remains neutral.

Question 21

I) How would you calculate the pH of a buffer solution?

Answer 21

[H+] = Ka x [acid]/[salt]

Question 22

C) Why is a buffer solution formed when
200cm^3 of 3.2moldm^-3 of HCOOH is mixed with
800cm^3 of 0.5moldm^-3 of NaOH

Answer 22

• HCOOH reacts with NaOH to form HCOO-/ HCOO-Na+

- Some weak acid/ HCOOH remains OR weak acid/HCOOH is in excess.

Question 23

(1 mark)
A student adds an excess of aqueous ethanoic acid to solid calcium carbonate and the resulting solution is able to act as a buffer.
Explain why the buffer solution has formed.

Answer 23

Solution contains CH3COOH AND CH3COO-