Acids, Bases and Buffers

Question 1

What are the the bronsted-lowry definition?

Answer 1

An acid is a proton donor.
A base is a proton acceptor

Question 2

Why is HCL a strong monobasic acid?

Answer 2

• Strong Acid – It fully dissociates into H+ and Cl− when dissolved in water.
• Monobasic – one mole of acid releases one mole of H+ ion

Question 3

What is an example of a dibasic and tribasic acid?

Answer 3

-H2SO4
-H3PO4

Question 4

What is an equation for the combination of H2O with an H+ ion?

Answer 4

H+ + H2O—> H3O+

Question 5

what type of bond forms when h+ and h2o combine?

Answer 5

dative covalent bond

Question 6

What are HCl and Cl called in this equation- HCl(aq) → H+ (aq) + Cl- (aq)?

Answer 6

HCl and Cl- are called a conjugate acid—base pair

Question 7

What is a conjugate acid-base pair

Answer 7

two species that can be interconverted by transfer of a proton

Question 8

In this equation- HCl(aq) → H+ (aq) + Cl- (aq), what happens?

Answer 8

• Here, HCl donates the proton and Cl- is the conjugate base.
• In the reverse reaction Cl- can accept a proton to form the conjugate acid, HCl.

Question 9

How would I explain this equation? CH3COOH (aq) + NH3(aq) CH3COO- (aq) + NH4+ (aq)
Acid 1 Base 2 Base 1 Acid 2

Answer 9

CH3COOH is an acid (1) as it donates H+; in the reverse reaction CH3COO- is the
conjugate base (1) as it accepts H+
• NH3 is a base (2) as it accepts H+; in the reverse reaction NH4+ is the conjugate
acid as it donates H+

Question 10

What does the reaction between magnesium carbonate with dilute nitric acid form?

Answer 10

MgCO3(s) + 2HNO3(aq) → Mg(NO3)2(aq) + CO2(g) +H2O(l)

Question 11

make sure to practice ionic equations

Question 12

What is the difference between strong and weak acids?

Answer 12

A strong acid completely dissociates into its ions in water
HCl(aq) → H+ (aq) + Cl- (aq)
Whereas a weak acid is only partially dissociated into its ions in water
CH3COOH(aq) CH3COO- (aq) + H+ (aq)
Organic acids are weak acids, not all inorganic acids are strong.

Question 13

Ethanoic acid reacts with an excess of magnesium carbonate. Write a
balanced equation for the reaction including state symbols.

Answer 13

2CH3COOH(aq) + MgCO3(s)  (CH3COO)2Mg(aq) + CO2(g) + H2O (l)

Question 14

What ion is common to all acidic solutions?

Answer 14

H+

Question 15

50.0 cm3 of 1.00 moldm-3 HCl reacts with excess powdered MgCO3. The volume
of CO2 produced is 0.600 dm3
50.0 cm3 of 1.00 moldm-3 CH3COOH reacts with excess powdered MgCO3.
The CO2 gas is given off at a lower rate.
Use your knowledge of theory of reaction rates and dynamic equilibrium to explain why CH3COOH(aq) reacts with MgCO3 at a slower rate than HCl
(aq)

Answer 15

HCl H+ + Cl−
• HCl is a strong acid and is fully dissociated.
• Eqm lies very far to the RHS and [H+] is high.
• CH3COOH H+ + CH3COO−
• CH3COOH is a weak acid and is partially dissociated.
• Eqm lies to the LHS and [H+] is low.
• [H+] is lower, less frequent collisions, less frequent successful
collisions per unit time.

Question 16

50.0 cm3 of 1.00 moldm-3 HCl reacts with excess powdered MgCO3. The volume of CO2 produced is 0.600 dm3
50.0 cm3 of 1.00 moldm-3 CH3COOH reacts with excess powdered MgCO3.
The CO2 gas is given off at a lower rate.
State and explain if the volume of CO2 produced in the second experiment
would be more, less or the same?

Answer 16

• Volume of CO2 will be the same.
• H+ ions react with MgCO3, eqm shifts to the RHS to increase [H+].
• More CH3COOH will dissociate until all of the acid has reacted.
• Moles of each acid are the same (volumes and concentrations
are the same)

Question 17

What is Ka?

Answer 17

Is a measure of the extent of dissociation of an acid.
A stronger acid will have larger Ka values but weak acids have relatively much
smaller.

Question 18

What is the equation for Ka (weak acid)?

Answer 18

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

Question 19

What is the equation of pKa?

Answer 19

pKa = - logKa
The inverse operation is also useful Ka= 10-pKa

Question 20

What does a lower Ka mean?

Answer 20

weaker the acid

Question 21

What does a lower pKa mean?

Answer 21

stronger the acid

Question 22

What is the pH definition?

Answer 22

pH = −log [H+]
used to find ph of strong acid

Question 23

For a strong monobasic acid (fully dissociated) what is will the [H+(aq)] be the same as?

Answer 23

the concentration of the acid.

Question 24

What are pH values normally quoted to?

Answer 24

2 dp

Question 25

What is pH best determined by?

Answer 25

using a pH meter rather than indicator paper

Question 26

How does we find the pH of a string acid when dilution has occurred?

Answer 26

We can use a dilution factor or amounts of substance to calculate a new concentration and then work out the new p

Question 27

A student added an equal volume of water to 25.0 cm3 of 1.00 mol dm-3 HCl. Calculate the pH of the new solution.

Answer 27

• Dilution factor = 2 (the total volume would now be 50cm3) • The dilute acid has a concentration of 1.00/2= 0.500 (it is 2 times less concentrated) • pH = -log 0.500 = 0.30

Question 28

How do you work out the pH upon reacting a strong acid with a strong base?

Answer 28

Full equation Ionic equation n(h+) n(base) n(h+) remaining new vol H+ (n/vol) ph

Question 29

What is HA?

Answer 29

weak acid

Question 30

for a weak acid not fully dissociated what can’t we say?

Answer 30

the eqm conc of protons is the same as that of the acid

Question 31

What is [H+] equal to?

Answer 31

[A-]

Question 32

What is the re written expression for Ka?

Answer 32

Ka = [H+]2/ [HA]

Question 33

How is [H+] found?

Answer 33

√ (Ka x [HA])

Question 34

Write the Ka expression for butanoic acid (CH3CH2CH2COOH). Find the pH of a 1.00 mol dm-3 solution, Ka = 1.51 x 10-5 mol dm-3

Answer 34

1.00 mol dm-3 solution, Ka = 1.51 x 10-5 mol dm-3. Ka = [CH3CH2CH2COO-][H+] [CH3CH2CH2COOH] If we assume that Ka = [H+]2 then [HA] [H+] = √ (Ka x [HA]) = √ (1.51 x 10-5 x 1.00) = 3.86 x 10-3 mol dm-3 pH = -log [H+] = -log(3.86 x 10-3) = 2.41

Question 35

What is the acid dissociation constant?

Answer 35

pKa

Question 36

How to work out % dissociation?

Answer 36

h+/ha x 100

Question 37

HA is a weak carboxylic acid found in the body. The Ka value of HA is 1.38x10-5 mol dm-3 A student analyses a sample of HA using the procedure outlined. • Dissolves 0.0113g of HA in water and makes the solution up to 0.500 dm3 • The student measures the pH of the resulting solution as 4.23 Determine the molar mass of HA, molecular formula and suggest a possible structure for HA HA has one carboxylic group and one OH group for an alcohol and contains C, H and O only. Show all your working

Answer 37

[H+] = 10-4.23 = 5.8884 x 10-5 mol dm-3 Ka = [H+]2/[HA] HA= (5.8884 x 10-5)2/1.38 x 10-5 = 2.51258 x 10-4 mol dm-3 n = cv = 2.51258 x 10-4 x 0.5 = 1.25629 x 10-4 moles n = m/ mr Mr = 0.0113/1.25629 x 10-4 =89.95 ≈ 90 g mol–1 COOH = 45 g mol–1 OH = 17 g mol–1 Total: CH2O3 = 62 g mol–1 Left: C2H4 = 28 g mol–1 Mol. Formula C3H6O3 = 90 g mol–1

Question 38

what is the dissociation of water equation

Answer 38

H2O <=>H+ + OH- ΔH= +v

Question 39

Explain why ΔH is positive?

Answer 39

It is an endothermic process as energy is being absorbed in order to break bonds.

Question 40

At room temp was is H+ and ph?

Answer 40

[H+] =1.00 x 10−7 mol dm-3, so the pH is 7.00.

Question 41

At other temps why is H+ diff?

Answer 41

because the equilibrium will have shifted, however, water can still be neutral even if the pH is not 7.00. • It is neutral because [OH–] = [H+]

Question 42

What is the dissociation equation for kw?

Answer 42

Kw = [H+][OH-]

Question 43

At room temperature what is H+ and OH?

Answer 43

[H+] = [OH-] = 1.00 x 10-7 mol dm-3

Question 44

What is the value and units of Kw?

Answer 44

1.00 x 10-14 mol2 dm-6

Question 45

For PH values that are whole numbers l why can you work out the H+ OH- concentrations?

Answer 45

as the indices for H+ OH- add up to -14

Question 46

When is water neutral?

Answer 46

when [H+] = [OH]

Question 47

How do we find the pH of a strong base?

Answer 47

[H+] = Kw/[OH-]

Question 48

Finding Ph upon reaction of a strong acid with a strong base- Worked During a titration a student added 10.0 cm3 of 0.100 mol dm-3 HCl, from a burette to 25.0 cm3 of 0.100 mol dm-3 NaOH in a conical flask. Determine the pH of the solution formed if the experiment was carried out at 298K.

Answer 48

Full equation: NaOH + HCl—>NaCl + H2O Ionic Equation: OH– + H+ —> H2O n(HCl ) = 10.0 x 0.100 = 1.00 x 10-3 mol 1000 n(H+) = 1.00 x 10-3 mol n(NaOH) = 25.0 x 0.100 = 2.50 x 10-3 mol 1000 n(OH-) = 2.50 x 10-3 mol n(OH-) remaining = 2.50 x 10-3 - 1.00 x 10-3 = 1.50 x 10-3 (alkali in excess) New volume = 10.0 + 25.0 = 35.0 cm3 [OH-] = 0.0015/ 35.0 x 1000 = 0.04286 mol dm-3 [H+] = Kw = 1.00 x 10-14/0.04286 = 2.33 x 10-13 mol dm-3 [OH-] pH = -log10[H+] = -log10[2.33 x 10-13] =12.63

Question 49

what is a buffer solution?

Answer 49

one that minimises changes in pH when small amounts of acid or alkali are added

Question 50

In what systems is it important to keep the pH constant in?

Answer 50

• Shampoo safety • Effectiveness of blood in transporting oxygen • Stability of molecules in living systems • Effectiveness of chromatography in separating mixtures

Question 51

what do buffers contain?

Answer 51

-two species that “remove” the added small amounts of acid or alkali added. -Species 1 is a weak acid (HA) and species 2 is its conjugate base(A-)

Question 52

What does a buffer solution do minimise the changes in pH?

Answer 52

Here we will consider a mixture which contains ethanoic acid and sodium ethanoate. The acid partially dissociates because it is weak. CH3COOH <=> CH3COO– + H+ • The salt is a source of the conjugate base which fully dissociates so we get a lot of ethanoate ions. • The result is that we have large amounts of both ethanoate ions (conjugate base) and of ethanoic acid in the mixture. • ethanoate ions and of ethanoic acid can counteract the addition of acid or alkali. If we add acid, the H+ ions react with the ethanoate ions to make ethanoic acid Equation 2: CH3COO– + H+ <=> CH3COOH Equilibrium in Equation 1 to shifts to the left, removing most of the H+ ions. If we add OH- ions, they react with H+ ions Equation 3: OH– + H+  H2O Equilibrium in Equation 1 to shift to the right. (Or we could say they react with CH3COOH to form the conjugate base and water). CH3COOH + OH– CH3COO– + H2O net effect is that the pH only change very slightly because we have a large reservoir of ethanoate ions and ethanoic acid to ‘absorb’ any acid or alkali added. There is a limit to how much acid or alkali you can add.

Question 53

What is the summary for the general action of a buffer?

Answer 53

-Write this equation for the acid concerned: HA H+ + A- [1] Addition of an acid Equilibrium will shift to the left The added acid reacts with A- Or you can write this as an equation H+ + A- → HA Addition of an alkali Equilibrium will shift to the right The added alkali reacts with H+ (or the named acid) Or you can write an equation H+ + OH- → H2O

Question 54

what are the methods used to make a buffer?

Answer 54

-mixing a weak acid with its salt (CH3COOH (acid)+ CH3COONa (salt) -partial neutralisation of a weak acid- СН3СООН + NaOH (SB)→ CH3COONa (Salt) + H2O WA in excess n(salt) formed = n(SB) n(WA) left = n(WA) - n(SB)

Question 55

How is a buffer made in the second method?

Answer 55

buffer is made by adding excess weak acid to a solution of an alkali like sodium hydroxide. In this way you end up with the same mixture as in a) a solution containing the weak acid and the salt of the weak acid.

Question 56

What is the Ka expression used to calculate the pH of a buffer?

Answer 56

[H+ ]=Ka x [HA]/[A-] (

Question 57

In a buffer solution why does [H+] not equal [A-]?

Answer 57

because the [A- ] has been added separately as one of the components.

Question 58

What does it mean If the buffer contains a 1:1 ratio of [HA]:[A−]?

Answer 58

the pH = pKa, because [H+] = Ka

Question 59

Calculating pH of a Buffer Solution: Mixture of a Weak Acid and its Salt

Answer 59

H+] = Ka x [HA]/[A-] = 1.70 x 10-5 x 0.500/0.100= 8.50 x 10-5 mol dm-3 pH = -log[H+] = -log(8.50 x 10-5) = 4.07

Question 60

What is anything ending in ate?

Answer 60

salt

Question 61

What happens at half neutralisation?

Answer 61

n(WA)=n(salt) Ka=[H+] pKa=pH

Question 62

what would happen to the pH of a solution Upon addition of potassium butanoate?

Answer 62

[H+] = Ka x [HA]/[A] [A-] increases [HA]/[A] decreases [H+] decreases pH increases

Question 63

What salt would needed to be added to phenol in order to make a buffer solution?

Answer 63

phenol with o- instead of oh

Question 64

What is the pH of a buffer solution if 25.0 cm3 of 0.500 moldm-3 sodium hydroxide is mixed with 75.0 cm3 of 0.500 mol dm-3 propanoic acid at 298K. The Ka value for propanoic acid at this temperature is 1.30 x 10-5 mol dm-3 -Making a Buffer: Calculate the concentration of propanoic acid and propanoateions present in the buffer solution.

Answer 64

-Initial n(CH3CH2COOH) = 75.0 X 0.500/1000 = 0.0375 mol Initial n(NaOH) = 25.0 X 0.500/1000 = 0.0125 mol CH3CH2COOH + NaOH → CH3CH2COONa + H2O initial moles 0.0375 0.0125 0 0 change in moles -0.0125 -0.0125 +0.0125 +0.0125 final moles 0.0250 0 0.0125 0.0125 Volume terms within [HA] and [A−] cancel out mathematically, n(HA) and n(A−) can be used to calculate [H+] Hence calculate the pH of the buffer solution: [H+] = Ka x [HA]/[A-] = 1.30 x 10-5 x 0.0250/0.0125 = 2.60 x 10-5 mol dm-3 pH = -log[H+] = -log(2.60 x 10-5) = 4.59

Question 65

-Calculating pH of a Buffer Solution: After Addition of a Base to a Buffer Solution A buffer solution contains 0.200 mol of propanoic acid and 0.180 mol of sodium propanoate in 1000 cm3 of solution. A 0.0150 mol sample of solid sodium hydroxide is then added to this buffer solution. Ka for propanoic acid = 1.30 x 10-5 mol dm-3 Calculate the pH after addition of the base.

Answer 65

Calculate the number of moles of propanoic acid and of propanoate ions present in the buffer solution after the addition of sodium hydroxide. CH3CH2COOH + NaOH → CH3CH2COONa + H2O initial moles 0.200 0.0150 0.180 change in moles -0.0150 -0.0150 +0.0150 final moles 0.185 0 0.195 Hence calculate the pH of the buffer solution after the addition of the sodium hydroxide. [H+] = Ka x [HA]/[A] [A-] increases [HA]/[A] decreases [H+] decreases pH increases

Question 66

What does moles left equal using a buffer adding strong base?

Answer 66

n(WA) left- initial n(WA)-n(SB) n(salt) left- initial n(salt) + n(SB)

Question 67

What does moles equal using a buffer adding strong acid?

Answer 67

n(WA) left- initial n(WA)+n(SB) n(salt) left- initial n(salt)-n(SB)

Question 68

1. A student adds 50.0 cm3 of 0.500 mol dm-3 butanoic acid to 25.0 cm3 of 0.250 mol dm-3 sodium hydroxide. A buffer solution forms. (a) Explain why a buffer solution forms.

Answer 68

Excess weak acid has been partially neutralised to form the conjugate salt

Question 69

A skin-care product has a buffered pH of 4.50. The buffer contains a solution of benzenecarboxylic acid (pKa 4.19, at 25oC) and its salt sodium benzenecarboxylate. Calculate the proportions of benzenecarboxylic acid (HA) and sodium benzenecarboxylate (A-) in the skin-care product. This is easiest to calculate as the fraction [A-]/[HA]

Answer 69

H+] = 10-pH therefore [H+] = 10-4.50 hence [H+] = 3.16x10-5 Ka = 10-pKa therefore [Ka] = 10-4.19 hence [Ka] = 6.46x10-5 Ka = [H+ ] [A- ]/[HA] Therefore [A- ]/[HA] = Ka/[H+] Hence [A-]/[HA] = 6.46x10-5 /3.16x10-5 = 2.04 It could also be expressed as the inverse of this ratio [HA]/[A-] = 0.490

Question 70

ALWAYS REMEMBER CHARGES SWAP

Question 71

What is H2CO3 an example of?

Answer 71

natural buffering ion which is responsible for the maintenance of blood pH and also the pH of lakes and streams. It is able to buffer because of this equation: H2CO3 (aq) H+ (aq) + HCO3- (aq)

Question 72

What happens if we add acid (H+) to H2CO3?

Answer 72

the position of the equilibrium shifts to the left Carbonic acid is formed (which can decompose to make carbon dioxide and water).

Question 73

What happens if we add an alkali (OH-) to H2CO3?

Answer 73

OH- (aq) this reacts with H+ (aq) and more H2CO3 dissociates, the equilibrium shifts to the right to restore the H+ (aq) ions.

Question 74

Healthy blood at a pH of 7.40 has a hydrogencarbonate: carbonic acid ratio of 10.5 : 1. A patient is admitted to hospital. The patient's blood pH is measured as 7.20. Calculate the hydrogencarbonate: carbonic acid ratio in the patient's blood.

Answer 74

Ka Method Ka = [H*][A]/[HA]| [A-1/[HA] = Ka / [H*] Healthy Blood [H+] = 10-7.40 = 3.9 x 10-8 mol dm-3 [A-]/[HA] = 10.5 / 1 Ка = (3.9 x 10-8 x 10.5)/1 = 4.18 × 10-7 mol dm-3 Patient's Blood [A-1/[HA] = 4.18 × 10-7/ 10-7.20 = 6.625 = 6.63: 1

Question 75

What is the enthalpy change of neutralisation?

Answer 75

the energy change when 1 mol of water is created by the neutralisation of an acid with an alkali under standard conditions.

Question 76

What is the end point? (titration curves)

Answer 76

minimum volume of reagent required for the indicator to change colour. The reagent in the flask has been neutralised completel

Question 77

What is the equivalence point?

Answer 77

the pH at the centre of the vertical section which occurs at equimolar portions of acid and base (at the end point)

Question 78

The pH at the equivalence point as the result of a “competition” between the acid and the base:

Answer 78

• strong acid vs strong base = 7.00 • weak acid vs weak base = 7.00 • strong acid vs weak base < 7.00 • weak acid vs strong base > 7.00

Question 79

What is an acid base indicator?

Answer 79

weak acid (HA). When in solution, the weak acid has a distinctly different colour to the conjugate base (A−)

Question 80

What will the indicator appear?

Answer 80

a different colour depending on if it is added to an acidic solution or a basic solution.

Question 81

What happens when you add H+ to an indicator?

Answer 81

Eqm shifts LHS to use additional H+ ions. Colour changes towards colour on LHS

Question 82

What happens when you add OH- to an indicator?

Answer 82

Eqm shifts RHS to use replace H+ ions consumed by OH– Colour changes towards colour on RHS