Acids and Bases

Question 1

What is an acid?

Answer 1

An acid is a proton (H⁺ ion) donor. They release hydrogen ions when they’re mixed with water; they’re always combined with H₂O to form hydroxonium ions, H₃O⁺ in water, though.

Question 2

What is a base?

Answer 2

A base is a proton acceptor; when they’re in solution, they ‘grab’ hydrogen ions from water molecules.

B_(aq) + H₂O_(l) > BH⁺_(aq) + OH^-_{(aq)<br></br> (where B is some random base)}

Question 3

What occurs in an acid-base equilibria? (when acids and bases react)

Answer 3

Proton transfer occurs; where an acid transfers a proton to a base.

Question 4

How do you calculate pH (concentration of hydrogen ions)?

(formula)

Answer 4

pH = - log₁₀ [H⁺]

(Where [H⁺] is the concentration of hydrogen ions in a solution measured in mol dm^-3)

If you know [H+] of a solution, you can calculate its pH.

E.g. A solution of HCl has a [H⁺] of 0.01 mol dm^-3. pH?
pH = - log₁₀ [H+] = - log₁₀ [0.01] = 2.

Question 5

How do you calculate hydrogen ion concentration from pH?

Answer 5

It is the inverse of the pH forumla:

[H⁺] = 10^-pH

E.g. A solution of H₂SO₄ has pH 1.52. What is [H⁺]?
[H⁺] = 10^-pH = 10^-1.52 =0.03 mol dm^-3 = 3 x 10^-2 mol dm^-3

Question 6

How do you calculate the pH of a solution of a strong acid from its concentration?

Answer 6

Strong acids dissociate (almost) completely in dilute aqueous solution (water); most H+ ions are released.
The reaction goes to completion.

Dissociation always produces a proton and a negative ion.

Monoprotic acids
Each mole of acid produces one more of hydrogen ions.
H+ concentration is the same as the acid concentration.

E.g., HCl: HCl(aq) > H⁺(aq) + Cl^-(aq)
So for 0.1 mol dm^-3 HCl, [H⁺] is 0.1 mol dm^-3.
∴ pH = -log₁₀ [H⁺] = -log₁₀ [0.01] = 1.0.

Diprotic acids
Each molecule of a strong diprotic (two hydrogens in acid molecule) acid releases 2 protons when it dissociates.

Thus diprotic acids produce two moles of hydrogen ions for each mole of acid.

E.g., H2SO4: H₂SO₄ > 2H⁺_(aq) + SO₄^2-_(aq)
So for 0.1 mol dm^-3 H₂SO₄, [H⁺] = 2 x 0.1 = 0.2 mol dm^-3.
∴ pH = -log₁₀ [H⁺] = -log₁₀ [0.2] = 0.70.

Question 7

How does water dissociate?

Answer 7

• Water dissociate very slightly; it is slightly ionised.
• There’s so much water compared to the amounts of H+ and OH- ions that the concentration of water is considered to have a constant value; thus a modified equilibrium constant is formed; K_w.

K_w = [H⁺] [OH^-]

Kw is the ionic product of water, and at 298K is equal to 1.0 x 10^-14 mol²dm^-6. Each H₂O that dissociates gives rise to one H⁺ and one OH^-, so in pure water at 298K:

[OH^-] = [H⁺]
∴ K_w = [H⁺]²
1.0 x 10^-14 = [H⁺]²
[H⁺] = 1.0 x 10^-7 moldm^-3 = [OH^-]

Question 8

How do you calculate the pH of a strong base from its concentration?

Answer 8

• Strong bases fully ionise in water; they donate one mole of OH^- ions per mole of base.
• Sodium hydroxide (NaOH) and potassium hydroxide (KOH) are examples of strong bases.
• ∴ concentration of OH^- ions = concentration of the base
• So for 0.02 mol dm^-3 NaOH solution, [OH-] is also 0.02 mol dm^-3.
• To work out pH, you use K_w (ionic product of water; K_w = [H⁺][OH^-]) substituting your value of [OH^-] and the given value of K_w, rearranging to get [H⁺].
• Then can work out pH.

So if you know [OH^-] for a strong aqueous base and K_w at a certain temperature, you can work out [H⁺] and then the pH.

• E.g.*
• *The value of K_w at 298K is 1.0 x 10^-14 mol³dm^-6. Find the pH of 0.1 moldm^-3 NaOH at 298K.**

1. [OH^-] = 0.1 mol dm^-3
2. [H⁺] = K_w/[OH^-] = 1.0 x 10^-14/0.1
   = 1.0 x 10^-13 mol dm^-3
3. pH = - log₁₀(1.0 x 10^-13)
   = 13.0

Question 9

How well do weak acids and bases dissociate in aqueous solution?

Construct an expression with units for the dissociation constant, K_a, of a weak acid.

Answer 9

• Weak acids and weak bases dissociate only slightly in aqueous solution.
• Examples of weak acids: methanoic acid, propanoic acid, ethanoic acid etc. (hydrocarbon first bit; carboxylic)
• ∴ [H⁺] isn’t the same as the acid concentration. You have to construct an acid dissociation constant, K_a, to find the pH.

The equilibrium of a weak aqueous acid, HA.

HA ⇄ H⁺_(aq) + A^-_(aq)

Applying the equilibrium law gives:

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

Can assume all H+ ions come from the acid,
so [H⁺] = [A^-].

∴

K_a = [H+]²
[HA]

• (^This general expression for weak acid does not apply to buffer solutions.)*
• E.g.*

Find the pH of a 0.02 mol dm^-3 solution of propanoic acid (CH₃CH₂COOH) at 298K. K_a for propanoic acid at this temperature is 1.30 x 10^-5 mol dm^-3.

1. Write expression for K_a for the weak acid, then rearrange to find [H⁺]².
   K_a = _ [H+]² _
   [CH₃CH₂COOH]

∴ [H⁺]² = K_a x [CH₃CH₂COOH]
= (1.30 x 10^-5) x 0.02
= 2.60 x 10^-7

2. [H⁺] = √2.60 x 10^-7
   = 5.10 x 10^-4 mol dm^-3
3. pH = -log₁₀ [H+] = -log₁₀ [5.10 x 10^-4]
   = **3.29 **

Question 10

What is pK_a?

Answer 10

• pK_a = – log₁₀ K_{<strong>a</strong>}
  (think of ‘p’ meaning ‘-log₁₀ of’)
• pK_a gives a measure of how strong a weak acid is; the smaller the value of pK_a, the stronger the acid.
• Calculated in exactly the same way as pH is calculated from [H⁺], and vice versa.
• e.g. K_a = 10^-pK_a

Question 11

What is the shape of a pH curve for a strong acid and strong base in an acid-base titration?
(base added from burette into acid)

What indicator would you select?

Answer 11

• Largest vertical equivalence point (‘perfect’ point at which sufficient base has been added to just neutralise the acid [or vice-versa])
• pH starts at around 1 as there’s an excess of strong acid.
• pH finishes up around pH 13, when you have an excess of strong base.

To choose an indicator, you must find one with the following criteria:

• Colour change must be sharp, not gradual, at the end point (volume of alkali/acid added when indicator just changes colour)
• Should give a distinct colour change
• Changes colour entirely on the vertical part of the pH curve; end point same as equivalence point.

For strong-acid, strong-base; both phenolphthalein and methyl orange change within the vertical, but phenolpthalein is usually preferred as its colour change is more easily seen.

Phenolpthalein: colourless > pink
pH 8.3 - 10
Methyl orange: red > yellow
pH 3.1 - 4.4

Question 12

What is the shape of a pH curve for a strong acid and weak base in an acid-base titration?

What indicator would you select?

(base added from burette into acid)

Answer 12

• Mid-sized vertical equivalence point section.
• pH starts around 1 due to an excess of strong acid
• pH finishes up around 9, when you have an excess of weak base.

To choose an indicator, you must find one with the following criteria:

• Colour change must be sharp, not gradual, at the end point (volume of alkali/acid added when indicator just changes colour)
• Should give a distinct colour change
• Changes colour entirely on the vertical part of the pH curve; end point same as equivalence point.

For strong-acid, weak-base, methyl orange will change sharply at the equivalence point but phenolphthalein would be of no use.

Phenolpthalein: colourless > pink
pH 8.3 - 10
Methyl orange: red > yellow
pH 3.1 - 4.4

Question 13

What is the shape of a pH curve for a weak acid and strong base in an acid-base titration?

What indicator would you select?

(base added from burette into acid)

Answer 13

• Mid-sized vertical equivalence point section.
• pH starts around 5 due to an excess of weak acid
• pH finishes up around 13, when you have an excess of strong base.

To choose an indicator, you must find one with the following criteria:

• Colour change must be sharp, not gradual, at the end point (volume of alkali/acid added when indicator just changes colour)
• Should give a distinct colour change
• Changes colour entirely on the vertical part of the pH curve; end point same as equivalence point.

For weak-acid, strong-base, methyl orange is not suitable as it does not change colour in the vertical portion of the curve and will change in the ‘wrong’ place, over the addition of too much BASSSSSSSS.
Phenolphthalein will change sharply at exactly 25cm³, the equivalence pont.

Phenolpthalein: colourless > pink
pH 8.3 - 10
Methyl orange: red > yellow
pH 3.1 - 4.4

Question 14

What is the shape of a pH curve for a weak acid and weak base in an acid-base titration?

What indicator would you select?

(base added from burette into acid)

Answer 14

• No vertical equivalence point section.
• pH starts around 5 due to an excess of weak acid
• pH finishes up around 9, when you have an excess of weak base.

Neither indicator is suitable; no indicator would be suitable as an indicator requires a vertical position of the curve over two pH units at the equivalence point to give a sharp change.

Generally isn’t done.

Question 15

Why is universal indicator shit for titrations?

Answer 15

Due to its gradual colour change; indicators should change colour over a pH range of around 2 units, centred around the value of pK_a for the indicator.

Question 16

Perform calculations for the titrations of
monoprotic acids with sodium hydroxide, based on experimental results.

Answer 16

Monoprotic

25 cm³ of 0.5 mol dm^-3 HCl was needed to neutralise 35cm³ of NaOH solution. Calculate the concentration of the sodium hydroxide solution.

• Write a balanced equation with what you know and what you need to know.

                HCl + NaOH → NaCl + H<sub>2</sub>O
           25 cm<sup>3  </sup>35 cm<sup>3</sup>
   0.5 mol dm<sup>-3     </sup>?

• 1 mole of HCl neutralises 1 mole of NaOH.
  Work out how many moles of HCl you have.

n(HCl) = c x v = 0.5 x (25 x 10^-3)
= 0.0125 moles.

• ∴ 0.0125 mol of HCl must neutralise 0.0125 mol of NaOH. Work out concentration.

c(NaOH) = n = 0.0125 = 0.36 mol dm ^-3.
v 35 x 10^-3

Question 17

Perform calculations for the titrations of
diprotic acids with sodium hydroxide, based on experimental results.

Answer 17

Diprotic

25 cm³ of ethanedioic acid, C₂H₂O₄, was completely neutralised by 20cm³ of 0.1 mol dm^-3 NaOH solution. Calculate the concentration of the ethanedioic acid solution.

• Write a balanced equation with what you know and what you need to know.

          C<sub>2</sub>H<sub>2</sub>O<sub>4</sub> + 2NaOH → Na<sub>2</sub>C<sub>2</sub>O<sub>4</sub> + 2H<sub>2</sub>O
            25 cm<sup>3</sup>   20 cm<sup>3</sup>
                ?        0.1 mol dm<sup>-3</sup>

• 2 moles of NaOH neutralises 1 mole of C₂H₂O₄.
  Work out how many moles of NaOH you have.

n(NaOH) = c x v = 0.1 x (20 x 10^-3)
= 0.002moles.

n(C₂H₂O₄) = 0.5 x n(NaOH)
= 0.5 x .002
= 0.001 moles of C₂H₂O₄

(It’s a diprotic acid, you need twice as many moles of base as moles of acid)

∴ 0.002 mol of NaOH must neutralise 0.001 mol of C₂H₂O₄. Work out concentration.
c(C₂H₂O₄) = n = 0.001 = 0.04 mol dm^-3.
v 25 x 10^-3

Question 18

What are buffers?

Answer 18

A buffer is a solution that resists changes in pH when small amounts of acid or alkali are added. Their pH remains almost constant.

Buffers are designed to keep the concentration of hydrogen and hydroxide ions almost unchanged. They are based on an equilibrium reaciton which will move in the direction ro remove either additional hydrogen ions or hydroxide ions, should they be added.

Question 19

How do acidic buffers work?

Answer 19

• Acidic buffers are made from weak acids; a buffer with a pH less than 7.
• They work because the dissociation of a weak acid is an equilibrium reaciton.
• Made by mixing a weak acid with one of its salts; often a sodium salt.

E.g.,
Ethanoic acid and sodium ethanoate.

CH₃COOH ⇄ H⁺ + CH₃COO^-

Ethanoic acid is a weak acid, thus the position of this equilibrium will be well to the left (only slightly dissociates; little product).
Adding sodium ethanoate to this adds lots of extra ethanoate ions (it fully dissociates into ions in water). This tips the position of the equilibrium even further to the left.

The solution will therefore contain these important things:

• lots of un-ionised ethanoic acid;
• lots of ethanoate ions from the sodium ethanoate;
• enough hydrogen ions to make the solution acidic.

Adding acid:

Buffer solution must remove most of the new H⁺ ions otherwise the pH would drop markedly; H⁺ ions combine with the ethanoate ions (CH₃COO-) to form ethanoic acid (CH₃COOH). Although the reaction is reversible, since the ethanoic acid is a weak acid, most of the new hydrogen ions are removed in this way.
Since most of the new H⁺ ions are removed, the pH won’t change very much - but because of the equilibria involved, it will fall a little bit.

Adding alkali:

OH^- ions react with ethanoic acid to produce ethanoate ions and water, removing the added OH^- to keep the pH fairly constant.

OH- ions also react with H⁺ ions (from dissociation of ethanoic acid) to form water. As soon as this happens, the equilibrium tips to replace the H⁺ ions. This keeps on happening until most of the OH^- ions are removed.

Not all of the hydroxide ions are removed - just most of them. The water formed re-ionises to a very small extent to give a few hydrogen ions and hydroxide ions.

Question 20

How do basic buffers work?

Answer 20

Basic buffers are made from a weak base and a salt of that base; a buffer with a pH more than 7.

E.g.,
Ammonia and ammonium chloride solution.

NH₃ + H₂O ⇄ NH4⁺ + OH^-

Ammonia is a weak base, thus the position of this equilibrium will be well to the left (only slightly dissociates; little product).
Adding ammonium chloride to this adds lots of extra ammonium ions (it fully dissociates into ions in solution). This tips the position of the equilibrium even further to the left.

The solution will therefore contain these important things:

• lots of unreacted ammonia;
• lots of ammonium ions from the ammonium chloride;
• enough hydroxide ions to make the solution alkaline.

Adding acid:

Buffer solution must remove most of the new H⁺ ions otherwise the pH would drop markedly; H⁺ ions react with OH^- ions (present from ammonia-water reaction) to form H₂O. As soon as this happens, the equilibrium tips to replace the OH^- ions. This keeps on happening until most of the H⁺ ions are removed.

H⁺ ions also collide with NH₃ molecules to form NH4⁺ (ammonium ions). Most, but not all, of the H⁺ ions will be removed. The ammonium ion is weakly acidic, and so some of the hydrogen ions will be released again.
(NH₃ + H⁺ ⇄ NH₄⁺)

Adding alkali:

If a small amount of base is added, the OH- ions from the alkali are removed by a simple reaction with ammonium ions, NH₄^+, forming NH₃ and H₂O.
(NH₄⁺ + OH^- ⇄ NH₃ + H₂O)

Because the ammonia formed is a weak base, it can react with the water - and so the reaction is slightly reversible. That means that, again, most (but not all) of the the hydroxide ions are removed from the solution.

Question 21

What are some applications of buffer solutions?

Answer 21

• Shampoo; most contain a pH 5.5 buffer, counteracting the alkaine soap in the shampoo.
• Biological washing powders; keep pH at the right level for the enzymes to work best.
• Blood; biological buffer system; buffered to pH 7.4; change as little as 0.5 may be fatal.

Question 22

How do you calculate the pH of acidic buffer solutions?

Answer 22

• Remember that an acid buffer can be made from a weak acid and one of its salts.
• Do not used simplified generic weak acid expression as [H+] ≠ [A^-]

E.g.
Calculate the pH of the buffer formed when 500 cm³ of 0.400 mol dm^-3 NaOH is added to 500 cm³ 1.00 mol dm^-3 HA. K_a = 6.25 x 10^-5.

Some of the weak acid is neutralised by the sodium hydroxide, leaving a solution containing A^- and HA, acting as the buffer.

                HA + NaOH → H<sub>2</sub>O + NaA  Intially: 0.500mol   0.200mol             0  Finally: 0.300mol   0 mol               0.200mol  (*0.200mol OH reacted with some HA; 0.5 mol - 0.2 mol gibing 0.300mol of HA, and the 0.200mol A<sup>-</sup> (NaA)*)

n(HA) = c x v = 1.00 x (500 x 10^-3)
= 0.500 mol.

n(NaOH) = n(OH^-) = c x v = 0.400 x (500 x 10^-3)
= 0.200 mol.

This leaves 1000cm³ of a solution containing 0.300 mol HA and 0.200 mol A^- since all the NaA (THE SALT) is dissociated to give A^- (hence the same molar).\

[HA] = 0.300 mol dm^-3
[A-] = 0.200 mol dm^-3

Ka = [H+][A-] ∴ 6.25 x 10^-5 = [H+][0.200]
[HA] [0.300]

[H+] = (6.25 x 10^-5)x [0.300] = 9.375 x 10^-5
[0.200]

pH = -log₁₀[9.375 x 10^-5] = 4.03

Question 23

Calculating the pH change when an acid or a base is added to a buffer.

Answer 23

You have 1.00 dm³ of buffer solution of ethanoic acid at concentration 0.10 mol dm^-3 and sodium ethanoate at concentration 0.10 mol dm^-3. Ka is 1.7 x 10^-5.
pH = 4.77.

We add 10.0 cm3 of HCl concentration 1.00 mol dm^-3 to the buffer. Virtually all added H+ ions will react with the ethanoate ions, [A-], to from molecules of ethanoic acid, [HA]. Thus amount of acid increases, salt decreasing:

Before: n(ethanoic acid) = 0.10 mol
n(sodium ethanouate) = 0.10 mol

After: n(acid) = 0.110mol (increased by 0.010 mol)
n(salt) = 0.090 mol (decreased by 0.010 mol)

∴ concentration of acid [HA] is now: 0.110
1 x 10³

∴ concentration of salt [A^-] is now: 0.090
1 x 10³

Ka = [H+][A-] ∴ 1.7 x 10^-5 = [H+][0.90/1000]
[HA] [0.110/1000]

[H+] = (1.7 x 10^-5) x [0.110/1000] = 2.08 x 10^-5
[0.90/1000]

pH = -log₁₀[2.08 x 10^-5] = 4.68.