12th Grade IB Chemistry

Buffers and Salts

IB Chemistry Power Points

Topic 18Acids and Bases

www.pedagogics.caBuffer SolutionsDEFINITION: A buffer solution contains a weak acid mixed with its conjugate base (or weak base and conjugate acid)Buffers resist changes in pH when a small amount of a strong acid or base is added to it.

HA H+ + A- 2If a small amount of a strong acid (H+) is added eqm shifts to the left as [H+] increases so system adjusts to increase [HA] and reduce [H+] again.

HA H+ + A- 3A small amount of a strong base will react with H+ to form H2O and eqm will shift to the right to increase [H+] again.

HA H+ + A-

4Making Buffer SolutionsAn example of a weak acid is ethanoic acid. This could be mixed with sodium ethanoate which will provide ethanoate ions (conjugate base).CH3COOH(aq) CH3COO-(aq)+H+(aq)

An example of a weak base is ammonia. This could be mixed with ammonium chloride to provide ammonium ions (conjugate acid).NH3(aq) + H2O(aq) NH4+(aq) +OH-(aq)In order for a buffer to work well the concentration of the acid/base and its salt must be much higher than the strong acid/base added.5Optimum BufferA buffer is most effective when the concentration of weak acid and its salt (the conjugate base) are equal and the pH is equal to pKa.

In practice it will work reasonably well with similar concentrations and the effective buffer range of any weak acid/ base is pKa 1.6Blood has buffering capacityBlood must maintain a pH of 7.4 so its enzymes can work. If 0.01 mol of H+ or OH- is added to blood it only changes pH by 0.1 unit.The eqm is:CO2(aq) + H2O(l) ) H+(aq) + HCO3-(aq)

7Buffer Calculation #1 Calculate the pH of a 1.00 dm3 buffer solution made by dissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3 ethanoic acid solution.Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations.

weak acid [CH3COOH] = 0.075 mol dm-3

conjugate base[CH3COO-] = 0.50 mol dm-3

Write the equilibrium equation and expression

CH3COOH H+ + CH3COO-

In this case, assume the equilibrium concentrations of the weak acid and the salt anion are the same as the given information (very little change when equilibrium is established.

pH = 5.6

8Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume.Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations.

weak base [NH3] = 0.100 mol dm-3

conjugate acid[NH4+] = ?

Write the equilibrium equation and expression

NH3+ H2O NH4+ + OH-

In this case, assume the equilibrium concentrations of the weak base is the same as the given information (very little change when equilibrium is established. Calculate [OH-] from desired pH

mass = 9.50 g

9Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution.Do the stoichiometry to determine what the concentrations are AFTER neutralization.

HCl + NH3 NH4+ + Cl-Write the equilibrium equation and expression

NH3+ H2O NH4+ + OH-

In this case, assume the equilibrium concentrations of the weak base and cation are the same as determined by stoichiometry (very little change when equilibrium is established.

pH= 9.6

10Salt HydrolysisA soluble salt is an ionic compound made of cations (ex. Na+) and anions (ex. Cl-) which completely dissociates into ions in aqueous solution.

Salts can affect the pH of a solution because the cations act as weak acids by bonding with OH- and the anions act as weak bases by accepting H+ ions.11Salt HydrolysisIf the cation comes from a strong base then it will have less acidic activity than one from a weak base. For example Na+ from NaOH is a weaker acid than NH4+ from NH3

If the anion comes from a strong acid then it will have less basic activity than one from a weak acid. For example Cl- from HCl is a weaker base than CH3COO- from CH3COOH.

Salts from a strong acid and strong base ex. NaCl will form a neutral solution.12Salt HydrolysisWould a solution of sodium ethanoate be acidic, basic or neutral?

Would a solution of ammonium chloride be acidic, basic or neutral?

Would a solution of potassium chloride be acidic, basic or neutral?13Salt HydrolysisWith salts of weak acids and weak bases the pH of the solution formed will reflect the relative strengths of the acid and base. Ex. Ammonium ethanoate is about neutral.

BE CAREFUL THOUGH . . .

Acidity of salts also depends on size and charge of the cation

14Salt HydrolysisSalts with small, highly charged cations are more acidic than large, low charge cations.

recall Period 3 chloride salts : NaCl, MgCl2 and AlCl3 which is most acidic

The aluminum ion and those of transition metals exist in water in hydrated form ie. [Al(H2O)6]3+ , [Fe(H2O)6]3+15Salt Hydrolysisthe e- attracting power of the ion weakens the O-H bond and stabilizes the resulting OH- ion.

As a result these ions are quite acidic in water.[Fe(H2O)6]3+ [Fe(OH)(H2O)5]2+(aq) + H+(aq)16Learning Check salt hydrolysisAcid Base Example Salt SolutionStrong Strong NaCl neutralWeak Weak ammonium ethanoate neutralStrong Weak ammonium chloride acidicWeak Strong sodium ethanoate basic17Buffers and Salts

IB Chemistry Power Points

Topic 18Acids and Bases

www.pedagogics.caAcid Base TitrationsIf a strong acid is added to a strong base gradually the pH will start off as 1.Once enough base is added that it is now in excess the pH will change very suddenly to about 13.The point at which this change is seen is when the amount of acid = amount of base.This is called the equivalence point.With this combination it occurs at pH 7 as the acid and base combine to make a neutral solution.19Strong Acid and Strong BaseMost indicators will work for this combination.

Weak Acid + Strong BaseA weak acid will have a pH of 3-5. When a strong base is added the pH will increase gradually as HA is converted to A-.This is called the buffering region as its acting like a buffer.When half the amount of base has been added it is called the half-neutralization point and [HA] = [A-] so [H+] = Ka x [HA]/[A-]Ka = [H+] and pKa = pH21Weak Acid + Strong BaseKa = [H+] and pKa = pHThis is the best way to determine the Ka for a weak acid.At the equivalence point the pH increases quickly to 13.The equivalence occurs when pH >7.Most suitable indicator is phenolphthalein

22Strong Acid + Weak BaseWhen a weak base is added to a strong acid the pH will remain around 1 until near the equivalence point when all the base has been converted into its conjugate acid.B(aq) + H+(aq) BH+(aq)At the equivalence point pH is