acid-base equilibria & common ions

Acid-base equilibria & common ions Consider solution containing HF (weak acid) and salt

NaF Species in solution: HF, H2O, Na+, F-

What effect does presence of NaF have on dissociation equilibrium of HF?

HF(aq) H+(aq) + F-(aq)

Use Le Chatelier’s principle; extra F- (from NaF) shifts equilibrium to the …..

Because of this, the [H+] will ….. What happens to the pH?

This is the Common Ion Effect

Equilibrium calculations involving common ions 34.6 g of NH4Cl is added to 3.98 L of a 0.0145 M solution of

NH3. Kb(NH3) = 1.8 x 10-5.

What is the pH of the original solution before the addition of NH4Cl?

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

So – what’s the pH of the solution after the addition of the NH4Cl (assume that the volume stays constant)?

Effect of adding common ions to weak acids?

What is the general statement that we can make about what happens to the acidity of a weak acid upon addition of a common ion?

Buffered solutions

Buffer: A solution that resists a change in its pH when either H+ or OH- ions are added

Very important example; blood (cells can only survive in very narrow pH range, thus constant blood pH is vital.)

Buffers can contain weak acids & salts or weak bases & salts (solutions can be buffered at almost any pH)

Strong acids / bases cannot be used…

But….just how exactly do buffers work?

How does buffering work? Buffer contains relatively large concs. of weak acid HA and its

conjugate base A-. Add OH- ions into solution, what happens? (OH- is a strong base

– it will look for H+) Weak acid HA is best source of H+.

OH- + HA H2O + A-

OH- ions cannot accumulate (replaced by A- ions)

Original bufferedsolution pH

OH- added Final pH of bufferclose to original pH

added OH- ions replaced by A- ions

How does buffering work? If OH- ions converted to A- ions, how does the pH stay so

stable? Look at eq. expression for [H+]; dissociation equilibrium for HA

Eq. [H+] (and thus pH) determined by ratio [HA] / [A-]. If [HA] and [A-] are large relative to [OH-] added,

change in [HA] / [A-] will be very small. Thus, [H+] and pH changes will also be very small.

The essence of buffering: [HA] and [A-] are very large relative to [OH-] added.

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

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

The essence of buffering

Original [HA]/[A]ratio

OH- added Final [HA]/[A]ratio close to original

Added OH- changes HA to A-,but [HA] and [A-] are large compared to [OH-] added

Buffer solution pH calculation Buffered solution contains 0.50 M HC2H3O2 (Ka = 1.8x10-5) and

0.50 M NaC2H3O2. Calculate pH of solution.

HC2H3O2(aq) C2H3O2-(aq) + H+(aq)

pH changes in buffered solutions Calculate change in pH that occurs when 0.010 mol solid

NaOH added to 1.0 L of buffered solution from last question.Compare this with pH change that occurs when 0.010 mol solid NaOH added to 1.0 L of H2O.

Expected: pH of buffered solution will change very little; pH of water will change by much larger amount.

How to handle these problems

Original bufferedsolution pH Modified pH

Step 2: Do equilibriumcalculations

Step 1: Do stoichiometrycalculations; assume reactionwith H+/OH- goes to completion

Adding H+, rather than OH-

Exactly the same thinking applied when H+ added to buffered solution of weak acid and conjugate base salt. A- (conjugate base) has high affinity for added H+, and will form weak acid HA; H+ + A- HA

H+ ions cannot accumulate (replaced by HA) Net change of A- to HA, but if [A-] and [HA] are very large

compared to [H+], little pH change will occur (as expected, in buffered solution).

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

Buffer Capacity Amount of H+/OH- a buffer can absorb without a considerable change

in pH

Consider two 1.0 L ammonia/ammonium buffer solutions:

Buffer 1: 1.0 M NH3, and 1.0 M NH4+

Ka = [NH3][H+] / [NH4+]

[H+] = Ka. [NH4+] / [NH3] = 5.6 x 10-10 M; pH = ?

What happens to the pH if 0.1 moles of NaOH is added?

After the reaction with NaOH, calculate the [H+] and thus the pH….

Buffer Capacity Buffer 2: 0.01 M NH3, and 0.01 M NH4

+

Ka = [NH3][H+] / [NH4+]

[H+] = Ka. [NH4+] / [NH3] = 5.6 x 10-10 M; pH = ?

What happens to the pH if 0.1 moles of NaOH is added?

After the reaction with NaOH, calculate the [H+] and thus the pH….

Buffer Capacity So - what general statement can be made

about buffer capacity with respect to the initial concentration of the weak acid and its conjugate base in the buffer?

High concentrations:

Low concentrations

Henderson-Hasselbalch equation Useful equation; allows for the calculation of buffer pH if the

concentration of weak acid (HA) and conjugate base (A-)are known.

A more convenient method of pH calculation – will give the same answer as with our previous method of pH calculation.

Buffered solution contains 0.50 M HC2H3O2 (Ka = 1.8x10-5) and 0.50 M NaC2H3O2. Calculate pH of solution (we did this a few slides ago, but let’s check that H-H equation gives us the same answer).

___pH = pKa + log [A-]

[HA]Henderson-Hasselbalch equation

Titrations / pH curves Experimental method to determine the concentration of an acid

(or base).

Think about this: If you were given 25 mL of a unknown concentration of HCl, a solution of 0.100 M NaOH and a burette, how could you determine the concentration of the HCl?

What would happen to the pH of the solution of HCl as the NaOH was added to it?

Titrations / pH curves

Titrations How do you know when all the acid has been neutralized? pH paper pH meter

Or…… Indicator can be used

Methyl orange; yellow in basic solution and red in acidic solution

Many other indicators available

Indicators

Bromothymol blue (yellow in acidic solution), slight green tint as base is added, and blue form in basic solution

Titration Calculations

Solubility Equilibria Fluoride – used to combat tooth decay One product of F- reaction at site of teeth is CaF2(s)

CaF2(s) Ca2+(aq) + 2F-(aq) (dissolving in water)

Consider equlibrium set up between these species:

CaF2(s) Ca2+(aq) + 2F-(aq)

Equilibrium Expression for this process? Ksp = [Ca2+][F-]2 (Why is CaF2 not included in expression?)

Ksp = Solubility Product Constant (solubility product)

Solubility vs Ksp

Ksp: Equilibrium Constant (solubility product)

Solubility: Equilibrium Position

Copper (I) bromide has a measured aqueous solubility of 2.0 x 10-4 mol/L at 25 °C. Calculate its Ksp value.

Equilibrium reaction? Equilibrium Expression? Ksp =

(equilibrium concentrations)

Initial Concentrations?