CO2(aq) + H2O(l) H+(aq) + HCO32-(aq)(It is not obvious how this is an example of equilibria involving complexes beyond the use of the "(aq)" in the chemical equations!
H+(aq) + CaCO3(s) Ca2+(aq) + HCO32-(aq)
NaF(s) Na+(aq) + F-(aq)The HF equilibrium is shifted according to the le Ch‚telier Principle towards the undissociated acid side:
HF(aq) H+(aq) + F-(aq)Another example, this time involving a base, is ammonium chloride added to "ammonium hydroxide" solution:
NH4Cl(s) NH4+(aq) + Cl-(aq)The effect is called the common ion effect and illustrates how an equilibrium does not discriminate between species of different origins. This suggests how to handle the calculations:
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
The equilibrium concentration of H+ in 1.0 M HF is 2.7x10-2 M and the percent dissociation of HF is 2.7%. What are the new values if the solution is also 1.0 M in NaF? Ka = 7.2x10-4.
Major solution species are: HF, F-, Na+ and H2O.
The only active equilibrium equation is:
HF(aq) H+(aq) + F-(aq)and:
Ka = [H+][F-]/[HF] = 7.2x10-4Starting concentrations:
[H+] = 0At equilibrium, the concentrations are:
[F-] = 1.0 M from NaF which is completely dissociated
[HF] = 1.0 M
[H+] = APlug them into the expression for Ka and solve for A making the usual approximation that A is very small compared to 1.0:
[F-] = 1.0 + A M from NaF which is completely dissociated
[HF] = 1.0 - A M
7.2x10-4 = (A)(1.0 + A)/(1.0 - A) ~ AThe percent dissociation = [H+]/[HF]o = (7.2x10-4/1.0)x100% = 0.072%
(Exercise 15.2) A buffered solution is made up to contain 0.50 M acetic acid (Ka = 1.8x10-5) and 0.50 M sodium acetate. What is its pH?
Answer:i.e. a somewhat acid solution. The interesting thing is to see what happens if, for example (Exercise 15.3) 0.01 mol of solid NaOH is added to a litre of this solution:
This calculation is exactly analogous to the preceding calculation:
[HC2H3O2] = 0.50 MEquilibrium:
[C2H3O2-] = 0.50 M
[H+] = 0
[HC2H3O2] = 0.50 - A MKa=1.8x10-5 = [H+][C2H3O2-]/[HC2H3O2] = (A)(0.5 + A)/(0.50 - A)
[C2H3O2-] = 0.50 + A M
[H+] = A M
so A = 1.8x10-5
and pH = 4.74
Answer:If 0.01 mol of NaOH had been added to a solution which did not contain the buffer, the pH would have gone to 12.0 (from 7.00); a change of 5.0 pH units!
Hypothetically, before any reaction occurs, the major solution species are:
HC2H3O2, Na+, C2H3O2-, OH- and H2OThe hydroxide ion present is a strong base and will react with the best source of protons (strongest acid) present, ie the acetic acid. This reaction can be treated as being essentially stoichiometric:
HC2H3O2(aq) + H+(aq) C2H3O2-(aq) + H2O(l)The concentration of acetic acid would go from 0.50 M (0.50 mol in 1.0 L) to 0.50-0.01 = 0.49 M, and the concentration of acetate will go from 0.50 to 0.50 + 0.01 = 0.51 M
This will alter the equilibrium problem:
[HC2H3O2] = 0.49 MEquilibrium:
[C2H3O2-] = 0.51 M
[H+] = 0
[HC2H3O2] = 0.49 - A MKa=1.8x10-5 = [H+][C2H3O2-]/[HC2H3O2] = (A)(0.51 + A)/(0.49 - A)
[C2H3O2-] = 0.51 + A M
[H+] = A M
so A = 1.7x10-5
and pH = 4.76
The net change was 0.02 pH units
Ka = [H+][A-]/[HA]which, when rearranged, gives:
[H+] = Ka[HA]/[A-]The pH is determined by the ratio of [HA] to [A-]. By making sure that these concentrations are large compared to the amount of acid or base which might be added to the solution, the pH will be prevented from changing very much.
Taking logs of both sides of the above equation and changing the signs gives:
-log([H+]) = -log(Ka) - log([HA]/[A-])
pH = pKa - log([HA]/[A-]) = pKa + log([base]/[acid])This is the Henderson - Hasselbalch Equation. It is valid to use the concentrations you calculate from the solution ingredients without worrying about the amount of HA that dissociates provided that this quantity is small - the usual situation:
Calculate the pH of a solution of 0.75 M lactic acid (HC3H5O3, Ka = 1.4x10-4) and 0.25 M sodium lactate. The pH will be controlled by the lactic acid/lactate equilibrium since H2O is a much weaker acid.Buffers can also be made from weak bases and their conjugate acid:
pH = pKa + log([A-]/[HA])pH = 3.85 + log(0.25/0.75) = 3.38
What is the pH of a solution containing 0.25 M NH3 (Kb = 1.8x10-5) and 0.40 M NH4Cl?(Exercise 15.6)
- Use equation:
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)Go through the usual calculation to get the concentration of OH- (1.1x10-5, hence the pOH (4.95) and hence the pH (9.05).
- Use the equation:
NH4+(aq) H+(aq) + NH3(aq)Remember that:Ka = Kw/KbCalculate Ka (5.6x10-10), and then use the Henderson-Hasselbalch equation
pH = 9.25 + log(0.25/0.40) = 9.05
Now add 0.10 mol of gaseous HCl to 1 L of the above ammonia/ammonium buffer solution. What is the new pH? Answer:
List the major species before equilibration:
NH3, NH4+, Cl-, H+, H2ORealize the ammonia will react with the H+ which came from the HCl, and assume a quantitative reaction:
NH3(aq)+H + (aq) NH4+(aq)Start:
[NH3] = 0.25 M[H+] = 0.10
[NH4+] = 0.40 M
[NH3] = 0.25 - 0.10 = 0.15 MFinally use the Henderson - Hasselbalch equation:
[NH4+] = 0.40 + 0.10 = 0.50 M
[H+] = 0
pH = pKa+log(0.15)/(0.50) = 8.73
This is just a little down from 9.05. (Note that 0.1 mol of the acid was buffered as apposed to 0.01 M of the base in the sodium acetate example so the change was bigger.
|Comparison of Buffer Capacities - Balanced Concentrations|
|Solution||[C2H3O2-] (M)||[HC2H3O2] (M)||Starting pH||pH with 0.01 M HCl|
The second scenario is adapted from the text where the weak acid to salt ratio is varied:
|Comparison of Buffer Capacities - Unbalanced Concentrations|
|Starting pH||pH with 0.01
The ideal buffer contains a roughly equal concentration of the weak acid or base and its salt, and a sufficient concentration of both to maximize the stabilization of the pH. This means choosing a buffer naturally close to the desired value. In the example (Exercise 15.8) the target pH is 4.30 and suggested buffers are listed below:
|Pick a Buffer to Control the pH near 4.3|
|Titration of a Strong Acid (50 ml of 0.200 M HNO3) with a Strong Base (0.100 M NaOH)|
Notice on the graph that:
|Titration of a Weak Acid (50.0 mL of 0.100 Acetic) with a Strong Base (0.100 M NaOH)|
Notice on the graph that:
Figure 15.4 shows curves for a series of hypothetical acids of differing pKa values. Notice that:
2.00 mmol of a weak monoprotic acid are dissolved in 100 mL of water and the solution titrated with 0.050 M NaOH solution. After 20.0 mL of the NaOH have been added the pH is 6.00. What is the pKa of the acid?
Deal first with the stoichiometric problem:HA(aq) + OH-(aq) A-(aq) + H2O(l)Before titration:
HA = 2.00 mmolAfter titration:
OH- = 20.0 mL x 0.0500 M = 1.00 mmol
A- = 0
HA = 2.00 - 1.00 mmol
OH- = 1.00 - 1.00 = 0
A- = 1.00
Now handle the equilibrium problem:
Major species: HA, A-, Na+, H2O
HA(aq) H+(aq) + A-(aq)Start concentrations:
[HA] = 1.00 mmol / (100.0 + 20.0) mL = 8.33x10-3 MEquilibrium concentrations:
[A-] = 1.00 mmol / (100.0 + 20.0) mL = 8.33x10-3 M
[H+] = 0
[HA] = 1.00 mmol / (100.0 + 20.0) mL = 8.33x10-3 - A MNow we know [H+] = antilog(-pH) = antilog(-6.00) = 1.0x10-6 M.
[A-] = 1.00 mmol / (100.0 + 20.0) mL = 8.33x10-3 + A M
[H+] = A
Ka = [H+][A-]/[HA]=1.0x10-6
Actually we should have noticed that the NaOH added contained 1.00 mmol of OH- whereas the original solution contained 2.00 mmol of acid therefore we were at the halfway point of the titration where [HA] = [A-]. At this point the pH = Ka!
Titrations of Weak Bases with Strong AcidsThis is similar to the weak acid/strong base case. Sample data are given below for the titration of 100 mL of 0.050 M NH3 solution with 0.10 M HCl:
Titration of a Weak Base (100 mL of 0.050 M NH3) with a Strong Acid (0.10 M HCl) NaOH added 0 10.0 25.0 50.0 60.0 pH 10.6 9.85 9.25 5.36 2.21
These data are plotted in Figure 15.5. Notice:
- The buffer region around pH = 9.25 at 25 mL acid added.
- The equivalence point at 50.0 mL.
- The final pH is determined by the excess H+.
Acid-Base IndicatorsWhile a pH meter can always be used to plot the course of a titration, it is possible to determine the equivalence point by picking a suitable indicator whose colour changes at the pH that would occur during the steep part of the curve.
Indicators are weak acids (or bases) whose conjugate bases (or acids) are different colours. Such an indicator is phenolphthalein (Figure 15.6) which has a pKa of 8.00. Remember that this is the pH at which half the phenolphthalein will be in the protonated (acid) form and the rest present as the conjugate base. The acid form of phenolphthalein is colorless and the basic form is pink.
Actually, our eyes can detect a colour change when about 10% of indicator changes form: the book example is bromothymol blue which is yellow in its acid form and blue when basic. In titrating an acid, the first tinge of greenishness can be seen when 10% of the basic form is present. Using the Henderson-Hasselbalch equation, the book shows that this corresponds to a point 1 pH unit to the acid side of its pKa. Similarly, going in the opposite direction, the first tinge of yellow in the blue basic form would be seen 1 pH unit to the basic side of the pKa value. Since pKa = 7 for bromthymol blue, its colour change will occur from pH = 6 to pH = 8. (Usually, the colour change is more easily seen going in one direction, for example it is much easier to detect the first tinge of pink with phenolphthalein going from acid to base than trying to detect the beginning of the disappearance of the pink colour going the other way. Similarly the yellow to blue direction is better for bromthymol blue.
For strong acid/ strong base titrations, there is a long vertical part to the curve around the equivalence point and many indicators will change in this range (Figure 15.8). For a titration of 100.0 mL of 0.1 M HCl with 0.1 M NaOH, the pH changes from 5.3 to 8.7 within 0.01 mL of the equivalence point. One drop is about 0.04 mL.
For weak acid or weak base titrations, the indicator must be much more carefully chosen. Figures 15.9 and 15.10 illustrate this.
Note: The method used to calculate the pH and generate the curves is the same as that in the text, though the full the quadratic solution is always used, that is, the "usual approximation" is not applied. However, the contribution to [H+] from the auto-ionization of water is ignored (as it is in the text). For very dilute solutions and/or very weak acids or bases, this leads to a discontinuity in the curve near the equivalence point which would not appear in a real experimental plot.
- Select the acid, base, their concentrations, which of them is in the burette, and choose an indicator.
Important. After altering the concentrations, press return or the altered values will not be registered.
- Click on the burette stopcock to start or stop the titration.
- Use the scroll bar to adjust the dropping rate
- Click in the graph area to clear it.
- Click on the burette column to set the starting level.
CaF2(s) Ca2+(aq) + 2F-(aq)It must be understood that equilibrium is a dynamic situation with solid going into solution and dissolved material redepositing. Over time, crystals in contact with a saturated solution may change their shape although the mass of the solid will not change.
The appropriate equilibrium constant is called the solubility product. For CaF2 it would be:
Ksp = [Ca2+][F-]2(Table 15.4 is a fairly extensive table of solubility products for rather insoluble ionic solids.) Note that the solubility constant is constant but the actual solubility will depend on the presence of common ions.
The calculation of Ksp from a known solubility is trivial:
Calculate the Ksp value for Bi2S3 which has a solubility of 1.0x10-15 M in pure water. Answer: The equation is:The calculation of solubility from the solubility constant can be more tricky:Bi2S3(s) 2Bi3+(aq) + 3S2-(aq)Ksp = [Bi3+]2[S2-]3The concentrations can be obtained directly from the chemical equation and plugged into the formula for Ksp.
[Bi3+] = 2x1.0x10-15 M
[S2-] = 3x1.0x10-15 MKsp = (2.0x10-15)2(3.0x10-15)3=1.1x10-73
The Ksp value for copper(II) iodate at 25 oC is 1.4x10-7. Calculate its solubility in pure water.
Answer: The equations are:
Cu(IO3)2(s) Cu2+(aq) + 2IO3-(aq)Ksp = [Cu2+][IO3-]2Let the equilibrium concentrations be:
[Cu2+] = A
[IO3-] = 2A1.4x10-7 = (A)(2A)2 = 4A3A = 3.3x10-3 M
|Solubility Data for Selected Compounds|
|Salt||Ksp||Solubility Equation||Solubility (mol/L)|
Consider the solubility of silver chromate (Ksp = 9.0x10-12) in a 0.100 M solution of silver nitrate.
Ag2CrO4 (s) 2Ag+(aq) + CrO42-(aq)Starting concentrations
[Ag+] = 0.100 M
[CrO42-] = 0
[Ag+] = 0.100 + 2A M
[CrO42-] = A
Ksp = 9.0x10-12 = [Ag+]2[CrO42-] = (0.100 + 2A)2(A)This will make a cubic equation but we can make the usual approximation since A will be very small compared to 0.100:
9.0x10-12 = (0.100)2(A) so A = 9.0x10-10 MCompare this with the solubility of Ag2CrO4 in pure water - 1.3x10-4 M.
Another example is Ag3PO4 where the presence of acid will convert PO43- to HPO42-. Silver ions can enter the solution since the silver hydrogen phosphate is more soluble.
Generally, if the anion of an insoluble compound has appreciable basic properties, the compound will tend to be more soluble in acidic solutions.
Mix 100.0 mL of 0.0500 M Pb(NO3)2 with 200.0 mL of 0.100 M NaI. Given that Ksp for PbI2 = 1.4x10-8 what will be the final concentrations of ions in the solution?
The first step is to determine the concentrations assuming no precipitation:
Assume the new volume is 300 mL and use M1.V1 = M2.V2
[Pb2+] = (100.0x0.0500)/300.0 = 1.67x10-2 MOnly the PbI2 will perhaps precipitate - other combinations are clearly soluble at the new concentrations. Calculation of Q will confirm this:
([NO3-] = 2x[Pb2+])
[I-] = (200.0x0.100)/300.0 = 6.67x10-2 M
([Na+] = [I-])
Q = [Pb2+][I-]2 = (1.67x10-2)(6.67x10-2)2 = 7.43x10-5 (> 1.4x10-8)The next step is to pretend that all the PbI2 that can be formed precipitates:
Amount of Pb2+ = 100.0 x 0.0500 = 5.00 mmolAfter formation of PbI2 and assuming it all precipitates for the moment:
Amount of I-=200.0 x 0.100 = 20.0 mmol
(ie an excess because only 10.0 mmol combine with the 5.00 mmol of Pb2+)
10.0 mmol of I- remain in 300 mL of the solutionThe next step exactly parallels the common ion problem. Start:
ie it is 10.0/300.0 = 3.33x10-2 M in I-.
[Pb2+] = 0At equilibrium:
[I-] = 3.33x10-2 M
[Pb2+] = ATherefore:
[I-] = 3.33x10-2 + 2A M
Ksp = 1.4x10-8=[Pb2+][I-]2 = (A)(3.33x10-2 + 2A)2
~ (A)(3.33x10-2)2[Pb2+] = A = 1.3x10-5 M and [I-] = 3.33x10-2 M
A solution contains 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If a source of I- is gradually added to the solution, which iodide will precipitate first? (Ksp(CuI) = 5.3x10-12, Ksp(PbI2) = 1.4x10-8.)
Calculate the greatest concentration of I- that can be present before precipitation of each iodide and compare:
Ksp = [Cu+][I-]max = (1.0x10-4)[I-]max = 5.3x10-8For PbI2
so [I-]max = 5.3x10-12 M
Ksp = [Pb2+][I-]max2 = (2x10-3)[I-]max2 = 1.4x10-8Clearly the copper compound precipitates first.
so [I-]max = 2.6x10-8 M
In one of the lab experiments, 5 mL of a saturated solution of sodium chloride (approx 5 M) was treated with 10 drops (about 0.4 ml) of 1 M hydrochloric acid, and in another test with 10 drops of 12 M (concentrated) hydrochloric acid. Rationalize the fact that a precipitate was not formed with the 1 M acid but did form with the 12 M acid. Answer:
The saturated sodium chloride solution has [Na+] = 5 M and [Cl-] = 5 M so Ksp = [Na+][Cl-] = 5 x 5 = 25
After the addition of the 10 drops of acid, the volume of the solution is 5.0 + 0.4 = 5.4 mL
The new [Na+] = 5 x 5.0/5.4 = 4.6 M
With the 1 M HCl, the final [Cl-] = (5 x 5.0/5.4) + (1 x 0.4/5.4) = 4.67 M
In this case Q = 4.6 x 4.67 = 21.5 which is less than 25, so no precipitate should form.
With the 12 M HCl, the final [Cl-] = (5 x 5.0/5.4) + (12 x 0.4/5.4) = 5.49
In this case Q = 4.6 x 5.9 = 25.3 which is greater than 25, so a precipitate is expected unless the solution supersaturates in the absence of any suitable nucleation sites.
H2O, NH3, Cl-, CN-Some less common ligands used in analytical chemistry:
EDTA, ethylene diamine, 8-hydroxyquinoline, dimethylglyoximeExamples of coordination numbers and complex ions:
Co(H2O)62+ 6-coordinate, pinkExample of stepwise formation or stability constants:
Ni(NH3)62+ 6-coordinate, green
CoCl42- 4-coordinate, blue
Ag(NH3)2+ 2-coordinate, colourless
Ag+(aq) + NH3(aq) Ag(NH3)+(aq) K1 = 2.1x103Provided the ligand is present in excess, the calculations are not complicated.
Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2+(aq) K2 = 8.2x103
What are the concentrations of the silver containing ions in a mixture of 100.0 mL of 2.0 M NH3 and 100.0 mL of 1.0x10-3 M AgNO3?
the major solution species are: Ag+, NO3-, NH3 and H2O but only the NH3/Ag+ system is of interest.
[Ag+]o = (100.0)(1.0x10-3)/(200.0) = 5.0x10-4 MAssume essentially complete reaction to give 5.0x10-4 M Ag(NH3)2+ because the equilibrium constants are large, and the consider these equilibria:
[NH3]o = (100.0)(2.0)/(200.0) = 1.00 M
Let [Ag(NH3)+] = A and assume negligible NH3 has been consumed:
K2 = 8.2x103 = [Ag(NH3)2+]/[Ag(NH3)+][NH3] =
(5.0x10-4 - A)/(A)(1.0 + A) ~ (5.0x10-4)/(A)(1.0)A = [Ag(NH3)+] = 6.1x10-8 MNow Let [Ag+] = A and still assume negligible NH3 has been consumed:
K1 = 2.1x103 = [Ag(NH3)+]/[Ag+][NH3] =
(6.1x108 - A)/(A)(1.0 + A) ~ (6.1x10-8)/(A)(1.0)A = [Ag+] = 2.9x10-11 Mi.e [Ag(NH3)2+] >> [Ag(NH3)+] >> [Ag+]
AgCl, PbCl and Hg2Cl2The Ag+ can be redissolved by adding NH3 and complexing it in the following sequence of steps:
AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 1.6x10-10 Ag+(aq) + NH3(aq) Ag(NH3)+(aq) K1 = 2.1x103The net reaction is:
Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2+(aq) K2 = 8.2x103
AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl-(aq)
K = [Ag(NH3)2+][Cl-]/[NH3]2 = Ksp.K1.K2=2.8x10-3Example:
Calculate the solubility of AgCl in 10 M NH3 solution.This is one factor that leads to the separation scheme for the above three species:
AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl-(aq)Start:
[NH3] = 10 MAt equilibrium:
[Ag(NH3)2+] =  0
[Cl-] = 0
[NH3] = 10 - 2A M
[Ag(NH3)2+] = A
[Cl-] = A(2.8x10-3)½ = ((A)(A)/(10 - 2A)2)½ = ((A)(A)/(10)2)½Note that the solubility of AgCl in pure water = (Ksp)½ = 1.3x10-5 mol/L
A = 0.48 mol/L