aspects of aqueous equilibria. aspects of aqueous equilibria: the common ion effect recall that...

Aspects of Aqueous EquilibriaAspects of Aqueous Equilibria

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect

Recall that salts like sodium acetate are strong electrolytes

NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)

…and that the C2H3O2- ion is a conjugate base of a weak acid

HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)

Ka = [H3O+] [C2H3O2

-]

[HC2H3O2]


Ka = [H3O+] [C2H3O2

-]

[HC2H3O2]

Now, lets think about the problem from the perspective of

LeChatelier’s PrincipleLeChatelier’s PrincipleWhat would happen if the concentration of the acetate ion were increased?

Q > K and the reaction favors reactant

Addition of CAddition of C22HH33OO22-- shifts equilibrium, shifts equilibrium,

reducing Hreducing H++


Since the equilibrium has shifted to favor the reactant, it would appear as if the dissociation of the weak acid(weak electrolyte) had decreased.


Ka = [H3O+] [C2H3O2

-]

[HC2H3O2]


So where might the additional C2H3O2-(aq) come from?

Remember we are not adding H+. So it’s not like we can add more acetic acid.


How about from the sodium acetate?




In general, the In general, the dissociation of a weak electrolyte (acetic acid)dissociation of a weak electrolyte (acetic acid)is decreased by adding to the solution a is decreased by adding to the solution a strong electrolyte thatstrong electrolyte thathas an ion in commonhas an ion in common with the weak electrolyte with the weak electrolyte

The shift in equilibrium which occurs is called theCOMMON ION EFFECTCOMMON ION EFFECT


Let’s explore the COMMON ION EFFECTCOMMON ION EFFECT in a little more detailSuppose that we add 8.20 g or 0.100 mol sodium acetate, NaC2H3O2, to 1 L of a 0.100 M solution of acetic acetic acid, HC2H3O2. What is the pH of the resultant solution?



Now you try it!


Calculate the pH of a solution containing 0.06 M formic acid (HCH2O, Ka = 1.8 x 10-4)and 0.03 M potassium formate, KCH2O.

Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of

HClHCl and 0.20 mol HF in 1.0 L

Note who the strong electrolyte is this time!


HF(aq) + H2O H3O+(aq) + F-(aq)

HCl+ H2O(aq) H3O+(aq) + Cl-(aq)

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion Effect The Common Ion Effect

Kb = [NH4

+] [OH-]

[NH3]

Now, lets think about the problem from the perspective of

LeChatelier’s PrincipleLeChatelier’s PrincipleBut this time lets deal with a weak base and a salt containing its conjugate acid.

Q > K and the reaction favors reactant

Addition of NHAddition of NH44+ shifts equilibrium,+ shifts equilibrium,

reducing OHreducing OH--

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

Calculate the pH of a solution produced by mixing 0.10 mol NH4Cl with 0.40 L of 0.10 M NH3(aq), pKb = 4.74?

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

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion Effect The Common Ion Effect

NH4Cl(aq) NH4+(aq) + Cl- (aq)

Aspects of Aqueous Equilibria: Common IonsAspects of Aqueous Equilibria: Common IonsGenerated by Acid-Base ReactionsGenerated by Acid-Base Reactions

The common ion that affects a weak-acid or weak-base equilibrium may be present because it is added as a salt, or the common ion can be generated by reacting an acid and base directly (no salt would be necessary….which is kind of convenient if you think about it)

Let’s take a look at a weak acid-strong base combination first:

HC2H3O2(aq) + OH- H2O + C2H3O2

Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxide

0.20 mol 0.10 mol

-0.10 mol -0.10 mol-0.10 mol

0

00.10 mol 0.10 mol

HC2H3O2(aq) + OH- H2O + C2H3O2

Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxide

0.20 mol 0.10 mol

-0.10 mol 0.10 mol-0.10 mol

0

00.10 mol 0.10 mol


0.10 M0.10 M

Let’s suppose that all this is occurring in 1.0 L of solution


0


Sample problem: Calculate the pH of a solution produced by mixing 0.60 L of 0.10 MNH4Cl with 0.40 L of 0.10 M NaOH

NH4Cl(aq) NH4+(aq) + Cl- (aq)

NH4+ + OH- NH3 + H2O

0.06 mol 0.04 mol

-0.04 mol 0.04 mol

0

0.04 mol0

-0.04 mol

0.02 mol

Don’t forgetto convert to

MOLARITIESMOLARITIES

0.02 M

NH4+ + H2O H3O+ + NH3

0.04 M0

Now you try it!


Calculate the pH of a solution formed by mixing 0.50 L of 0.015 M NaOH with0.50 L of 0.30 M benzoic acid (HC7H5O2, Ka = 6.5 x 10-5)

[H +] = [HX]

[X-]Ka

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS

A buffered solution is a solution that resists change in pH upon addition of smallamounts of acid or base.

HERE’S HOW IT WORKS !HERE’S HOW IT WORKS !

Suppose we have a salt: MX M +(aq) + X- (aq)

And we’ve added the salt to a weak acidcontaining the same conjugate base as the salt, HX:

HX +H2O H3O+ + X-

And the equilibrium expression for thisreaction is

Ka = [H+ ] [ X-]

[HX]

Note that the concentration of the H+ isdependent upon the Ka and the ratio between the HX and X- (the conjugateacid-base pair)

CONTINUED ON NEXT SLIDECONTINUED ON NEXT SLIDE


Two important characteristics of a buffer are buffering capacity and pH. Bufferingcapacity is the amount of acid or base the buffer can neutralize before the pH beginsto change to an appreciable degree.

•This capacity depends on the amount of acid and base from which thebuffer is made

•The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and hence the pH, to change

•The pH of the buffer depends upon the Ka

[H +] = [HX]

[X-]Ka

-log[H +] = [HX]

[X-] -log Ka

pH = pKa - log [HX]

[X-]

Henderson-Hasselbalch Equation

pH = pKa + log [X-]

[HX]



0.1 M 0 0.1

0.1 M 0.1 M 0.1 M

-x x x

x0.1 - x 0.1 + x

1.8 x 10-5 = x(0.1 + x )

0.1 - x

x = 1.8 x 10-5

pH = 4.74


pH = 4.74 + log [.1]

[.1]


Note that these are initialconcentrations


A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2

forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) afterafter 0.020 mol NaOH is added0.020 mol NaOH is added, (b) after adding 0.020 mol HCladding 0.020 mol HCl is added.

HC2H3O2(aq) + OH- H2O + C2H3O2-(aq)

0.1 M 0.02 M0.02 M

-0.02 M-0.02 M-0.02 M-0.02 M

0.1 M


0.1 M 0.1 M 0.1 M

0.02 M0.02 M

0.12 M0.00 M0.00 M0.08 M


pH = 4.74 + log [.12]

[.08]


pH = 4.92

Step 1

Step 2

Note that the OH- reacts with the HX


A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2

forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH0.020 mol NaOH is added, (b) after adding 0.020 mol HCl is addedafter adding 0.020 mol HCl is added.

C2H3O2-(aq) + H+ HC2H3O2

Step 1

0.10 M 0.10 M0.02 M0.02 M

-0.02 M-0.02 M

0.00 M0.00 M

-0.02 M-0.02 M 0.02 M0.02 M

0.12 M0.08M


pH = 4.74 + log [.08]

[.12]


pH = 4.56

Step 2

Now consider, for a moment, what would have happened if I had added 0.020 molof NaOH or 0.02 mol HCl to .1 M HC2H3O2.


0.1 M 0 0

-x x x

x0.1 - x x

1.8 x 10-5 = x2

0.1 - x

x = 0.0013

pH = 2.9


HC2H3O2(aq) + OH- H2O + C2H3O2-(aq)

0.1 M 0.02 M0.02 M

-0.02 M-0.02 M-0.02 M-0.02 M

0.00

0.02 M0.02 M

0.02 M0.00 M0.00 M0.08 M


pH = 4.74 + log [.02]

[.08]

pH = 4.13


C2H3O2-(aq) + H+ HC2H3O2

0.00 0.10 M0.02 M0.02 M

pH = 1.7


Complete dissociation

pH = -log [0.02]


Sample exercise: Consider a litter of buffered solution that is 0.110 M in formic acid(HC2H3O )and 0.100 M in sodium formate (NaC2H3O ). Calculate the pH of the buffer(a) before any acid or base are added, (b) after the addition of 0.015 mol HNO3, (3) afterthe addition of 0.015 mol KOH

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Titration CurvesTitration Curves

HCl(aq) + NaOH(aq) H2O + NaCl

Stoichiometrically equivalent quantities of acid and base have reacted

Titration of a weak acid and a strong base results in pH curves that look similarto those of a strong acid-strong base curve except that the curve (a) begins at a higherpH, (b) the pH rises more rapidly in the early part of the titration, but more slowly near the equivalence point, and (3) the equivalence point pH is not 7.0

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Titration Curves of Both Strong and Weak AcidsTitration Curves of Both Strong and Weak Acids

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Calculating pH’s from TitrationsCalculating pH’s from Titrations

Calculate the pH in the titration of acetic acid by NaOH after 30.0 ml of 0.100 M NaOH has been added to 50 mL of 0.100 acetic acid

HC2H3O2 (aq) + OH- H2O(l) + C2H3O2

0.005 mol 0.003 mol 0

0.003 mol

0.003 mol

-0.003 mol-0.003 mol

-0.002 mol 0

pH = 4.74 + log [.0370]

[.0250]

pH = 4.91

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Determining the KDetermining the Kaa From the Titration Curve From the Titration Curve

pKa = pH = 4.74

Na2CO3 2Na+(aq) + CO32-

H+(aq) + CO32- HCO3

- (aq)

H+(aq) + HCO- H2CO3 (aq)

Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Titrations of Polyprotic AcidsTitrations of Polyprotic Acids

aspects of aqueous equilibria. aspects of aqueous equilibria: the common ion effect recall that...

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