Chapter 15
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Chapter 15. Applying equilibrium. The Common Ion Effect. When the salt with the anion of a weak acid is added to that acid, Lowers the percent dissociation of the acid. Ex: NaF and HF mixed together F- is the common ion HF ↔ H + + F -

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Chapter 15

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Chapter 15

Chapter 15

Applying equilibrium


The common ion effect

The Common Ion Effect

  • When the salt with the anion of a weak acid is added to that acid,

  • Lowers the percent dissociation of the acid.

  • Ex: NaF and HF mixed together

    • F- is the common ion

      HF ↔ H+ + F-

      Adding F- shifts the equilibrium to the left and therefore fewer H+ ions present


Applying equilibrium

The same principle applies to salts with the cation of a weak base.

The calculations are the same as last chapter.

NH4Cl and NH3 (NH4+ is the common ion)

NH3 + H+ NH4+


Buffered solutions

Buffered solutions

  • A solution that resists a change in pH.

  • Consists of either a weak acid and its salt or a weak base and its salt.

  • We can make a buffer of any pH by varying the concentrations of these solutions.

  • Same calculations as before.

  • Calculate the pH of a solution that is .50 M HAc and .25 M NaAc (Ka = 1.8 x 10-5)


Applying equilibrium

Na+ is a spectator and the reaction we are worried about is

HAc H+ + Ac-

  • Choose x to be small

Initial 0.50 M 0 0.25 M

-x

x

x

Change

Final

0.50-x

x

0.25+x

  • We can fill in the table


Applying equilibrium

HAc H+ + Ac-

Initial 0.50 M 0 0.25 M

  • Do the math

  • Ka = 1.8 x 10-5

Change

-x

x

x

Final

0.50-x

x

0.25+x

x (0.25)

x (0.25+x)

=

1.8 x 10-5 =

(0.50)

(0.50-x)

  • Assume x is small

x = 3.6 x 10-5

  • Assumption is valid

  • pH = -log (3.6 x 10-5) = 4.44


Adding a strong acid or base

Adding a strong acid or base

  • Do the stoichiometry first.

    • Use moles not molar

  • A strong base will grab protons from the weak acid reducing [HA]0

  • A strong acid will add its proton to the anion of the salt reducing [A-]0

  • Then do the equilibrium problem.

  • What is the pH of 1.0 L of the previous solution when 0.010 mol of solid NaOH is added?


Applying equilibrium

HAc H+ + Ac-

0.25 mol

0.50 mol

  • In the initial mixture M x L = mol

  • 0.50 M HAc x 1.0 L = 0.50 mol HAc

0.26 mol

0.49 mol

  • 0.25 M Ac- x 1.0 L = 0.25 mol Ac-

  • Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole

  • Because it is in 1.0 L, we can convert it to molarity


Applying equilibrium

HAc H+ + Ac-

0.25 mol

0.50 mol

0.26 M

0.49 M

  • In the initial mixture M x L = mol

  • 0.50 M HAc x 1.0 L = 0.50 mol HAc

  • 0.25 M Ac- x 1.0 L = 0.25 mol Ac-

  • Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole

  • Because it is in 1.0 L, we can convert it to molarity


Applying equilibrium

HAc H+ + Ac-

0.25 mol

0.50 mol

0.26 M

0.49 M

  • Fill in the table

HAc H+ + Ac-

Initial 0.49 M 0 0.26 M

-x

x

x

Final

0.49-x

x

0.26+x


Applying equilibrium

HAc H+ + Ac-

Initial 0.49 M 0 0.26 M

  • Do the math

  • Ka = 1.8 x 10-5

-x

x

x

Final

0.49-x

x

0.26+x

x (0.26)

x (0.26+x)

=

1.8 x 10-5 =

(0.49)

(0.49-x)

  • Assume x is small

x = 3.4 x 10-5

  • Assumption is valid

  • pH = -log (3.4 x 10-5) = 4.47


Notice

Notice

  • If we had added 0.010 mol of NaOH to 1 L of water, the pH would have been.

  • 0.010 M OH-

  • pOH = 2

  • pH = 12

  • But with a mixture of an acid and its conjugate base the pH doesn’t change much

  • Called a buffer.


General equation

General equation

  • Ka = [H+] [A-][HA]

  • so [H+] = Ka [HA] [A-]

  • The [H+] depends on the ratio [HA]/[A-]

  • taking the negative log of both sides

  • pH = -log(Ka [HA]/[A-])

  • pH = -log(Ka)-log([HA]/[A-])

  • pH = pKa + log([A-]/[HA])


This is called the henderson hasselbach equation

This is called the Henderson-Hasselbach equation

  • pH = pKa + log([A-]/[HA])

  • pH = pKa + log(base/acid)

  • Works for an acid and its salt

  • Like HNO2 and NaNO2

  • Or a base and its salt

  • Like NH3 and NH4Cl

  • But remember to change Kb to Ka


Applying equilibrium

  • Calculate the pH of the following

  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

  • pH = 3.38


Applying equilibrium

  • Calculate the pH of the following

  • 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)

  • Ka = 1 x 10-14 1.8 x 10-5

  • Ka = 5.6 x 10-10

  • remember its the ratio base over acid

  • pH = 9.05


Prove they re buffers

Prove they’re buffers

  • What would the pH be if .020 mol of HCl is added to 1.0 L of both of the preceding solutions.

  • What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of each of the proceeding.

  • Remember adding acids increases the acid side,

  • Adding base increases the base side.


Prove they re buffers1

Prove they’re buffers

  • What would the pH be if .020 mol of HCl is added to 1.0 L of preceding solutions.

  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 H+ + C3H5O3-

Initially 0.75 mol 00.25 mol

After acid 0.77 mol0.23 mol

Compared to 3.38 before acid was added


Prove they re buffers2

Prove they’re buffers

  • What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of the solutions.

  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 H+ + C3H5O3-

Initially 0.75 mol 00.25 mol

After acid 0.70 mol0.30 mol

Compared to 3.38 before acid was added


Prove they re buffers3

Prove they’re buffers

  • What would the pH be if .020 mol of HCl is added to 1.0 L of preceding solutions.

  • 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+H+ + NH3

Initially 0.40 mol 00.25 mol

After acid 0.42 mol0.23 mol

Compared to 9.05 before acid was added


Prove they re buffers4

Prove they’re buffers

  • What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L each solutions.

  • 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+H+ + NH3

Initially 0.40 mol 00.25 mol

After acid 0.35 mol0.30 mol

Compared to 9.05 before acid was added


Buffer capacity

Buffer capacity

  • The pH of a buffered solution is determined by the ratio [A-]/[HA].

  • As long as this doesn’t change much the pH won’t change much.

  • The more concentrated these two are the more H+ and OH- the solution will be able to absorb.

  • Larger concentrations = bigger buffer capacity.


Buffer capacity1

Buffer Capacity

  • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of each of the following:

  • 5.00 M HAc and 5.00 M NaAc

  • 0.050 M HAc and 0.050 M NaAc

  • Ka= 1.8x10-5

  • pH = pKa


Buffer capacity2

Buffer Capacity

  • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 5.00 M HAc and 5.00 M NaAc

  • Ka= 1.8x10-5

HAcH+ + Ac-

Initially 5.00 mol 05.00 mol

After acid 5.04 mol4.96 mol

Compared to 4.74 before acid was added


Buffer capacity3

Buffer Capacity

  • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 0.050 M HAc and 0.050 M NaAc

  • Ka= 1.8x10-5

HAcH+ + Ac-

Initially 0.050 mol 0 0.050 mol

After acid 0.090 mol 0 0.010 mol

Compared to 4.74 before acid was added


Buffer capacity4

Buffer capacity

  • The best buffers have a ratio [A-]/[HA] = 1

  • This is most resistant to change

  • True when [A-] = [HA]

  • Makes pH = pKa (since log 1 = 0)


Titrations

Titrations


Titrations1

Titrations

  • Millimole (mmol) = 1/1000 mol

  • Molarity = mmol/mL = mol/L

  • Makes calculations easier because we will rarely add liters of solution.

  • Adding a solution of known concentration until the substance being tested is consumed.

  • This is called the equivalence point.

  • Where moles of acid = moles of base


Applying equilibrium

  • Strong acid with strong Base

  • Equivalence at pH 7

7

pH

mL of Base added


Applying equilibrium

  • Weak acid with strong Base

  • Equivalence at pH >7

>7

7

pH

mL of Base added


Applying equilibrium

  • Strong base with strong acid

  • Equivalence at pH 7

7

pH

mL of acid added


Applying equilibrium

  • Weak base with strong acid

  • Equivalence at pH <7

7

<7

pH

mL of acid added


Strong acid with strong base

Strong acid with Strong Base

  • Do the stoichiometry.

  • mL x M = mmol

  • There is no equilibrium .

  • They both dissociate completely.

  • The reaction is H+ + OH- HOH

  • Use [H+] or [OH-] to figure pH or pOH

  • The titration of 20.0 mL of 0.10 M HNO3 with 0.10 M NaOH


Weak acid with strong base

Weak acid with Strong base

  • There is an equilibrium.

  • Do stoichiometry.

    • Use moles

  • Determine major species

  • Then do equilibrium.

    Calculate the pH after 0.0mL, 15.0mL, 25.0 mL, and 30.0mL of 0.10 M NaOH are added to 25.0mL of 0.10M of Nicotinic acid Ka= 1.4 x 10 -5


Summary

Summary

  • Strong acid and base just stoichiometry.

  • Weak acid with 0 ml of base - Ka

  • Weak acid before equivalence point

    • Stoichiometry first

    • Then Henderson-Hasselbach

  • Weak acid at equivalence point- Kb-Calculate concentration

  • Weak acid after equivalence - leftover strong base. -Calculate concentration


  • Example

    Example

    • A 25.0 mL sample of benzoic acid (0.120M) is titrated w/ NaOH (.210 M) Ka= 6.3 x 10 -5

      • Calculate # moles of Benzoic acid

      • Calculate volume of NaOH needed to reach equiv. pt

      • Calc. the pH before any base is added

      • Calc. pH after 10ml base is added

      • Find the pH at equivalence pt

      • Calculate the pH after 15.3mL of OH- is added


    Summary1

    Summary

    • Weak base before equivalence point.

      • Stoichiometry first

      • Then Henderson-Hasselbach

  • Weak base at equivalence point Ka. -Calculate concentration

  • Weak base after equivalence – left over strong acid. -Calculate concentration


  • Applying equilibrium

    Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10-4) with 0.10 M NaOH


    Indicators

    Indicators

    • Weak acids that change color when they become bases.

    • weak acid written HIn

    • Weak base

    • HIn H+ + In-clear red

    • Equilibrium is controlled by pH

    • End point - when the indicator changes color.

    • Try to match the equivalence point


    Indicators1

    Indicators

    • Since it is an equilibrium the color change is gradual.

    • It is noticeable when the ratio of [In-]/[HI] is 1/10 or [HI]/[In-] is 10/1

    • Since the Indicator is a weak acid, it has a Ka.

    • pH the indicator changes is

    • pH=pKa +log([In-]/[HI]) = pKa +log(1/10)

    • pH=pKa - 1


    Indicators2

    Indicators

    • pH=pKa + log([HI]/[In-]) = pKa + log(10)

    • pH=pKa+1

    • Choose the indicator with a pKa 1 more than the pH at equivalence point if you are titrating with base.

    • Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with acid.


    Sample problem

    Sample Problem

    • Two drops of indicator HIn (Ka= 1.0 x 10 -9), hwere HIn is yellow and In- is blue are placed in 100.0 mL of 0.10 M HCl.

      • What color is the solution initially?

      • The solution is titrated with 0.10 M NaOH. At what pH will the color change (greenish yellow) occur?

      • What color will the solution be after 200.0mL of NaOH has been added?


    Applying equilibrium

    • What color will the solution be after 200.0mL of NaOH has been added?


    Solubility equilibria

    Solubility Equilibria

    Will it all dissolve, and if not, how much?


    Applying equilibrium

    • All dissolving is an equilibrium.

    • If there is not much solid it will all dissolve.

    • As more solid is added the solution will become saturated.

    • Soliddissolved

    • The solid will precipitate as fast as it dissolves .

    • Equilibrium


    General equation1

    General equation

    • M+ stands for the cation (usually metal).

    • Nm-stands for the anion (a nonmetal).

    • MaNmb(s) aM+(aq) + bNm- (aq)

    • Remember the concentration of a solid doesn’t change. So, what’s the K expression?

    • Ksp = [M+]a[Nm-]b

    • Called the solubility product for each compound.


    Watch out

    Watch out

    • Solubility is not the same as solubility product.

    • Solubility product is an equilibrium constant.

    • it doesn’t change except with temperature.

    • Solubility is an equilibrium position-usually expressed as a Molarity for how much can dissolve.


    Calculating solubility

    Calculating Solubility

    Ex: Calc. Solubility of CaC2O4 if the value of Ksp= 4.8 x 10 -5 mol/L

    CaC2O4 (s) ↔ Ca+2(aq) + C2O4 -2 (aq)

    Ksp = [Ca+2] [C2O4-2]

    Ksp= [s] [s] = [s]2 = 4.8x10-5 = s2

    s= 6.9 x 10 -3 M


    Calculating k sp

    Calculating Ksp

    • The solubility of copper(I) bromide is

      2.0 x 10 -4 mol/L. Calc. Ksp value.

      CuBr(s) ↔ Cu+1(aq) + Br-1 (aq)

      I .0002M0M 0M

      C - .0002M+.0002M +.0002M

      E 0.0002M .0002M

      Ksp = [.0002] [.0002] = 4.0 x 10 -8

      (units usually omitted)


    Applying equilibrium

    The solubility of Calcium Carbonate is 8.7 x 10 -9 mol/L. Calc. Ksp.


    Calculating k sp1

    Calculating Ksp

    • The solubility of iron(II) oxalate FeC2O4 is 65.9 mg/L

    • The solubility of Li2CO3 is 5.48 g/L


    Common ion effect

    Common Ion Effect

    • If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.

    • Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M in a solution of 0.010 M Na2SO4.

    • Calculate the solubility of CaF2

      Ksp= 4.0 x 10 -11 in a 0.025 M NaF solution.


    Agcl and nacl in solution

    AgCl and NaCl in solution

    • Cl- is the common ion

    • When [Ag+] [Cl-] < Ksp, no precipitate will be formed.

    • When [Ag+] [Cl-] > Ksp, a precipitate will be formed.


    Ph and solubility

    pH and solubility

    • OH- can be a common ion.

    • More soluble in acid.

    • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.

    • CaC2O4 Ca+2 + C2O4-2

    • H+ + C2O4-2 HC2O4-

    • Reduces [C2O4-2] in acidic solution.


    Sample problem1

    Sample problem

    • What is the pH in a saturated solution of Ca(OH)2?Ksp = 5.5 x 10-6 for Ca(OH)2.


    Q and precipitation

    Q and Precipitation

    • not necessarily at Equil.

    • Q= Ion product

    • Q =[M+]a[Nm-]b

    • If Q>Ksp a precipitate forms. (supersaturated)

    • If Q<Ksp No precipitate. (unsaturated)

    • If Q = Ksp equilibrium. (saturated)


    Example1

    Example1

    Chemical analysis gave [Pb2+] = 0.012 M, and [Br-] = 0.024 M in a solution. From a table, you find Ksp for PbBr2 has a value of 4x 10-5. Is the solution saturated, oversaturated or unsaturated?


    Example 2

    Example 2

    • A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M) precipitate and if so, what is the concentration of the ions?


    Selective precipitations

    Selective Precipitations

    • Used to separate mixtures of metal ions in solutions.

    • Add anions that will only precipitate certain metals at a time.

    • Used to purify mixtures.

    • Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.


    Selective precipitation

    Selective Precipitation

    • Then add OH-solution [S-2] will increase so more soluble sulfides will precipitate.

    • Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3


    Selective precipitation1

    Selective precipitation

    • Follow the steps

    • First with insoluble chlorides (Ag, Pb, Ba)

    • Then sulfides in Acid.

    • Then sulfides in base.

    • Then insoluble carbonate (Ca, Ba, Mg)

    • Alkali metals and NH4+ remain in solution.


    Complex ion equilibria

    Complex ion Equilibria

    • A charged ion surrounded by ligands.

    • Ligands are Lewis bases using their lone pair to stabilize the charged metal ions.

    • Common ligands are NH3, H2O, Cl-,CN-

    • Coordination number is the number of attached ligands.

    • Cu(NH3)42+ has a coordination # of 4


    The addition of each ligand has its own equilibrium

    The addition of each ligand has its own equilibrium

    • Usually the ligand is in large excess.

    • And the individual K’s will be large so we can treat them as if they go to completion.

    • The complex ion will be the biggest ion in solution.


    Applying equilibrium

    • Calculate the concentrations of Ag+, Ag(S2O3)-2, and Ag(S2O3)2-3 in a solution made by mixing 150.0 mL of 0.010 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3

    • Ag+ + S2O3-2 Ag(S2O3)- K1=7.4 x 108

    • Ag(S2O3)- + 2S2O3-2 Ag(S2O3)2-3 K2=3.9 x 104

      (Hint: set up ice problem, determine L.R.)

      This problem is found on pg. 768-769 in your book. Check your work!


    Complex ions and solubility

    Complex Ions and Solubility

    The dissolving of solid AgCl in excess NH3 is represented as…

    AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ (aq) + Cl-(aq)

    However, this is a result of several steps

    AgCl(s) ↔ Ag+ + Cl- Ksp= 1.6 x 10-10

    Ag+ + NH3 ↔ Ag(NH3)+ K1 = 2.1 x 103

    Ag(NH3)+ + NH3 ↔ Ag(NH3)2+ K2= 8.2 x 103


    Agcl s 2nh 3 aq ag nh 3 2 aq cl aq

    AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ (aq) + Cl-(aq)

    So the Equilibrium Expression is written as…. K= [Ag(NH3)2+] [Cl-]

    [NH3]2

    = Ksp x K1 x K2

    = (1.6 x 10-10) ( 2.1 x 103) (8.2 x 103)

    = 2.8 x 10-3


    Summary2

    Summary

    3 Factors that affect solubility:

    - Common Ions

    - pH

    - Complexing Agents/Ions

    Mixtures of Ions can be separated by selective precipitation (you know this as Quantitative Analysis)

    - Steps listed on pg.765 or in your lab


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