Outline:Outline: 3/12/073/12/07 Chem. Dept. Seminar Wed @ 4pm 2 more lectures until Exam 2… Chemistry Advising – Today @ 4pm

Today: More Chapter 18 Polyprotic acid titrations

Solubility Product (Ksp)

Worksheet #8 practice…

#1a. pH = 2.5

[H2A] = 0.997 M

[HA-] = 0.00316 M

[A2-] = 1 10-8 M

#1b. pH = 10.5#2a. pH = 9.74

#2b. pH = 11.1

Quiz # 7

Please put away all books/papers

If you don’t have a calculator, just set up the problems fully…

Quiz # 7

Please turn your papers over and Please turn your papers over and pass them to the right…pass them to the right…

Quiz #7 : Buffers #1pH = pKpH = pKaa + log ([base]/[acid]) + log ([base]/[acid])

4.00 = 4.75 + log ([base]/[1.0M])4.00 = 4.75 + log ([base]/[1.0M])

[base] = 10[base] = 100.75 0.75 = 0.178 M= 0.178 M

[base] = 0.178 mol/1.0L[base] = 0.178 mol/1.0L

= 14.6 g CH= 14.6 g CH33COONa / 1.0LCOONa / 1.0L

Quiz #7 : Weak Base #2Pyr + HPyr + H22O = pyrHO = pyrH++ + OH + OH

KKbb = 10 = 108.72 8.72 = 1.91= 1.91101099

0.0150.015x +x +xx +x +x

1.911.91101099 = = xx2 2 / 0.015/ 0.015

xx = 5.34 = 5.34 101066 = [OH = [OH]]

pHpH = 14 – log( 5.34 = 14 – log( 5.34 101066) = 8.73 ) = 8.73

Quiz #7 : Titration #3

pH = pKpH = pKaa + log ([base]/[acid]) + log ([base]/[acid])

pH = 3.74 + log ([0.0005]/[0.0020])pH = 3.74 + log ([0.0005]/[0.0020])

= 3.14= 3.14

HA + OHHA + OH A A + H + H22O (titration)O (titration)

0.0025 + 0.0005 0.0025 + 0.0005 0 + 0 (init) 0 + 0 (init)

0.0020 + 0 0.0020 + 0 0.0005 (equil) 0.0005 (equil)

Titration of Polyprotic Acids

Weak acid: Ka1 = x2/[H2A]

Titration of Polyprotic Acids

Buffer:

pKa2+log[A]/[HA]Buffer:

pKa1+log[HA]/[H2A]

Titration of Polyprotic Acids

log((Ka1× Ka2)0.5) See page 785

Try example 18 – 12 (page 787)

Sulfurous Acid, HSulfurous Acid, H22SOSO33, has two acidic , has two acidic

hydrogen atoms, with pKhydrogen atoms, with pKaa values of 1.85 values of 1.85

and 7.20. Construct a titration curve for and 7.20. Construct a titration curve for the titration of 125 mL of 0.150 M the titration of 125 mL of 0.150 M sulfurous acid with 0.800 M NaOH.sulfurous acid with 0.800 M NaOH.

changes color as the titration passes the changes color as the titration passes the stoichiometric point if : stoichiometric point if :

pKpKinin≈ pH≈ pHstoichiometric pointstoichiometric point

The Solubility-Product Constant, Ksp

• Consider

• for which

• Ksp is the “solubility product”. (BaSO4 is ignored because it is a pure solid so its concentration is constant.)

BaSO4(s) Ba2+(aq) + SO42-(aq)

]SO][Ba[ -24

2spK

• The solubility product is another example of equilibrium calculations

• Solubility product calcs depend on the common ion effect (LeChâtelier).

• They have particular applications with metal ions and pH calculations (environmental applications).

Solubility EquilibriaSolubility Equilibria

Types of Equilibrium Constants:

Lots of different names…. Keq, KH , Ksp, Ka , Kb, Kf , Kc, Kp…

All the same idea!

Solubility Equilibria

Insoluble compoundsInsoluble compounds: solubility is less : solubility is less than 0.01 mol of dissolved material per than 0.01 mol of dissolved material per liter of solution, Kliter of solution, Kspsp << 1 << 1

Slightly solubleSlightly soluble: 10: 10-5-5 < K < Kspsp < 10 < 10-2-2

SolubleSoluble: K: Kspsp > 10 > 10-2-2

• The solubility product is another example of equilibrium calculations

• Solubility product calcs depend on the common ion effect (LeChâtelier).

• They have particular applications with metal ions and pH calculations (environmental applications).

Solubility EquilibriaSolubility Equilibria

The Common Ion Effect

• Solubility is decreased when a common ion is added (Le Châtelier again)

• as F- (from NaF, say) is added, the equilibrium shifts left, therefore CaF2(s) is formed (precipitation occurs).

• As NaF is added to the system, the solubility of CaF2 decreases.

Factors that Affect SolubilityFactors that Affect Solubility

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