Ch. 15: Applications of Aqueous Equilibria 15.4 Titrations and pH curves

Titration  used to find the concentration of a solution using a solution of known concentration  _____________ solution with unknown concentration solution with unknown concentration  _____________ solution with known concentration solution with known concentration  equivalence point is usually marked by the end point of an indicator

pH curve  also called a titration curve  plotting of pH of the analyte as a function of the amount of titrant added  millimole (mmol) 1/1000 th of a mole 1/1000 th of a mole = molarity = molarity

Strong Acid and Strong Base  Titrate 50 mL of 0.2 HNO 3 with 0.1 M NaOH A. no NaOH mL HNO 3 B mL NaOH mL HNO 3 H + + OH -  H 2 O H + + OH -  H 2 O I C E

Strong Acid and Strong Base C. 20 mL NaOH and 50 mL HNO 3 D. 50 mL NaOH and 50 mL HNO 3 H + + OH -  H 2 O H + + OH -  H 2 O I C E I C E

Strong Acid and Strong Base E. 100 mL NaOH and 50 mL HNO 3 F. 150 mL NaOH and 50 mL HNO 3 H + + OH -  H 2 O H + + OH -  H 2 O I C E I C E

Strong Acid and Strong Base very gradual changes in pH until close to the equivalence pt.

Weak Acid and Strong Base  Titrate 50 mL of 0.10M HC 2 H 3 O 2 with 0.1 M NaOH A. no NaOH mL HC 2 H 3 O 2 HC 2 H 3 O 2  H + + C 2 H 3 O 2 - HC 2 H 3 O 2  H + + C 2 H 3 O 2 - I C E

Weak Acid and Strong Base B. 10mL NaOH and 50 mL HC 2 H 3 O 2 OH - + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 - I C E HC 2 H 3 O 2  H + + C 2 H 3 O 2 - HC 2 H 3 O 2  H + + C 2 H 3 O 2 -I C E

Weak Acid and Strong Base C. 50mL NaOH and 50 mL HC 2 H 3 O 2 OH - + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 - I C E C 2 H 3 O H 2 O  HC 2 H 3 O 2 + OH- C 2 H 3 O H 2 O  HC 2 H 3 O 2 + OH-I C E

Weak Acid and Strong Base D. 75mL NaOH and 50 mL HC 2 H 3 O 2 OH - + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 - I C E there are two bases to consider: which will be most important? Can just use the OH - to determine the pH

Comparing Shapes

Weak Base and Strong Acid

Ch. 15: Applications of Aqueous Equilibria 15.5 Indicators

Acid-Base Indicators  used to mark end of titration (____________________)  only works if you choose one whose end point is the same as eq. point  usually are complex molecules that act as weak acids  change colors when H+ is removed phenolphthalein

Indicators HIn  H + + In - yellow blue  If more H+ (acid) is added to the solution, the reaction shifts to the left to make it yellow  If more OH- (base) is added to the solution, the reaction shifts to the right to make it blue Bromthymol Blue

Calculating the pH at end point  Assume that color change is visible to the eye when ratio of [In - ]/[HIn] is 1/10  that means that one tenth of the indicator must have changed forms  Using the H-H equation and the one tenth rule, the pH of end point or color change is:

Only shows where the color change occurs

Choosing an Indicator The weaker the acid being titrated, the smaller the vertical area around eq. pt, that means less flexibility in choosing an indicator.

Example  Estimate the pH of a solution in which bromcresol green is blue and thymol blue is yellow.  A M HCl solution is made. 2 drops of methyl orange are added. What color is the solution?

Ch. 15: Applications of Aqueous Equilibria 15.6: Solubility Equilibria and Solubility Products

Solubility  As a salt dissolves in water and ions are released, they can collide and re- from the solid  Equilibrium is reached when the rate of ________________ equals the rate of _____________________ CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq)  ______________ solution When no more solid can dissolve at equilibrium When no more solid can dissolve at equilibrium

Solubility Product  Solubility product: K sp CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq)  Ksp=  Why do we leave out the CaF 2 ?  Adding more solid will not effect the amount of solid that can dissolve at a certain temperature  It would increase both reverse and forward reaction rates b/c there is a greater amount

K sp Values

Example 1  CuBr has a solubility of 2.0x10 -4 mol/L at 25°C. Find the K sp value. The solubility tells us the amount of solute that can dissolves in 1 L of water The solubility tells us the amount of solute that can dissolves in 1 L of water Use ICE chart: solubility tells you x value Use ICE chart: solubility tells you x value Ksp= Ksp= CuBr(s) ↔ Cu + (aq) + Br - (aq) CuBr(s) ↔ Cu + (aq) + Br - (aq) I C E

Example 2  The K sp value for Cu(IO 3 ) 2 is 1.4x10 -7 at 25°C. Calculate its solubility. Solve for solubility = x value using ICE chart Solve for solubility = x value using ICE chart K sp =[Cu 2+ ][IO 3 - ] 2 K sp =[Cu 2+ ][IO 3 - ] 2 Cu(IO 3 ) 2 (s) ↔ Cu 2+ (aq) + 2IO 3 - (aq) Cu(IO 3 ) 2 (s) ↔ Cu 2+ (aq) + 2IO 3 - (aq) I C E

Comparing Solubilities  You can only compare solubilities using K sp values for compounds containing the same number of ions CaSO 4 > CuCO 3 > AgI CaSO 4 > CuCO 3 > AgI K sp values: 6.1x10 -5 > 2.5x > 1.5x10 -6  Why can we use K sp values to judge solubility?  Can only compare using actual solubility values (x) when compounds have different numbers of ions

Common Ion Effect  Solubility of a solid is lowered when a solution already contains one of the ions it contains  Why?

Example 3  Find the solubility of CaF 2 (s) if the K sp is 4.0 x and it is in a M NaF solution. K sp =[Ca 2+ ][F - ] 2 K sp =[Ca 2+ ][F - ] 2 CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq) CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq) I C E

pH and solubility  pH can effect solubility because of the common ion effect  Ex: Mg(OH) 2 (s) ↔ Mg 2+ (aq) + 2OH - (aq) How would a high pH effect solubility? How would a high pH effect solubility? High pH = _______ [OH-]  solubility ____________ High pH = _______ [OH-]  solubility ____________  Ex: Ag 3 PO 4 (s) ↔ 3Ag + (aq) + PO 4 3- (aq) What would happen if H+ is added? What would happen if H+ is added? H + uses up _____ to make phosphoric acid H + uses up _____ to make phosphoric acid Eq. shifts to _______ - Solubility __________ Eq. shifts to _______ - Solubility __________

Ch. 15: Applications of Aqueous Equilibria 15.7: Precipitation and Qualitative Analysis

Precipitation  Opposite of dissolution  Can predict whether precipitation or dissolution will occur  Use Q: ion product Equals K sp but doesn’t have to be at ________________ Equals K sp but doesn’t have to be at ________________  Q > K: more reactant will form, ____________ until equilibrium reached  Q < K: more product will form, ____________

Example 1  A solution is prepared by mixing mL of 4.00x10 -3 M Ce(NO 3 ) 3 and mL 2.00x10 -2 M KIO 3. Will Ce(IO 3 ) 3 precipitate out? Calculate Q value and compare to K (on chart) Calculate Q value and compare to K (on chart) Ce(IO 3 ) 3 (s) ↔ Ce 3+ (aq) + 3IO 3 - (aq) Ce(IO 3 ) 3 (s) ↔ Ce 3+ (aq) + 3IO 3 - (aq) Q=[Ce 3+ ][IO 3 - ] 3 Q=[Ce 3+ ][IO 3 - ] 3

Example 2  A solution is made by mixing mL of 1.00x10 -2 M Mg(NO 3 ) 2 and mL of 1.00x10 -1 M NaF. Find concentration of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp =6.4x10 -9 ) MgF 2 (s)  Mg 2+ (aq) + 2F - (aq)  Need to figure out whether the concentrations of the ions are high enough to cause precipitation first  Find Q and compare to K

Example 2  Q > K so shift to left, precipitation occurs  Will all of it precipitate out??  No- we need to figure out how much is created using stoichiometry we need to figure out how much is created using stoichiometry then how much ion is left over using ICE chart then how much ion is left over using ICE chart  Like doing acid/base problem

Example 2 Mg 2+ (aq) + 2F - (aq) ↔ MgF 2 (s) Mg 2+ (aq) + 2F - (aq) ↔ MgF 2 (s) I C E MgF 2 (s) ↔ Mg 2+ (aq) + 2F - (aq) MgF 2 (s) ↔ Mg 2+ (aq) + 2F - (aq)I C E How much will be used if goes to completion? Some of it dissolved- how much are left in solution?

Example 2

Qualitative Analysis  Process used to separate a solution containing different ions using solubilities  A solution of 1.0x10 -4 M Cu + and 2.0x10 -3 M Pb 2+. If I - is gradually added, which will precipitate out first, CuI or PbI 2 ? 1.4x10 -8 =[Pb 2+ ][I - ] 2 = (2.0x10 -4 )[I - ] 2 1.4x10 -8 =[Pb 2+ ][I - ] 2 = (2.0x10 -4 )[I - ] 2 [I - ]= : [I - ]> than that to cause PbI 2 to ppt [I - ]= : [I - ]> than that to cause PbI 2 to ppt 5.3x =[Cu + ][I - ] = (1.0x10 -4 )[I - ] 5.3x =[Cu + ][I - ] = (1.0x10 -4 )[I - ] [I - ]= : [I - ]> than that to cause CuI to ppt [I - ]= : [I - ]> than that to cause CuI to ppt Takes a much lower conc to cause CuI to ppt so it will happen first Takes a much lower conc to cause CuI to ppt so it will happen first

Qualitative Analysis

Ch. 15: Applications of Aqueous Equilibria 15.8: Complex Ion Equilibria

Complex Ion Equilibria  _________________ Charged species containing metal ion surrounded by ligands Charged species containing metal ion surrounded by ligands  __________ Lewis bases donating electron pair to empty orbitals on metal ion Lewis bases donating electron pair to empty orbitals on metal ion Ex: H 2 O, NH 3, Cl-, CN-, OH- Ex: H 2 O, NH 3, Cl-, CN-, OH-  ___________________________ Number of ligands attached Number of ligands attached

Complex Ion Equilibria  Usually, the conc of the ligand is very high compared to conc of metal ion in the solution  Ligands attach in stepwise fashion Ag + + NH 3  Ag(NH 3 ) + Ag + + NH 3  Ag(NH 3 ) + Ag(NH 3 ) + + NH 3  Ag(NH 3 ) 2+ Ag(NH 3 ) + + NH 3  Ag(NH 3 ) 2+

Example 3  Find the [Ag + ], [Ag(S 2 O 3 ) - ], and [Ag(S 2 O 3 ) 2 3- ] in solution made with mL of 1.00x10 -3 M AgNO 3 with mL of 5.00 M Na 2 S 2 O 3. Ag + + S 2 O 3 2-  Ag(S 2 O 3 ) - K 1 =7.4x10 8 Ag + + S 2 O 3 2-  Ag(S 2 O 3 ) - K 1 =7.4x10 8 Ag(S 2 O 3 ) - + S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- K 2 =3.9x10 4 Ag(S 2 O 3 ) - + S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- K 2 =3.9x10 4  Because of the difference in conc between ligand and metal ion, the reactions can be assumed to go to completion

Example 3 Ag + + 2S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- Ag + + 2S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- I C E