Aqueous Equilibria Buffers, Titration and Solubility Chapter 17

Common Ion Effect Common Ion: ion produced by more than one species in an equilibrium equation -solution of HF and its salt NaF -major species: -common ion:

Common Ion Effect What would happen if additional F- was added (can be in the form of addl. NaF) -think about the strength of the acid and Le Chatelier's Principle

Common Ion Effect NH 3 + H 2 O NH OH - -add NH 4 Cl -what is the common ion, what shift, pH?

Common Ion Problem 1a What is the pH of a solution made by adding 0.30mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0L of solution?

Common Ion Problem 1b What is the pH of the acetic acid without the salt added?

Common Ion Problem 2 Calculate the pH of a solution containing 0.085M nitrous acid and 0.10M potassium nitrite.

Common Ion Problem 3 Calculate the fluoride ion concentration and pH of a solution that is 0.20M in HF and 0.10M in HCl.

Common Ion Problem 4 Calculate the formate ion concentration and pH of a solution that is 0.050M in formic acid (HCOOH, K a = 1.8x10 -4 ) and 0.10M in HNO 3.

Buffered Solutions Buffered solution: resists a change in pH when either OH- or H+ is added. -made of a weak acid and its salt OR a weak base and its salt -think of buffer like a “sponge”, it removes H+ or OH- from solution -neutralizes the added OH- or H+ -add a strong base: -add a strong acid:

Making a Buffer 1.Combine a weak acid with its salt HC 2 H 3 O 3 and NaC 2 H 3 O 2 2. Combine a weak base with its salt CH 3 NH 2 and CH 3 NH 3 Cl

Making a Buffer 3. Combine a weak acid and 1/2M strong base 2.0M HC 2 H 3 O 2 and 1.0M NaOH 4. Combine a weak base and 1/2M strong acid 2.0M NH 3 and 1.0M HCl

Buffer Concept Problem Given a buffer of HF and NaF… 1.What happens when HCl is added? 2. What happens when NaOH is added?

Henderson-Hasselbalch Equation H-H is used to determine the pH of a buffer pH = pK a + log([base]/[acid]) pOH = pK b + log([acid]/[base])

Calculating the pH of a Buffer 1 What is the pH of a buffer that is 0.12M in lactic acid and 0.10M in sodium lactate?

Calculating the pH of a Buffer 2 Calculate the pH of a buffer composed of 0.12M benzoic acid and 0.20M sodium benzoate.

Preparing a Buffer 1 How many moles of NH 4 Cl must be added to 2.0L of 0.10M NH 3 to form a buffer whose pH is 9.00? (assume addition of NH 4 Cl does not change the total volume)

Preparing a Buffer 2 Calculate the concentration of sodium benzoate that must be present in a 2.0M solution of benzoic acid to produce a pH of 4.00.

Buffering Capacity The amount of H + or OH - a buffer can absorb without a significant change in pH -capacity is determined by the magnitude of [HA] and [A - ] (or [B] and [BH + ] -higher molarities can absorb more H + or OH -

Buffering Capacity Concept Which would be the best buffer? Why? a.1.0M HC 2 H 3 O 2 and 1.0M NaC 2 H 3 O 2 b.0.50M HC 2 H 3 O 2 and 1.0M NaC 2 H 3 O 2 c.1.0M HC 2 H 3 O 2 and 0.5M NaC 2 H 3 O 2 d.0.50M HC 2 H 3 O 2 and 0.50M NaC 2 H 3 O 2

Buffer Review/Practice 1 What is the pH of a buffer made of 2.00 g of benzoic acid (C 6 H 5 CO 2 H) and 2.00 g of sodium benzoate in 1.00 L of solution.

Buffer Review/Practice 2 What is the pH of a 1.00L buffer solution that is made of 15.0g of sodium hydrogen carbonate (NaHCO3) and 18.0g of sodium carbonate

Buffer Review/Practice 3 How would you prepare 1.0L of a buffer solution with acetic acid and sodium acetate that has a pH of 4.30?

Buffer Review/Practice Side Note the Henderson-Hasselbalch equation assumes that the equilibrium concentrations are equal to the initial concentrations -pH buffer = pKa of the weak acid -in the previous problem, the pKa was greater than the pH of the buffer -- what did you observe about the ratio of concentrations of the acid and the conjugate base? -in the next problem, you determine how to prepare a buffer solution where the pKa is less than the pH -what will the ratio of acid/conj. base be?

Buffer Review/Practice 4 How would you prepare 1.0L of a buffer solution using acetic acid and sodium acetate so the pH = 5.0?

Buffer Review/Practice 5 Write a chemical equation for the acetic acid/sodium acetate buffer from the previous problems. Identify the acid and conjugate base. What is the purpose of the acid? What is the purpose of the base?

Titration volumetric analysis commonly used to determine the amount of acid or base in a solution -a solution of known concentration (the titrant) is delivered from a buret into a solution fo unknown concentration until the titrant is consumed -this is the stoichiometric point or equivalence point -equal moles of acid and base

Titration -an acid-base titration is often monitored by plotting the pH of the solution being titrated as a function of the volume of the titrant added -pH curve or a titration curve -in our titration last semester, we used an indicator that changed color (end point) at approximately the equivalence point

pH Curves equivalence point: -[H+] = [OH-] end point: -where indicator changes color -phenolphthalein changes color at pH=9 -does that represent a significant difference in amount of NaOH used?

pH Curves HCl + NaOH → NaCl + H 2 O H + + OH - → H 2 O Fast way to determine molarity for SA/SB titration n A M A V A = n B M B V B

initial pH = pH of weak acid between initial and equiv. = H-H -pH = pKa -½ mol of OH- = 1 mol HA equivalence point -equal mol OH- and HA pH due to CB hydrolysis above equiv. point = pH of strong base Weak Acid - Strong Base Titration

the weaker the acid, the greater the pH at the equiv. pt. -pH is due to conjugate base Weak Acid - Strong Base Titration

Titration of Polyprotic Acids -multiple equivalence points graph shows the titration of 50.0ml 0.10M H 3 PO 3 with 0.10M NaOH

Indicators

weak acid - complex molecules HIn H + + In - phenolphthalein - HIn = colorless In - = pink pH = pKa +/- 1 -acidic to basic pH = pKa - 1 -basic to acidic pH = pKa + 1

Solubility Equilibria dissolution or precipitation of ionic compounds -heterogeneous (we have been dealing with same phase solutions) BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) -extent of dissolution depends on solubility-product constant (K sp ) -Appendix D -greater value of K sp = more dissolves

Does Ksp = Solubility? No! Solubility is the amount of solute that is dissolved to form a saturated solution (g solute/L solution) -molar solubility: mol/L Ksp is an equilibrium constant -measure of how much of a solid can dissolve

Relative Solubilities K sp values can be used to compare solubility of salts only if the salts produce the same number of ions -larger K sp = most soluble

K sp Concept Question 1 Which 2 can be compared: calcium sulfate, calcium hydroxide, calcium fluoride, calcium phosphate Of those 2, which is the most soluble in water at 25C

K sp Concept Question 2 Which of the following compounds has the greatest molar solubility in water a.AgCl b.AgBr c.AgI

Calculations with K sp

K sp Calculation 1 - finding K sp Calculate K sp for copper(I)bromide that has a solubility of 2.0x10 -4 M

K sp Calculation 2 - finding K sp Calculate K sp for Bi 2 S 3. The solubility is 1.0x at 25 C

K sp Calculation 3 - finding solubility Calculate the solubility at 25C for copper (II) iodate. K sp = 1.4x10 -7

K sp and Common Ions -an ionic solid is dissolved in an aqueous solution which contains an ion in common with the dissolving salt -treat like you did common ions/buffers: put the concentration of the already dissolved salt in the “I” row of the ICE chart

K sp Calculation 4 - common ion Calculate the solubility of Ag 2 CrO 4 in 0.100M solution of AgNO 3 (aq). The K sp of Ag 2 CrO 4 is 9.0x10 -12

pH and Solubility the pH of a solution can greatly affect the solubility -dissolving a solute in a solution that has an excess of H + or OH - Example: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) -dissolving Mg(OH) 2 in a basic solution will decrease its solubility -equilibrium will shift left -dissolving Mg(OH) 2 in an acidic solution will increase its solubility -OH - will react with the H + and will be removed from the reaction -equilibrium will shift right

pH and Solubility Many compounds have increased solubility when dissolved in an acidic solution -they contain a strong conjugate base which will react with the H+ in the acid Example: Dissolve Ca(NO 2 ) 2 in an acidic solution

pH and Solubility Dissolve AgI in an acidic solution. Will solubility increase?

Precipitation -Formation of a solid from solution -memorize some solubility rules (group I, nitrate, ammonium) -Use Q and K sp to predict whether a precipitate will form when 2 solutions are mixed -insoluble compounds Q = initial concentrations K sp = equilibrium concentrations -K sp <Q: precipitation occurs -K sp =Q: equilibrium condition (saturated solution) -K sp >Q: solid dissolves, increasing ion concentration until Q=K

Precipitation Example 1 Add 750.0mL of 4.00x10 -3 M Ce(NO 3 ) 3 to mL of 2.00x10 -2 M KIO 3. Will Ce(IO 3 ) 3 precipitate? The Ksp of Ce(IO 3 ) 3 is 1.9x

Selective Precipitation Remove cations or anions from solution by selecting reagents that will cause them to form precipitates Example: Pb(NO 3 ) 2 (aq) -to remove Pb 2+ : add KI 2KI(aq) + Pb(NO 3 ) 2 → 2KNO 3 (aq) + PbI 2 (aq)