III. Titrations In an acid-base titration, a basic (or acidic) solution of unknown [ ] is reacted with an acidic (or basic) solution of known [ ]. An indicator is a substance used to visualize the endpoint of the titration. At the equivalence point, the number of moles of acid equals the number of moles of base.

III. Visual of a Titration

III. Titration/pH Curves A titration or pH curve is a plot of how the pH changes as the titrant is added. It is possible to calculate the pH at any point during a titration. Multiple pH’s can be calculated, and the results plotted to create the theoretical titration curve.

III. Find the Equivalence Point! The keys to these types of problems are writing the titration equation and finding the equivalence point of the titration. The calculation then depends on what region of the titration curve you are in: 1)Before titration begins 2)Pre-equivalence 3)Equivalence point 4)Post-equivalence

III. Illustrative Problem Sketch the pH curve for the titration of 25.0 mL of M HCl with M NaOH.

III. Illustrative Problem Solution 1)Write the titration equation.  HCl (aq) + NaOH (aq)  H 2 O (l) + NaCl (aq) 2)Calculate the equivalence point. What volume of NaOH is needed to completely react with HCl?

III. Illustrative Problem Solution 3)Calculate initial pH before titration.  Since HCl is strong, M HCl has [H 3 O + ] = M, and pH = )Calculate the pH of some points in the pre-equivalence region.  As NaOH is added, the neutralization reaction OH - (aq) + H 3 O + (aq)  H 2 O (l) takes place.  We calculate the pH after addition of 5.00 mL of NaOH.

III. Illustrative Problem Solution To calculate the pH after 5.00 mL, we need to calculate initial moles of acid and the number of moles of base. We put these moles into a reaction chart.

III. Illustrative Problem Solution H 3 O + + OH -  2H 2 O Initial mol mol0 mol Change mol mol Final mol0 mol mol The H 3 O + leftover is in a larger volume so we calculate its concentration.

III. Illustrative Problem Solution We do the same thing for some other points: 10.0 mL, 15.0 mL, and 20.0 mL. Results summarized below. Volume (ml)Mol H 3 O + [H 3 O + ] (M)pH

III. Illustrative Problem Solution 5)Calculate the pH at the equivalence point.  For a strong-strong titration, pH always equals 7.00 at the equivalence point! 6)Calculate the pH of some points in the post-equivalence region.  In this region, the pH depends on the excess OH - added.

III. Illustrative Problem Solution To find the excess added, calculate how many mL past the equivalence point have been added, convert to moles, and divide by total volume. For 30.0 mL: pH can then be found from pOH.

III. Illustrative Problem Solution Again, calculate for additional points like 35.0, 40.0, and 50.0 mL. Results summarize below. Volume (mL)[OH - ] (M)pOHpH

III. Illustrative Problem Solution Now we plot the data points and sketch the pH titration curve!

III. Sample Problem A L sample of M KOH is being titrated with M HI. What is the pH after 10.0 mL, 25.0 mL, and 30.0 mL of the titrant have been added?

III. Weak Acid/Base Titrations The situation becomes a little more complicated when a weak acid/base is titrated with a strong base/acid. Again, the keys are to identify the titration reaction and the equivalence point. The method of calculating the pH will then depend on the region of the titration curve.

III. Four Different Regions For a weak acid titrated with a strong base, there are 4 regions as well: 1)Before titration: only HA in solution, so it’s a weak acid problem! 2)Pre-equivalence: a mixture of HA and A -, so it’s a buffer! 3)Equivalence point: only A - in solution, so it’s a weak base problem! 4)Post-equivalence: adding excess OH -, so it’s a dilution problem! For a weak base titrated with a strong acid, everything is just rewritten w/ conjugates!

III. Illustrative Problem A mL sample of M CH 3 COOH is being titrated with M NaOH. Calculate the pH before the titration begins, after 3.00 mL of the titrant have been added, at the equivalence point, and after mL of the titrant have been added. Note that K a = 1.75 x for acetic acid.

III. Illustrative Problem Solution 1)First, we need to write the titration eqn.  CH 3 COOH (aq) + OH - (aq)  CH 3 COO - (aq) + H 2 O (l) 2)Next, we calculate the equivalence point.

III. Illustrative Problem Solution 3)Before titration begins, it’s just a weak acid problem. CH 3 COOH + H 2 O  H 3 O + + CH 3 COO - Initial M---00 Change-x---+x Equil x---xx Solving this (with simplification), we get [H 3 O + ] = x 10 -4, so pH =

III. Illustrative Problem Solution 4)After 3.00 mL of M NaOH have been added, we will have a mixture of CH 3 COOH and CH 3 COO - in solution.  Since it’s a buffer, we can use the Henderson-Hasselbalch eqn.

III. Illustrative Problem Solution In the H-H equation, we need a ratio of CH 3 COO - to CH 3 COOH. After adding 3.00 mL, we are 3.00/10.00 to equivalence. We can use a relative concentration chart. CH 3 COOH + OH -  CH 3 COO - + H 2 O Relative Initial Change-3.00/ / Relative Final7.00/ /

III. Illustrative Problem Solution Now we just plug the relative final row into the H-H equation.

III. Illustrative Problem Solution 5)At the equivalence point, all CH 3 COOH has been converted to CH 3 COO -.  Initial moles of CH 3 COOH = moles of CH 3 COO - at the equivalence point, but the volume has increased.  Must calculate [CH 3 COO - ] at equiv. pt.

III. Illustrative Problem Solution Now we solve a weak base problem. CH 3 COO - + H 2 O  OH - + CH 3 COOH Initial M---00 Change-x---+x Equil x---xx Using K b = x and the simplification, x = x Thus, pOH = and pH = Note that pH does not equal 7.00!!

III. Illustrative Problem Solution 6)At mL of added titrant, we are 0.20 mL past equivalence, and the pH depends only on excess OH -. Thus: Of course, this means that pOH = and pH =

III. Sample Problem A mL sample of M pyridine is being titrated with M HCl. What’s the pH after 4.63 mL of the HCl has been added? Note that pyridine has a K b of 1.69 x

III. Sample Weak/Strong Curves Important aspects about pH curves for weak/strong titrations:  At equivalence, pH does not equal  At ½ equivalence, pH = pK a.

III. Polyprotic Acid Titration If the K a ’s are different enough, you will see multiple equivalence points. Since protons come off one at a time, 1 st equiv. pt. refers to the 1 st proton, 2 nd to the 2 nd, etc.

III. Detecting the Equiv. Pt. During a titration, the equivalence point can be detected with a pH meter or an indicator. The point where the indicator changes color is called the endpoint. An indicator is itself a weak acid that has a different color than its conjugate base.

III. Phenolphthalein

III. Indicators The indicator has its own equilibrium:  HIn (aq) + H 2 O (l)  In - (aq) + H 3 O + (aq) The color of an indicator depends on the relative [ ]’s of its protonated and deprotonated forms.  If pH > pK a of HIn, color will be In -.  If pH = pK a of HIn, color will be in between.  If pH < pK a of HIn, color will be HIn.

III. Selecting an Indicator

IV. Solubility In 1 st semester G-chem, you memorized solubility rules and regarded compounds as either soluble or insoluble. Reality is not as clear cut – there are degrees of solubility. We examine solubility again from an equilibrium point of view.

IV. Solubility Equilibrium If we apply the equilibrium concept to the dissolution of CaF 2(s), we get:  CaF 2(s)  Ca 2+ (aq) + 2F - (aq) The equilibrium expression is then:  K sp = [Ca 2+ ][F - ] 2 K sp is the solubility product constant, and just like any other K, it tells you how far the reaction goes towards products.

IV. Some K sp Values

IV. Calculating Solubility Recall that solubility is defined as the amount of a compound that dissolves in a certain amount of liquid (g/100 g water is common). The molar solubility is obviously the number of moles of a compound that dissolves in a liter of liquid. Molar solubilities can easily be calculated using K sp values.

IV. K sp Equilibrium Problems Calculating molar solubility is essentially just another type of equilibrium problem. You still set up an equilibrium chart and solve for an unknown. Pay attention to stoichiometry! Fe(OH) 3(s)  Fe 3+ (aq) + 3OH - (aq) Initial---00 Change---S3S Equil.---S3S

IV. Sample Problem Which is more soluble: calcium carbonate (K sp = 4.96 x ) or magnesium fluoride (K sp = 5.16 x )?

IV. The Common Ion Effect The solubility of Fe(OH) 2 is lower when the pH is high. Why?  Fe(OH) 2(s)  Fe 2+ (aq) + 2OH - (aq)  Le Châtelier’s Principle! common ion effect: the solubility of an ionic compound is lowered in a solution containing a common ion than in pure water.

IV. Sample Problem Calculate the molar solubility of lead(II) chloride (K sp = 1.2 x ) in pure water and in a solution of M NaCl.

IV. pH and Solubility As seen with Fe(OH) 2, pH can have an influence on solubility. In acidic solutions, need to consider if H 3 O + will react with cation or anion. In basic solutions, need to consider if OH - will react with cation or anion.

IV. Sample Problems a)Which compound, FeCO 3 or PbBr 2, is more soluble in acid than in base? Why? b)Will copper(I) cyanide be more soluble in acid or base? Why? c)In which type of solution is AgCl most soluble: acidic, basic, or neutral?

IV. Precipitation K sp values can be used to predict when precipitation will occur. Again, we use a Q calculation.  If Q < K sp, solution is unsaturated. Solution dissolve additional solid.  If Q = K sp, solution is saturated. No more solid will dissolve.  If Q > K sp, solution is supersaturated, and precipitation is expected.

IV. Sample Problems a)Will a precipitate form if mL M Pb(NO 3 ) 2 is mixed with mL M MgSO 4 ? b)The concentration of Ag + in a certain solution is M. What concentration of SO 4 2- is needed to precipitate out the Ag + ? Note that K sp = 1.2 x for silver(I) sulfate.

V. Complex Ions In aqueous solution, transition metal cations are usually hydrated.  e.g. Ag + (aq) is really Ag(H 2 O) 2 + (aq).  The Lewis acid Ag + reacts with the Lewis base H 2 O. Ag(H 2 O) 2 + (aq) is a complex ion.  A complex ion has a central metal bound to one or more ligands.  A ligand is a neutral molecule or an ion that acts as a Lewis base with the central metal.

V. Formation Constants Stronger Lewis bases will replace weaker ones in a complex ion.  e.g. Ag(H 2 O) 2 + (aq) + 2NH 3(aq)  Ag(NH 3 ) 2 + (aq) + 2H 2 O (l)  For simplicity, it’s common to write Ag + (aq) + 2NH 3(aq)  Ag(NH 3 ) 2 + (aq)  Since this is an equilibrium, we can write an equilibrium expression for it.

V. Formation Constants K f is called a formation constant. Unlike other equilibrium constants we’ve seen, K f ’s are large, indicating favorable formation of the complex ion. Ag + (aq) + 2NH 3(aq)  Ag(NH 3 ) 2 + (aq)

V. Sample Formation Constants

V. Calculations w/ K f ’s Since K f ’s are so large, calculations with them are slightly different. We assume the equilibrium lies essentially all the way to the right. This changes how we set up our equilibrium chart.

V. Illustrative Problem Calculate the concentration of Ag + ion in solution when g silver(I) nitrate is added to a mL solution that is 0.20 M in KCN.

V. Illustrative Problem Solution 1)First, we must identify the complex ion.  In solution, we will have Ag +, NO 3 -, and K +, and CN -.  The complex ion must be made from Ag + and CN -.  Looking at table of K f ’s, we find that Ag(CN) 2 - has K f = 1 x

V. Illustrative Problem Solution 2)Next, we need concentrations.  Already know that [CN - ] = 0.20 M.  We calculate the [Ag + ].

V. Illustrative Problem Solution 3)Now we set up our equilibrium chart.  Since K f is so big, we assume the reaction essentially goes to completion. Ag + (aq) + 2CN - (aq)  Ag(CN) 2 - (aq) Initial M0.20 M0 Change≈ ≈ ≈ Equil.x

V. Illustrative Problem Solution 4)Finally, we solve for x. Thus, [Ag + ] = 5 x It is very small, so our approximation is valid. Note that book would use 0.20 for [CN - ] in the calculation.

V. Sample Problem A mL sample of a solution that is M in NiCl 2 is mixed with a mL sample of a solution that is M in NH 3. After the solution reaches equilibrium, what concentration of Ni 2+ (aq) remains?

V. Complex Ions & Solubility Formation of complex ions enhances the solubility of some normally insoluble ionic compounds. Typically, Lewis bases will enhance solubility. e.g. Adding NH 3 to a solution containing AgCl (s) will cause more AgCl (s) to dissolve. AgCl (s)  Ag + (aq) + Cl - (aq) K sp = 1.77 x Ag + (aq) + 2NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7

V. Complex Ion Formation

V. Metal Hydroxides All metal hydroxides can act as bases.  e.g. Fe(OH) 3(s) + 3H 3 O + (aq)  Fe 3+ (aq) + 6H 2 O (l) Some metal hydroxides can act as acids and bases; they are amphoteric.  In addition to the above, Al(OH) 3(s) can also absorb hydroxide.  Al(OH) 3(s) + OH - (aq)  Al(OH) 4 - (aq)

V. Aluminum Hydroxide In acidic solutions:  Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) As OH - is added, solution becomes neutral.  Al(H 2 O) 5 (OH) 2+ (aq) + 2OH - (aq)  Al(H 2 O) 3 (OH) 3(s) + 2H 2 O (l) In basic solutions:  Al(H 2 O) 3 (OH) 3(s) + OH - (aq)  Al(H 2 O) 2 (OH) 4 - (aq) Thus, solubility is very pH dependent.

V. Aluminum Hydroxide

V. Amphoteric Hydroxides There are not that many metal hydroxides that are amphoteric. Only Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sn 2+.