Acid-Base Equilibria Sections (Unit 18A)

15.1 Common Ion Effect Suppose we have a solution containing the weak acid hydrofluoric acid (K a =7.2x10 -4 ) and its salt NaF. Remember that when a strong electrolyte dissolves, it completely dissociates. –What is the common ion? –How does this ion effect pH?

HF (aq)↔ H + (aq) +F - (aq) Added F - ions from NaF Equilibrium shift. Form more HF. Fewer H + ions present. Less acidic.

NH 3 (aq) + H 2 O↔ NH 4 + (aq) +OH - (aq) Added NH 4 + ions from NH 4 Cl Equilibrium shift. Form more NH 3 and H 2 O. Fewer OH - ions present. Less basic.

Equilibrium Calculations Use the same procedures for finding pH of a solution with a weak acid or base as last unit (use ICE). The only difference is the initial concentration of the common ion will NOT be zero. Example on pg. 740

15.2 Buffered Solutions Buffered solutions resist a change in pH regardless of whether OH - or H + is added. –Most common example is our blood, which maintains a constant pH so that cells can survive. –Contain a weak acid and its common ion salt, such as HF and NaF. –Contain a weak base and its common ion salt, such as NH 3 and NH 4 Cl.

1.Buffered solutions are simply weak acids or bases with a common ion. 2.pH calculations for buffered solutions are exactly the same procedures as last unit. 3.When a strong acid or base is added to a buffered solution, deal with stoichiometry first, then equilibrium calculations.

How does buffering work? Buffer solution of a weak acid (HA) and its conjugate base (A - ).

K a = [H + ][A - ] or [H + ]= K a [HA] [HA] [A - ] –If you add hydroxide ions, OH - + HA → A - + H 2 O –HA is converted to A - ; [HA]/[A - ] decreases; [H + ] decreases; pH decreases.

Buffered Solutions –If amounts of HA and A - originally present are large compared to OH -, change in [HA]/[A - ] is small. Therefore, [H + ] and pH remain constant.

K a = [H + ][A - ] or [H + ]= K a [HA] [HA] [A - ] –If you add protons, H + + A - → HA –A - converted to HA ; [HA]/[A - ] increases; [H + ] increases; pH increases. –If [A - ] and [HA] are large compared to [H + ] added, then there will be little change in the pH.

Buffer Calculations Calculating [H + ]: [H + ]= K a [HA] [A - ] Henderson-Hasselbalch Equation -log[H + ]=-log(K a ) –log ([HA]/[A - ]) pH=pK a -log([HA]/[A - ]) pH=pK a +log([A - ]/[HA]) pH=pK a +log(base/acid)

See the Summary of the Most Important Characteristics of Buffered Solutions on page 692. Examples 15.21, 15.23, and odd on pages

15.3 Buffering Capacity Buffering capacity is the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. –pH depends on the [A - ]/[HA] ratio –Buffering capacity depends on the magnitudes of [HA] and [A - ]

Optimal Buffering –When [HA]=[A - ], so that [A - ]/[HA]=1. –pK a =pH (or as close as possible)

Examples odd on pg. 741

15.4 Titrations and pH Curves Titration: used to determine the amount of acid or base in a solution, involving a solution of known concentration (titrant), indicator, and unknown solution. pH curve or titration curve: a plot of the pH of an acid- base titration as a function of the amount of titrant added.

Strong Acid-Strong Base Titrations Net Ionic Reaction: H + (aq) + OH - (aq)→ H 2 O (l) Computing [H + ]: –Determine mmol of H + at that point and divide by total volume of solution –1 mmol= (1 mol/1000) = mol *used because titrations involve small quantities –Molarity = mol solute = (mol solute/1000) = mmol solute L solution (L soltn/1000) mL soltn

Case Study 50.0 mL of M HNO 3 is titrated with M NaOH. We will calculate the pH of the solution at selected points during the course of the titration. –A. No NaOH has been added. [H + ]=0.200 M and pH=0.699

–B mL of M NaOH has been added. [H + ]=9.00 mmol/( )= 0.15 M and pH=0.82 –C mL (total) of M NaOH has been added. [H + ]=8.00 mmol/( )= 0.11 M and pH=0.942 H + +OH - → H2OH2O Before50.0 mL x M =10.0 mmol 10.0 mL x M =1.00 mmol Reaction-1.00 mmol 1.00 mmol After9.00 mmol left0 mmol1.00 mmol H + +OH - → H2OH2O Before50.0 mL x M =10.0 mmol 20.0 mL x M =2.00 mmol Reaction-2.00 mmol 2.00 mmol After8.00 mmol left0 mmol2.00 mmol

–D mL (total) of M NaOH has been added. Proceed exactly as for points B and C. pH=1.301

–E mL (total) of M NaOH has been added. Stoichiometric point or equilvalence point (pH=7.00) –F mL (total) of M NaOH has been added. [OH - ]=5.00 mmol/( )= M, pOH=1.60, pH=12.40 H + +OH - → H2OH2O Before50.0 mL x M =10.0 mmol mL x M =10.0 mmol Reaction-10.0 mmol 10.0 mmol After0 mmol 10.0 mmol H + +OH - → H2OH2O Before50.0 mL x M =10.0 mmol mL x M =15.0 mmol Reaction-10.0 mmol 10.0 mmol After0 mmol5.0 mmol left10.0 mmol

–G mL (total) of M NaOH has been added Proceed the same as point F. pH = 12.60

Characteristics of Strong-Strong pH Titration Curves pH changes are gradual until the titration is close to the equivalence point, due to the large amount of original H + or OH - ions depending on the titration. At the equivalence point, pH=7.00.

Titrations of Weak Acids with Strong Bases Calculating the pH curve: 1.Stoichiometry problem (BRA): The rxn of OH - with the weak acid runs to completion, and [remaining acid] and [formed conjugate base] are determined. 2.Equilibrium problem (ICE): The position of the weak acid equilibrium is determined; pH is calculated.

Case Study 50.0 mL of 0.10 M acetic acid (K a =1.8x10 -5 ) is titrated with 0.10 M NaOH. We will calculate pH at various points. A. No NaOH has been added. –Complete this the same as last unit. –pH=2.87 (check it yourself) B mL of 0.10 M NaOH has been added. OH - +CH 3 COOH → CH 3 COO - + H 2 O Before10.0 mL x 0.10 M =1.0 mmol 50.0 mL x 0.10 M =5.0 mmol 0 mmol Reaction-1.0 mmol +1.0 mmol After0 mmol4.0 mmol left1.0 mmol

Since only CH 3 COOH and CH 3 COO - are the only species left in solution: –K a =1.8x10 -5 = [H + ][CH 3 COO - ] = (0.017)x [CH 3 COOH] –X=7.2 x M = [H + ] –pH=4.14 CH 3 COOH → CH 3 COO - + H + Initial 4.0 mmol ( ) mL 1.0 mmol ( ) mL 0 Change-x+x Equilibrium0.067-x ≈ x ≈ 0.017x

–C mL (total) of 0.10 M NaOH has been added This is the halfway point. [CH 3 COOH]=[CH 3 COO - ] [H + ]=K a pH=pK a –D mL (total) of 0.10 M NaOH has been added. This is the equivalence point because 5.0 mmol of OH - will react with the 5.0 mmol of CH 3 COOH originally present. Species left is CH 3 COO -. –CH 3 COO - + H 2 O (l) ↔ CH 3 COOH (aq) + OH - –Find K b. –Determine pH.

–E mL (total) of 0.10 M NaOH has been added At this point, excess OH - has been added. pH is determined by the excess OH - because it is a strong base [OH - ]=1.0 mmol / ( )mL= 9.1x10 -3 M pOH=2.04 pH=11.96 OH - +CH 3 COOH → CH 3 COO - + H 2 O Before60.0 mL x 0.10 M =6.0 mmol 50.0 mL x 0.10 M =5.0 mmol 0 mmol Reaction-5.0 mmol +5.0 mmol After1.0 mmol in excess4.0 mmol left5.0 mmol

Comparisons: Near the beginning, pH increases more rapidly than strong/strong. Weak curve levels off near the halfway point due to buffering. Weak curve, the pH at the equivalence point > 7 due to basicity of the conjugate base. Shapes of both curves are the same after the equivalence point. *Remember that equivalence point is determined by stoichiometry (enough titrant has been added to react exactly with acid or base), not the pH.

The amount of acid present determines the equivalence point, not its strength. pH value at the equivalence point is affected by the acid strength. –Weak acid=high pH Notice the difference in the shapes of the curves.

Titrations of Weak Bases with Strong Acids 1 st : Think about the major species in solution. 2 nd : Decide whether a reaction occurs that runs to completion and do stoichiometric calculations. 3 rd : Choose the dominant equilibrium and calculate pH. -See Case Study on page 709

Note that the pH at the equivalence point < 7 since the solution contains the weak acid NH 4 +. Examples odd on pages

15.5 Acid-Base Indicators 2 methods for determining equivalence points: –Use a pH meter to plot the titration curve. The center of the vertical region indicates the equivalence point. –Use an acid-base indicator, which marks the end point by changing colors. Although equivalence point≠end point, careful selection of the indicator will ensure error is negligible.

Common Acid-Base Indicators are Complex Molecules HIn: represents the acid form of a complex molecule that is a particular color or colorless. In - : represents the conjugate base of the complex molecule that shows as a different color.

Assume that the ratio of [In - ]/[HIn] must be 1/10 for the human eye to detect a color change. a.) Yellow acid form of bromthymol blue; b.) a greenish tint is seen when the solution contains 1 part blue and 10 parts yellow; c.) blue basic form

Henderson-Hasselbalch equation useful to determine pH at the indicator color change. –Acid Solution being titrated. pH=pK a + log ([In - ]/[HIn]) pH=pK a + log (1/10)= pK a -1 –Basic solution being titrated (indicator initially in the form of In - ) pH=pK a + log ([In - ]/[HIn]) pH=pK a + log(10/1)=pK a +1

The useful pH range for an indicator is pK a (indicator)± 1.

Choosing an Indicator The indicator end point (when it changes colors) should be as close as possible to the titration equivalence point. –What indicator would you use for the titration of mL of M HCl with M NaOH? Equivalence point occurs at pH of 7.00 (strong/strong) Titrating an acid; pH=pK a -1 –pK a =pH+1=8 –K a =1x10 -8

Indicators for strong acid-strong base titrations –Color changes will be sharp, occurring with the addition of a single drop of titrant. –There is a wide choice of suitable indicators.

Titration of Weak Acids –Smaller vertical area around the equivalence point. –Less flexibility in choices of indicators. –Choose an indicator whose useful pH range has a midpoint as close as possible to the equivalence point.

Examples 15.65, 15.69, and on pages THE END!!