Chem. 31 – 4/20 Lecture

Announcements I• Exam 2

– Result: average = 75– Distribution similar to

Exam 1 (except a few more high and low scores)

• Lab Reports– AA regular report due

today– Soda Ash report due

4/27– Will be posting

information about Formal Report soon

100s 90s 80s 70s 60s 50s <500

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Axis Title

Announcements II• Today’s Lecture

– Chapter 8: Acid Base Chemistry• Weak Acid Problems• Weak Base Problems• Buffer Solutions

Acid – Base Equilibria (Ch. 8)

• Weak Acid Problems:– e.g. What is the pH and the concentration of

major species in a 2.0 x 10-4 M HCO2H (formic acid, Ka = 1.80 x 10-4) solution ?

– Can use either systematic method or ICE method.

– Systematic method will give correct answers, but full solution results in cubic equation

– ICE method works most of the time– Use of systematic method with assumptions

allows determining when ICE method can be used

Acid – Base Equilibria

• Weak Acid Problem – cont.:– Systematic Approach (HCO2H = HA to make

problem more general where HA = weak acid)• Step 1 (Equations) HA ↔ H+ + A-

H2O ↔ H+ + OH-

• Step 2: Charge Balance Equation: [H+] = [A-] + [OH-] 2 assumptions possible: ([A-] >> [OH-] – assumption used

in ICE method or [A-] << [OH-])• Step 3: Mass Balance Equation: [HA]o = 2.0 x 10-4 M

= [HA] + [A-]• Step 4: Kw = [H+][OH-] and Ka = [A-][H+]/[HA]• Step 5: 4 equations (1 ea. steps 2 + 3, 2 equa. step

4), unk.: [HA], [A-] [H+], [OH-]

Acid – Base Equilibria• Weak Acid Problem – cont.:

– Assumption #1: [A-] >> [OH-] so [A-] = [H+]– Discussion: this assumption means that we

expect that there will be more H+ from formic acid than from water. This assumption makes sense when [HA]o is large and Ka is not that small (valid for [HA]o>10-6 M for formic acid)

– ICE approach (Gives same result as systematic method if assumption #1 is made)

– (Equations) HA ↔ H+ + A-

Initital 2.0 x 10-4 0 0 Change - x +x +x Equil. 2.0 x 10-4 – x x x

Acid – Base Equilibria

• Weak Acid Problem – Using ICE Approach– Ka = [H+][A-]/[HA] = x2/(2.0 x 10-4 – x)x = 1.2 x 10-4 M (using quadratic equation)Note: sometimes (but not in this case), a 2nd

assumption can be made that x << 2.0 x 10-4 to avoid needing to use the quadratic equation

[H+] = [A-] = 1.2 x 10-4 M; pH = 3.92[HA] = 2.0 x 10-4 – 1.2 x 10-4 = 8 x 10-5 M

Note: a = fraction of dissociation = [A-]/[HA]total

a = 1.2 x 10-4 /2.0 x 10-4 = 0.60

Acid – Base Equilibria

• Weak Acid Problem – cont.:– When is Assumption

#1 valid (in general)?– When both [HA]o and

Ka are high or so long as [H+] > 10-6 M

– More precisely, when [HA]o > 10-6 M and Ka[HA]o > 10-12

– See chart (shows region where error < 1%)

– Failure also can give [H+] < 1.0 x 10-7 M

Where ICE Method Works

1.00E-07

1.00E-06

1.00E-05

1.00E-04

1.00E-03

1.00E-02

1.00E-01

1.00E+00

1E-10 1E-08 1E-06 0.0001 0.01 1

Ka

[HA

]o

Assupmption #1 Works

Fails

Acid – Base Equilibria• Weak Base Problem:

– As with weak acid problem, ICE approach can generally be used (except when [OH-] from base is not much more than [OH-] from water)

– Note: when using ICE method, must have correct reaction

– Example: Determine pH of 0.010 M NH3 solution (Ka(NH4

+) = 5.7 x 10-10, so Kb = Kw/Ka = 1.75 x 10-5)

– ReactionNH3 + H2O NH4

+ + OH-

– Go over on board

Acid – Base Equilibria

• Weak Acid/Weak Base Questions:– A solution is prepared by dissolving 0.10 moles

of NH4NO3 into water to make 1.00 L of solution. Show how to set up this problem for determining the pH using the ICE method.

– A student is solving a weak base problem for a weak base initially at 1.00 x 10-4 M using the ICE method and calculates that [OH-] = 2.4 x 10-8 M. Was the ICE method appropriate?

– The pH of an unknown weak acid prepared to a concentration 0.0100 M is measured and found to be 3.77. Calculate a and Ka.

Acid – Base Equilibria

• Buffer Solutions:– A buffer solution is designed so that a small

addition of acid or base will only slightly change the pH

– Most buffer solutions have a weak acid and its conjugate base both present

– Example: Determine pH of a mix of 0.010 M HCO2H and 0.025 M Na+HCO2

- solution (ignoring activity)

– Go to board to show if ICE approach is needed

• Buffer Solutions:– Question: Was the ICE Problem set up

needed?– Answer: No. The assumption of x <<

[HA], [A-] is valid for all “traditional” buffers

– Traditional Buffer• Weak acid (3 < pKa < 11)• Ratio of weak acid to conjugate base in

range 0.1 to 10• mM+ concentration range

Acid – Base Equilibria

Acid – Base Equilibria

• Buffer Solutions:– Since ICE not needed, can just use Ka

equation– Ka = [H+][A-]/[HA] = [H+][A-]o/[HA]o

(always valid) (valid for traditional buffer)

– But log version more common– pH = pKa + log([A-]/[HA])– Also known as Henderson-Hasselbalch

Equation

Acid – Base Equilibria

• Buffer Solutions:– Ways to make buffer solution:

• Mix weak acid and conjugate base (done in making reference solution for soda ash lab)

• Add strong base to weak acid (weak acid must be in excess) – this converts some of the weak acid to its conjugate base

• Add strong acid to weak base (weak base must be in excess) – this converts some of weak base to its conjugate acid