Chem. 1B – 9/22 Lecture

Announcements I

• Exam 1– On Oct. 1 (week from next Thurs.)– Some example exams posted (my last Exam 2

for this class is closest to this material)– Also Mr. Spark’s website has an example exam

posted (see link on my website)

• Mastering Chemistry– Chapter 15A assignment due Thurs.– Longer than previous one– Some questions are a little different than

examples given so far

Announcements II

• Today’s Lecture – Chapter 15/16 Topics– Acid-Base Properties of Ions and Salts– More Problem Practice– Polyprotic Acids– Relating Acid Strength to Molecular Structure– Chapter 16 – Section 16.2: Buffer Solutions

Chem 1B – Aqueous Chemistry

Acid-Base Properties of IonsExample Question: Determine if the ionic

compounds are acidic or basic in the following examples:

1. NH4CN (one left from last time)

Chem 1B – Aqueous ChemistrySome Practice

1. Which solution will have a greater fraction of ionization? 0.10 M HClO vs. 0.10 M HFKa(HClO) = 2.9 x 10-8 Ka(HF) = 3.5 x 10-4

2. An unknown base is dissolved in water so that its initial molarity is 0.050 M. The pH is measured and found to be 10.13. What is its Kb value?

3. The Kb for NH3 is 1.76 x 10-5. What is the pH of a solution initially made to 0.10 M NH4Cl?

Chem 1B – Aqueous Chemistry

Polyprotic Acids• Generic Example: H2A – has two protons that

can be lost through acid reactions (diprotic)• Some Examples:

– H2SO4 (sulfuric – first H+ loss is strong acid)

– H2SO3 (sulfurous)

– H2CO3 (carbonic)

– H3PO4 (phosphoric – triprotic)

• Reaction of generic diprotic example1) H2A(aq) ↔ H+(aq) + HA-(aq) K = Ka1

2) HA- (aq) ↔ H+(aq) + A2-(aq) K = Ka2

Chem 1B – Aqueous Chemistry

Polyprotic Acids – in Problems• Solving polyprotic acid problems can be

challenging (the concentrations of the products from the first reaction affect the equilibrium in the second reaction)

• To simplify the problem, we assume the two reactions occur independently (valid if Ka1 >> Ka2)

• Example Problem: calculate [H2CO3], [HCO3

-], pH, and [CO32-] for a 1.0 x 10-3 M

solution of H2CO3

Chem 1B – Aqueous Chemistry

Polyprotic Acids – Salts of• While conjugate bases of monoprotic weak

acids can only be basic, conjugate bases of polyprotic acids may be acidic or basic

• Example: from H2CO3 (carbonic acid), we have HCO3

- and CO32- as conjugate bases

(from 1st and then 2nd weak acid reactions)1) H2CO3 (aq) ↔ H+(aq) + HCO3

-(aq) K = Ka1

2) HCO3-(aq) ↔ H+(aq) + CO3

2-(aq) K = Ka2

• Salts allow us to “start” in the intermediate or basic form

Chem 1B – Aqueous Chemistry

Polyprotic Acids – Salts of – cont.• The most basic form (CO3

2-) can only be basic (it has no H+ to lose), while the intermediate form (HCO3

-) can react as an acid or as a base

Acid reaction: HCO3-(aq) ↔ H+(aq) + CO3

2-(aq)

Base reaction: HCO3-(aq) + H2O(l) ↔ H2CO3(aq) + OH-(aq)

• To determine the acidity of the intermediate form, we must compare K values for the acid and base reactions

Acid reaction: K = Ka2 = 4.7 x 10-11

Base reaction: K = Kw/Ka1 = 2.2 x 10-8 so basic

Chem 1B – Aqueous Chemistry

Polyprotic Acids – Salts of – cont.• Rank the following salts from most acidic

to most basic:KHSO4 Na3PO4 KHCO3 KHC2O4

Acid Ka1 Ka2 Ka3

H2SO4 >> 1 1.2 x 10-2

H3PO4 7.11 x 10-3 6.32 x 10-8 4.5 x 10-13

H2CO3 4.45 x 10-7 4.69 x 10-11

H2C2O4 5.60 x 10-2 5.42 x 10-5

Chem 1B – Aqueous Chemistry

Molecular Structure – Acidity Relationship• Acid strength depends on ability for H

bond to break and on stability of conjugate base formed

• More stable conjugate bases means stronger acid

• For example, what makes ethanol (C2H5OH) neutral while acetic acid (CH3CO2H) is acidic?

H

O-

CH3H

acetate: stabilized by delocalized electrons

OCH3

O

-

ethanol anion (not very stable)

Chem 1B – Aqueous Chemistry

Molecular Structure – Acidity Relationship• General Rules for Simpler Structures:

– Binary Acids: e.g. HCl• more electronegative element makes for stronger

acid• longer (and weaker bond) makes for stronger acid

(HCl is stronger than HF due to bond strength)

– Oxyacids: e.g. HClO2

• more oxygens make acid stronger (HClO4 is a strong acid, HClO is a very weak acid)

Chem 1B – Aqueous Chemistry

Buffers (Chapter 16)• We have discussed some mixtures briefly

(e.g. strong acid + weak acid)• One particular type of mixture: acid +

conjugate base (or base + conjugate acid) makes a solution called a buffer

• Buffers are desirable because they keep the pH nearly constant even if an acid or base is added

• Buffers are very important in Biology because many enzymes (a protein catalyst) will only work over a narrow pH range

Chem 1B – Aqueous Chemistry

Buffers (Chapter 16)• Example: Determine pH of a mix of 0.010 M

HCHO2 and 0.025 M Na+CHO2- solution

Chem 1B – Aqueous Chemistry

Buffers (Chapter 16)• 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

Chem 1B – Aqueous Chemistry

Buffers (Chapter 16)• 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

Chem 1B – Aqueous Chemistry

Buffers (Chapter 16)• Addition of small amounts of acid to

a buffer:– Example: let’s say we have a buffer

made to be 0.050 M NH3 + 0.100 M NH4Cl in 1.00 L

– Calculate the pH– Now lets add 0.005 moles of HCl. What

is the new pH?