1. Chem. 1B – 9/22 Lecture

2. 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

3. 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

4. Chem 1B – Aqueous Chemistry Acid-Base Properties of Ions Example Question: Determine if the ionic compounds are acidic or basic in the following examples: 1.NH 4 CN (one left from last time)

5. Chem 1B – Aqueous Chemistry Some Practice 1.Which solution will have a greater fraction of ionization? 0.10 M HClO vs. 0.10 M HF K a (HClO) = 2.9 x 10 -8 K a (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 K b value? 3.The K b for NH 3 is 1.76 x 10 -5. What is the pH of a solution initially made to 0.10 M NH 4 Cl?

6. Chem 1B – Aqueous Chemistry Polyprotic Acids Generic Example: H 2 A – has two protons that can be lost through acid reactions (diprotic) Some Examples: –H 2 SO 4 (sulfuric – first H + loss is strong acid) –H 2 SO 3 (sulfurous) –H 2 CO 3 (carbonic) –H 3 PO 4 (phosphoric – triprotic) Reaction of generic diprotic example 1)H 2 A(aq) ↔ H + (aq) + HA - (aq) K = K a1 2)HA - (aq) ↔ H + (aq) + A 2- (aq) K = K a2

7. 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 K a1 >> K a2 ) Example Problem: calculate [H 2 CO 3 ], [HCO 3 - ], pH, and [CO 3 2- ] for a 1.0 x 10 -3 M solution of H 2 CO 3

8. 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 H 2 CO 3 (carbonic acid), we have HCO 3 - and CO 3 2- as conjugate bases (from 1 st and then 2 nd weak acid reactions) 1)H 2 CO 3 (aq) ↔ H + (aq) + HCO 3 - (aq) K = K a1 2)HCO 3 - (aq) ↔ H + (aq) + CO 3 2- (aq) K = K a2 Salts allow us to “start” in the intermediate or basic form

9. Chem 1B – Aqueous Chemistry Polyprotic Acids – Salts of – cont. The most basic form (CO 3 2- ) can only be basic (it has no H + to lose), while the intermediate form (HCO 3 - ) can react as an acid or as a base Acid reaction: HCO 3 - (aq) ↔ H + (aq) + CO 3 2- (aq) Base reaction: HCO 3 - (aq) + H 2 O(l) ↔ H 2 CO 3 (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 = K a2 = 4.7 x 10 -11 Base reaction: K = K w /K a1 = 2.2 x 10 -8 so basic

10. Chem 1B – Aqueous Chemistry Polyprotic Acids – Salts of – cont. Rank the following salts from most acidic to most basic: KHSO 4 Na 3 PO 4 KHCO 3 KHC 2 O 4 AcidK a1 K a2 K a3 H 2 SO 4 >> 11.2 x 10 -2 H 3 PO 4 7.11 x 10 -3 6.32 x 10 -8 4.5 x 10 -13 H 2 CO 3 4.45 x 10 -7 4.69 x 10 -11 H2C2O4H2C2O4 5.60 x 10 -2 5.42 x 10 -5

11. 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 (C 2 H 5 OH) neutral while acetic acid (CH 3 CO 2 H) is acidic? acetate: stabilized by delocalized electrons ethanol anion (not very stable)

12. 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. HClO 2 more oxygens make acid stronger (HClO 4 is a strong acid, HClO is a very weak acid)

13. 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

14. Chem 1B – Aqueous Chemistry Buffers (Chapter 16) Example: Determine pH of a mix of 0.010 M HCHO 2 and 0.025 M Na + CHO 2 - solution

15. 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 < pK a < 11) Ratio of weak acid to conjugate base in range 0.1 to 10 mM+ concentration range

16. Chem 1B – Aqueous Chemistry Buffers (Chapter 16) Buffer Solutions: –Since ICE not needed, can just use K a equation –K a = [H + ][A - ]/[HA] = [H + ][A - ] o /[HA] o (always valid) (valid for traditional buffer) –But log version more common –pH = pK a + log([A - ]/[HA]) –Also known as Henderson-Hasselbalch Equation

17. 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 NH 3 + 0.100 M NH 4 Cl in 1.00 L –Calculate the pH –Now lets add 0.005 moles of HCl. What is the new pH?