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Chem. 1B – 9/22 Lecture Announcements I • Next Stuff that is due – Mastering (15b – on Saturday, 16a - buffers on Tues.) – Quiz 4 (Experiment 2 + Polyprotic Acids + Buffers) – Exam 1 (Thursday) – Experiment 7 due (Wed./Thurs.) • Exam #1 – Next week on Thursday (9/29) – Same format as last year – Covers material covered through buffers (covering today) Announcements II • Exam 1 – cont. – Possible Help Session Tues., 4:00 to 5:00 time frame in Sequoia 452 – PAL lead session Monday 4-9; Sequoia 443 – Everyone here but Lab Sect. 7 (in Sequoia 426) – Need to bring Scantron Form SC982-E • Today’s Lecture – Buffers (Chapter 16) – Exam 1 Review (part I – Chapter 14; rest on Tues.) – Titrations (Chapter 16) Chem 1B – Aqueous Chemistry Buffers (Chapter 16) • • • • We have discussed some mixtures briefly (e.g. strong acid + weak acid) One particular type of mixture: weak acid + conjugate base (or weak 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 (protein catalysts) 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 HC2H3O2 and 0.025 M Na+C2H3O2- solution (Ka of HC2H3O2 = 1.8 x 10-5) 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? Chem 1B – Aqueous Chemistry Buffers (Chapter 16) • Change of pH and Buffer Capacity – A buffer is designed to minimize the change in pH from addition of base or acid – What is the most effective pH? • (show next slide) – Can we add too much acid or base to a buffer? – Does the absolute concentration of acid/base affect pH? Why is a higher concentration better? – Buffer Capacity is the ability of the buffer to absorb acid or base without the pH changing significantly Chem 1B – Aqueous Chemistry Buffers (Chapter 16) • Most Effective pH Range – NH3 + NH4Cl in 1.00 L Example [NH3] + [NH4+] = 0.150 M – Addition of 0.005 moles HCl [NH3] (M) [NH4+] (M) pH DpH 0.005 0.145 7.78 -2.75 0.030 0.120 8.64 -0.09 0.050 0.100 8.94 -0.07 0.075 0.075 9.24 -0.06 0.100 0.050 9.55 -0.07 0.120 0.030 9.84 -0.08 0.145 0.005 10.71 -0.32 Best region (equal moles of WA and CB) Chem 1B – Aqueous Chemistry Buffers (Chapter 16) • A Second Way to Make Buffers (not on Exam 1) – To make a traditional buffer, we need both an acid and its conjugate base, but this also can be “made” through other combinations such as: – A weak acid and a strong base or a weak base and a strong acid – This is not covered in the text for buffers, but is needed for titrations (section 16.4) – Example: What is the pH when we add 10.0 mL of 0.50 M KOH with 34.1 mL of 0.19 M HC2H3O2? Chem 1B – Aqueous Chemistry Buffers (Chapter 16) • One Final Question: – How many mL of 1.00 M HCl must be added to 200.0 mL of 0.065 M NH3 to make a pH 10.00 buffer? Exam 1 Review • General Advice – Multiple choice questions: • Read full question and all answers before selecting (some answers may “sound right” until you see the actual right answer) • If it is taking time to calculate or if you are uncertain how to proceed, skip that problem and get back to later (circle question so you know you still need to do it) – General Question (12 pts): • Will be multiple parts and involve calculations (e.g. like determination of equilibrium concentrations on quiz) • Need to show work (don’t just punch numbers into your calculator and write answer) • I will give partial credit for parts done correctly Exam 1 Review • Topics – Chapter 14 – Understand how equilibrium relates to kinetics – Understand how to get equilibrium equations from equilibrium reactions (chemical equations) – Understand what the K value tells you about equilibrium conditions – Know the difference and be able to convert between KP and KC (for given reaction with R and T given) – Know how equilibrium reaction “manipulation” affects K values (e.g. reversing reaction leads to Knew = 1/Kold) Exam 1 Review • Topics – Chapter 14 – cont. – Be able to determine K from all equilibrium concentrations or from initial and equilibrium conditions – Be able to determine equilibrium concentrations from “at equilibrium” conditions and K values or from initial conditions and K values – Be able to determine whether reaction will proceed to products or reactants from K and initial concentrations Exam 1 Review • Topics – Chapter 14 – cont. – Know how to apply Le Châtelier’s principle to systems initially at equilibrium with the following changes: • reactant/product addition/subtration • increase/decrease in volume (including dilutions) • changes in T (with DH known) Chem 1B – Aqueous Chemistry Titrations (Chapter 16) • Review from 4.8 – Titrations are a way to accurately tell when we have reached a stoichiometric point in a reaction – Generic reaction: aA + bB → products – The equivalence point is defined where (moles A)/(moles B) = a/b – mL of titrant (either A or B) are carefully measured (usually to determine the concentration of B or A) Chem 1B – Aqueous Chemistry Titrations (Chapter 16) • Review from 4.8 – Example question: An unknown H2SO4 solution is pipeted (25.00 mL) into a flask. It is titrated with KOH until reaching an endpoint (where the equivalence point is observed). It requires 39.1 mL of 0.150 M KOH. What is the concentration of H2SO4 in the unknown solution? Chem 1B – Aqueous Chemistry Titrations (Chapter 16) • Titrations – pH behavior – Review material covered how we can use titrations, but not the exact behavior during the titration – In Chapter 16, we cover, through calculations, the pH of the full titration curve – The most valuable titrations are accurate (observed end point gives the equivalent point) and precise (reproduceable) – This occurs when change in pH in a titration is rapid Chem 1B – Aqueous Chemistry Titrations (Chapter 16) • Strong Acid – Strong Base Titration – How does pH change as NaOH is added? – 3 regions to titrations (different calculations in each region): • before equivalence point • at equivalence point • after equivalence point – Show pH at 5 mL, 12.5 mL, and 15 mL 0.100 M NaOH 0.050 M HCl, 25 mL Chem 1B – Aqueous Chemistry Titrations (Chapter 16) Titration Plot 0.100 M NaOH Titration Plot 14.00 12.00 pH 10.00 8.00 0.050 M HCl, 25 mL 6.00 4.00 2.00 0.00 0 5 10 15 V(HCl) 20 25