Chem 222 #18 Ch 10, Review Mar 10, 2005
Announcement Next Tuesday Review Class (Lecture note will be probably uploaded on this Saturday) Midterm Exam on the next Thursday this room in the same time (50 mins). Come on time.
Quiz 4 (from 16-14) Balance the equation MnO 4- + Mo 3+ Mn 2+ + MoO 2+ 2
Complete Half Reactions MnO 4 - Mn 2+ 1) Count Charges of Mn (x) in MnO 4 (oxygen: -2) Total Charge -1 = x + 4(-2) x = +7 Mn(VII)O 4 - Mn(II) 2+ 2) Add (7-2) electrons Mn(VII)O 4- + 5e - Mn(II) 2+ Mo 3+ MoO 2 2+ 2 = y + 2(Q1) y = 6 Mo(III) 3+ Mo(VI)O 2 2+ + [Q2]e -
Mn(VII)O 4- + 5e - Mn(II) 2+ 3 Mo(III) 3+ Mo(VI)O 2+ 2 + [Q2]e - 5 + 3Mn(VII)O 4- + 5Mo(III) 3+ 3Mn(II) 2+ + 5 Mo(VI)O 2 2+ 3) Add H 2 O to balance oxygen Left: 3 4 =12 Right: 5 2 =10 3Mn(VII)O 4- + 5Mo(III) 3+ 3Mn(II) 2+ + 5 Mo(VI)O 2+ 2 +[Q3]H 2 O 4) Add H +
10-5 Buffers (p189) The buffers is a mixture of an acids and its conjugated base. There must be comparable amounts (factor 1/10-10) of the conjugated base and acids. ph dependence of enzyme Reaction rate
Mixing a weak acid and its conjugated base HA H + +A - pk a = 4.00 K a = 10 [Q1] F- x x x F = 0.01 M x 2 /(F-x) = K a x =3.1 10-3 Fraction of dissociation α =[A - ]/{[A - ] +[HA]} = x/f = 0.031 =3.1 % Remember Le Chatelier s principle? (p103) What happens if you add A - to H 2 O HA H + +A - How much fraction of A - reacts with water in a solution containing 0.10 M of A -? A - +H 2 O HA + OH - pk b =[Q2] F y y y y 2 /(F -y) = K b y =3.1 10-6 & α = 3.1 10-5
Conjugated Base & Ka Kb = Kw Acid HA H + + A - K a + A - +H 2 O HA + OH - K b H 2 O H + + OH - K w See P101
Calculation of buffer composition is simple If you add x moles of A - and y moles of AH to 1 L of water, [A - ] = x M, [AH] = (Q1) M Henderson-Hasselbalch Equation ph = pk a + Log [ A ] [ HA]
How to get Henderson Equation? Henderson-Hasselbalch Equation Expecting ph is also simple K a = [H + ][A - ]/[HA] -Log K a = -Log[H + ] - Log{Q1} pk a = ph - Log {Q1} ph = pk a + Log [ A ] [ HA] Remember the derivation K b = {Q2}/[B] K a?
10-4 Weak base equilibria B + H 2 O BH + + OH - [Q1] x x [ BH ][ OH [ B] ] = x F + 2 x = K b Ex. (p188) Find the ph of 0.10 M Ammonia with pk b = 4.756. NH 3 + H 2 O NH 4+ + OH - (F-x) x x X 2 /(F-x) = K b =1.75 10-5 x = ([ Q2 ] 1.75 10-5 ) 1/2 = 1.3 10-3 M [OH - ] = (Q1 ) [H + ] = (Q2)/[OH - ]
Ex. Effect of adding acid to a buffer We prepare 1L of solution containing 0.102 M of tris and 0.0296 M of trishydrochloride with pka of 8.075. ph = pk a + Log[B]/[BH + ] ph = (Q1) +Log(Q2 /0.0296) = 8.612 Let s add 12 ml of 1M HCl to the solution
Ex. How to prepare a buffer solution (p194) How many milliliters of 0.500 M NaOH should be added to 10.0 g (0.0635 moles) of tris hydrochloride to give a ph of 7.60 in a final volume of 250 ml? pk b of tris is 5.925. BH+ + OH- B Initial 0.0635 x Final 0.0635 x x pka= (Q2) pkb [Q1] = pka + Log[B]/[BH + ] [B]/[BH + ] = 10 {(Q1) pka} x / (0.0635 x) = 10 {(Q1) pka}
Preparation of a 0.02M Tris- HCl buffer, ph 8.0, 1 Liter, 0.5M NaCl gradient http://ntri.tamuk.edu/fplc/prep.html
Exp 15 (Ch 7 p139) Titrate a mixture of Cl - and I - with Ag + Because K sp is lower for I -, I - is consumed first as AgI. At equivalence point, Moles Ag+ = Moles I-
Expected Titration Curve V e1 V e1 C Ag = Moles of I - V e2 How about Cl -?
Apparatus to measure [Ag + ] Read mv 53 mv pag + ±1 Ag + +e - Ag aa +ne- bb E + = E0 - (0.05916/n)Log{[B] b /[A] a } = E 0 (0.05916/n)Log{1/[Ag + ]}
Mol, Molarity, Molality? 1 mol = 6.02 10 23 molecules Concentration Molarity: M = molarity = Molality: m = molality = mol solute L solution mol solute kg solvent Weight percent = kg solute kg solution Q1. Tell the definition of solution, solute, and solvent? Solution: homogenous mixture of substances in a liquid for
Unit Conversion Wt % Molarity Obtain how much mol of substance exists in 1L of solution M =mol/l = (Wt %/100) 1000d[g/mL] /MW[g/mol] d is the density of the solution Molality Wt % Assume in 1L of solution A g of solute and B kg of solvent is included. Molality (m) A/B = A/B Wt % 100 A/(1000B + A) = 100 (A/B)/(1000 + A/B) = 100 m/(1000 + m)
3-2 Significant Number in Arithmetic (p47) Addition and Subtraction If the numbers to be added or subtracted have equal number of digits, the answer goes to the same decimal places (1.362±0.001) 10-4 + (3.111±0.001) 10-4 ------------------------------ 4.472±0.001 10-4
Addition and Subtraction #2 If the numbers to be added or subtracted does not have the same number of significant figures, we are limited by least certain one 121.80 for MW of KrF 2 (Q1. Why are the numbers not significant?) Addition and Subtraction #2 In adding or subtracting numbers expressed in scientific notation, all the numbers should be expressed in the same exponent as
Multiplication and Division In multiplication and division, we are normally limited to the numbers of digits contained in the number with fewest significant figures. The power of 10 Has no influence on the number of figures (a ± e a ) (b ± e b ) ~ ab ± be a ± ae b = ab(1 ± e a /a ± e b /b)
Propagation of Uncertainty Suppose you wish to perform the following arithmetic, in which the experimental uncertainties, designated e 1, e 2, and e 3, are given in parentheses. The arithmetic answer is 3.06. But what is the uncertainty associated with this result?
Uncertainty in addition and subtraction (a + δ 1 ) + (b + δ 2 ) = (a+b) + δ 1 + δ 2 < δ SUM > = {<(δ 1 ± δ 2 ) 2 >} 1/2 = {< (δ 12 ± 2 δ 1 δ 2 + δ 22 )>} 1/2 = {< δ 12 > ± 2 < δ 1 > < δ 2 > + < δ 22 > } 1/2 = {< δ 12 > + < δ 22 > } 1/2 where δ 1 and δ 2 are measurement errors for Measurement 1 and 2, respectively, and < δ 1 > = < δ 2 > = 0 and e k = <δ k2 > 1/2 (k =1,2) Ex. Suppose that the initial reading is 0.05 (0.02) ml and the final Reading is 17.88 (0.02) ml. The volume delivered is the difference: How much is e?
Multiplication and Division Ex. Convert absolute uncertainties to percent relative uncertainties %e 4 = 4.0 % corresponds to e 4 = 0.040 5.64 = 0.23 (a + δ a )(b + δ b ) ~ ab + a δ b + b δ a = ab + ab(δ b /b) + ab(δ a /a) e ab = { (ab) 2 (e a /a) 2 + (ab) 2 (e b /b) 2 } 1/2 = ab {(e a /a) 2 + (e b /b) 2 } 1/2 %e ab = 100 e ab / ab = 100{(e a /a) 2 + (e b /b) 2 } 1/2 = {(%e a ) 2 + (%e b ) 2 ) Mixed Operation: See p53
4-6 Q Test for Bad Data how to justify excluding an inconsistent data point Consider the five results 12.53, 12.56, 12.47, 12.67, and 12.48. Q1. Is 12.67 a bad point? Q test can tell you how to distinguish a bad point from good points Range: Max Min Gap: Difference between the questionable point and the nearest point Q calc = 0.11/0.20 = 0.55 If Q calc > Q table the point should be discarded.