Chem 222 #9 Ch 7 Sep 21, 2004

Chem 222 #9 Ch 7 Sep 21, 2004

Announcement The answer for quiz 2 with a grading guide line is at the web site The due date for report of Exp 5 is this W/R

Strengths of Acids and Bases Strong Acids and Bases A strong acid or base is completely dissociated in aqueous solution. HCl(aq) H + + Cl - (~ 100%) KOH(ac) K + + OH - (~ 100%) Remember all the acids and bases in Table 6-2. Q. Calculate the concentration of H + and the ph in the following solutions at 25 C: a. 0.056 M HCl, b. 0.25 M KOH, c. 5.60 M HCl. K w 1.01 10-14 a. [H + ] 0.056 M ph -log[h + ] -log(0.056) 1.25 2 b. [OH - ] 0.25 M [H+] K w /[H + ] a. ph -log[h + ] -log(5.60) - 0.748 2

Weak acids and bases HA H + + A - a weak acid 0.056 M - x x x K a [H + ][A - ]/[HA] x 2 /(0.056 x) B + H 2 O BH + + OH - a weak base K b [BH + ][OH - ]/[B]

7-1 Titration At equivalent point The color changes indicates end point, which is close to the equivalent point The titrant functions as an indicator Indicator BB (Blue Yellow) MR (Yellow Red) ph meter (ph is detected by voltage; see P 323)

Standardization of KMnO 4 Suppose that 0.3562g of Na 2 C 2 O 4 is dissolved in a 250.0mL volumetric flask. If 10.00 ml of this solution require 48.36 ml of KMnO 4 solution for titration, what is the molarity of KMnO 4 solution? Oxalate concentration C OX is Weight of Na C2O4 / MW olume of solution 2 0.3562g /(134.00g 0.2500L C OX 1.063 3 mm / mol) Moles of Moles of C KMnO4 0.8794 7 Oxalate KMnO4 C OX OX KMnO4 mmol 2 5 C C OX MnO4 OX MnO4 5 moles 2 moles 0.0010633 M 10mL 48.36mL 2.000 5.000

Analysis of unknown C KMNO4 0.8794 mm Calcium in a 5.00 ml urine sample was precipitated, was redissolved, and then required 16.17mL of standard MnO 4- solution. Find the concentration of Ca 2+ in the urine. Ca 2+ : C 2 O 4 2- : MnO 4-5 : 5: 2 Moles of Ca Moles of KMnO4 C C Ca MnO4 Ca MnO4 5 moles 2 moles Ca (5/2) (C MnO4 MnO4 )/ Ca (5/2) (0.8794 mm 16.17 ml)/5ml

Ex. (p132) (Related to Exp 8) y g (1.372 y) g Na 2 CO 3 moles 2 + NaHCO 3 moles HCl moles {y/105.99 + (1.372 y)/84.01} C HCL HCl 29.11mL 0.734 M 0.02138 moles (1/104.99 1/84.01)y + 1.372/84.01 0.02138 Do not forget to weigh your sample! Without 1.372 g you can never obtain the answer.

Spectrophotometric Titration (Lab. Exp 19) Absorbance: A εbc (p411) Path length Concentration Fe 3+ Binding Site of Transferrin Apo-transferrin + 2Fe 3+ (Fe 3+ ) 2 transferrin Show absorbance at 465 nm! Titration of 2.00 ml of apotransferrin With 1.79 10-3 M ferric nitrilotriacetate

203 µl Titration of 2.00 ml of apotransferrin With 1.79 10-3 M ferric nitrilotriacetate Apo-transferrin + 2Fe 3+ (Fe 3+ ) 2 transferrin 3+ Moles of Fe Moles of protein C C Fe protein Fe protein 2 moles 1moles C protein CFeFe 1.79mM 203µ L 2 2 2.00mL protein To construct the above graph the dilution effect must be corrected. Corrected Abs Observed Abs total initial

Ex. The absorbance measured after adding 125 ul of ferric nitrilotriacetate to 2.000 ml of apotransferrin was 0.260. Calculate the corrected absorbance that should be plotted in the Figure 7-5. Corrected Abs Observed Abs total ini 0.260 2.125mL 2.000mL

7-4 The precipitation Titration Curve The titration curve is a graph showing how the concentration of one of the reactions varies as a titrant is added Because concentration of the reagent varies over many orders, it is most useful to plot pfunction: px -log 10 [X]

Q. Consider the titration of 25.00 ml of 0.1000 M I - with 0.05000 M Ag + I - + Ag + AgI(s) (7-9) Suppose we titrant are monitoring Ag + with an electrode. (7-9) is the reverse of the dissolution of AgI(s) I - +Ag + K sp [I - ][Ag + ]8.3 10-17 Almost 100 % forms AgI(s) Q1. What volume of Ag + titrant is needed to reach the equivalence point? C I I C Ag Age Age C I I /C Ag (0.1000 M)25.00 ml/0.05000 M 50.00 ml

How does [Ag + ] change before the end point? Q. What is [Ag + ] when Ag 49.00 ml? At the equivalence point at Ag e 50.00 ml. A. Before the equivalence point, we have an excess of I -. [Ag + ] is obtained from [Ag + ] K sp /[I - ] 8.3 10-17 /[I - ] [I - ] (Moles of Original I-) - (Moles of Total volume CII CAgAg + I Ag Ag+ ) If Ag 49.00 ml Moles of I - (25.00 ml) (0.1000 M) - (49.00 ml)(0.05000 M) 2.500 mmol - 2.450 mmol 0.050 mmol total (25.00 ml + 49.00 ml) 74.00 ml [I - ] [Q1]/74.00 ml 6.7 6 10-4 M See P135 [Ag + ] [Q2]/6.76 10-4 M 1.2 3 10-13 pag + -log[ag + ] 12.91

At the equivalence point, what is [Ag + ]? At equivalence point, Age C Ag I C I Little l - and Ag + are left, and [Ag + ] [l - ]. We define [Ag + ] x. Then, K sp 8.3 10-17 [Ag + ][I - ] x 2 x 9.1 10-8 M pag + -log(9.1 10-8 ) 7.04 1 Q. What is [Ag + ] when Ag 51 ml

After the end point, Q. What is [Ag + ] when Ag 51 ml Now, all the I - ions are consumed, we have an excess of Ag +. Moles of Ag + (C Ag Ag C I I ) (51.00 ml)(0.05000 M) - (25.00 ml) (0.1000 M) 1.00 ml 0.05000 M 0.05 mmol [Ag+] 0.05 mmol/(total volume) 0.05 mmol/(51 ml + 25 ml) 6. 6 10-4 M pag+ 4 log6. 6 3.1 8

Titration curve Excess I - Excess Ag + For 1:1 stoichiometry of reactants, the equivalence point can be indicated by

Ion dependence (pag + ) e -Log(K sp 1/2 )

12-1 Strong acid-strong base Titration NaOH + HCl H 2 O + NaCl Excess OH - Excess H + ph a (ml) Excess OH - Calculate [OH - ] [H + ] Kw/[OH - ] Equivalence Point K w x 2 ph 7.0 Excess H + Calculate [H + ] (C HCL HCl C NaOH NaOH )/( HCl + NaOH )

P141 Calculation of titration from spread sheet Known [M + ] [X - ] Ksp/[M + ] Then [M + ] & [X - ] M

Home work Read Ch 13 1-3 Ch7: 7-A, 7-B, 7-4, 7-7, 7-8, 7-9, 7-10, 7-14, 7-21, 7-23