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Chem. 31 – 2/2 Lecture. Announcements Due Wednesday –Turn in corrected diagnostic quiz –HW Set 1.1 – just additional problem Quiz on Wednesday (covering.

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Presentation on theme: "Chem. 31 – 2/2 Lecture. Announcements Due Wednesday –Turn in corrected diagnostic quiz –HW Set 1.1 – just additional problem Quiz on Wednesday (covering."— Presentation transcript:

1 Chem. 31 – 2/2 Lecture

2 Announcements Due Wednesday –Turn in corrected diagnostic quiz –HW Set 1.1 – just additional problem Quiz on Wednesday (covering Ch. 1) Today’s Lecture –Stoichiometry –Titrations –Start to Ch. 3

3 Stoichiometry Remember: there are two (common) ways to deliver a known amount (moles) of a reagent: –Mass (using formula weight) –Volume (if molarity is known) Titrations = A practical way of using stoichiometry with precise measure of added volume

4 Titrations Definitions Titrant: – Reagent solution added out of buret (concentration usually known) Analyte solution: –Solution containing analyte Equivalence Point: –point where ratio of moles of titrant to moles of analyte is equal to the stoichiometric ratio titrant analyte solution for: Al 3+ + 3C 2 O 4 2- → Al(C 2 O 4 ) 3 3- n(Al 3+ )/n(C 2 O 4 2- ) = 3/1 at equivalence pt.

5 Titrations Practical Requirements The equilibrium constant must be large –Precision of titration will depend on size of K and concentration of analyte –Typically K ~ 10 6 is marginal, K > 10 10 is better The reaction must be fast It must be possible to “observe” the equivalence point –observed equivalence point = end point

6 Titrations Other Definitions Standardization vs. Analyte Titrations –To accurately determine an analyte’s concentration, the titrant concentration must be well known –This can be done by preparing a primary standard (high purity standard) –Alternatively, the titrant concentration can be determined in a standardization titration (e.g. vs. a known standard) Rationale: –many solutions can not be prepared accurately from available standards Example: –determination of [H 2 O 2 ] by titration with MnO 4 - –neither compound is very stable so no primary standard –instead, [MnO 4 - ] determined by titration with H 2 C 2 O 4 (made from primary standard) in standardization titration –then, H 2 O 2 titrated using standardized MnO 4 -

7 Titrations Other Definitions Direct vs. Back Titration –In a direct titration, the titrant added slowly to the analyte until reaching an end point –In a back titration, a reagent is added to the analyte in excess, and then the excess reagent (beyond what was needed to react) is titrated to an end point –Often done to get sharper endpoint

8 Titrations Back Titration Example Titration to determine moles of Na 2 CO 3 in a sample: First, direct titration: Na 2 CO 3 + 2HCl → H 2 CO 3 + NaCl (we will do following AA lab) Na 2 CO 3 HCl not that sharp

9 Titrations Back Titration Example Titration to determine moles of Na 2 CO 3 in a sample: Now, via back titration: excess HCl added to sample Na 2 CO 3 + HCl → NaCl + H 2 CO 3 +heat → NaCl + H 2 O + CO 2 (g) Excess HCl titrated with NaOH to NaCl + H 2 O Na 2 CO 3 HCl After heating only NaCl and excess HCl left NaOH Excess HCl Very Sharp

10 Titrations What Makes a Titration Sharp? A sharp titration has a large slope (absolute value) Slope at endpoint seen in plot of - log[analyte] vs. V(titrant) With a sharp titration, errors or uncertainties in V(equivalence point) are small V(titrant) -Log[analyte] SHARP TITRATION NON-SHARP TITRATION [reactant] at eq. point V(eq. pt.) uncertainties in log[analyte] small uncertainty in V results larger unc. in V

11 Titrations Some Questions 1.List two requirements for a titration to be functional. 2.In a back titration, what is actually being titrated? (a) analyteb) reagent added c) excess reagent d) secondary reagent) 3.Why might one want to standardize a prepared solution of 0.1 M NaOH rather than prepare it to exactly 0.100 M? NaOH is a hygroscopic solid that also absorbs CO 2.

12 Titrations Back Titration Example Sulfur dioxide (SO 2 ) in air can be analyzed by trapping in excess aqueous NaOH (see 1). With addition of excess H 2 O 2, it is converted to H 2 SO 4 (see 2), using up additional OH - (see 3). 1.SO 2 (g) + OH - (aq) → HSO 3 - 2.HSO 3 - + H 2 O 2 (aq) → HSO 4 - + H 2 O 3.HSO 4 - + OH - (aq) → SO 4 2- + H 2 O 208 L of air is trapped in 5.00 mL of1.00 M NaOH. After excess H 2 O 2 is added to complete steps 1 to 3 (above), the remaining NaOH requires 21.0 mL of 0.0710 M HCl. What is the SO 2 concentration in  mol/L?

13 Chapter 3 – Error and Uncertainty Error is the difference between measured value and true value or error = measured value – true value Uncertainty –Less precise definition –The range of possible values that, within some probability, includes the true value

14 Measures of Uncertainty Explicit Uncertainty: Measurement of CO 2 in the air: 399 + 3 ppmv The + 3 ppm comes from statistics associated with making multiple measurements (Covered in Chapter 4) Implicit Uncertainty: Use of significant figures (399 has a different meaning than 400 and 399.32)

15 Significant Figures (review of general chem.) Two important quantities to know: –Number of significant figures –Place of last significant figure Example: 13.06 4 significant figures and last place is hundredths Learn significant figures rules regarding zeros

16 Significant Figures - Review Some Examples (give # of digits and place of last significant digit) –21.0 –0.030 –320 –10.010

17 Significant Figures in Mathematical Operations Addition and Subtraction: –Place of last significant digit is important (NOT number of significant figures) –Place of sum or difference is given by least well known place in numbers being added or subtracted Example: 12.03 + 3 Hundredths placeones place = 15.03 Least well known = 15

18 Significant Figures in Mathematical Operations Multiplication and Division –Number of sig figs is important –Number of sig figs in Product/quotient is given by the smallest # of sig figs in numbers being multiplied or divided Example: 3.2 x 163.02 2 places5 places = 521.664= 520 = 5.2 x 10 2


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