CHEM 2364 : Study Study Study For Anayltcal Exam 2

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azuremammoth471

15 Mar 2019

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Question 2. (20 pts) provide a detailed block diagram for one of the following, and clearly. Choose only one. block diagram for an inductively-coupled plasma atomic emission spectrophotometer (icp-aes) block diagram of a flame atomic absorption spectrophotometer (aas) block diagram for a gas chromatography system (gc) block diagram for a high-performance liquid chromatography system (hplc) Flame aas monochromator flame detector readout sample wavelength selector light source hollow cathode lamp) Light source: for aas, the source of light is a hollow cathode lamp, which has a thin metal coating of the element of interest (such as copper, magnesium, iron, etc. ). The metal is ionized when a voltage is applied to the lamp, so that the ionized metal is present in the light path. photomultiplier tube computer. Sample: the sample is a liquid, which is sprayed into the gas stream that is burned as a flame. Flame: the sample is ionized in a flame, typically an acetylene burner.

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I am not asking for strictly answers, I actually need to knowhow to figure this out.. So whoever has the patience to help mefigure out what I am doing wrong, I promise to give youfull set of stars for your time. I keepfeeling like I missed a step when doing these and they dont lookcorrect.

Please explain how to calculatethe expected pH of the following solutions and tell me the pointwhere my calculations went wrong and/or if they arecorrect...

****Abbreviations: WB= weak base;WA= weak acid; SB= strongbase...***

Soultion1: 10 mL of 150 mMWB, HEPES,Na+ (pKa= 7.55)

Kb= 3.548 x 10 ^-7 = [OH-]^2 / 0.15M ------>moles OH- = 2.307 x 10^-4

However, since no acid or base is added, Ididn't think the 10 mL volume will affect the pH so Idid:

pOH= -log(2.307 x 10^-4)= 3.637 ------> pH=10.36

Solution 1 + StrongBase: 0.1 mL of 1.25M NaOH added to 10 mL of 150 mM WB,HEPES,Na+ (pKa= 7.55).

(0.0001 L x 1.25 M) = 0.000125 moles SB

molesWB + moles SB= (2.307 x 10^ -4 ) + (0.000125) = 3.557 x10^ -4 molesbase

......do I divide this by the total volume (10.1mL= 0.0101 L)and take the -log for pOH or do I simply take that value anddetermine pOH?

Solution 2 (dilutedSolution 1) 10 mL of30 mM WB ( HEPES,Na+ pKa=7.55) (The stock solution wasoriginally 150 mM but had a 1:5 dilution, with a new concentrationof 30 mM)

Kb= 3.548 x 10 ^-7 = [OH-]^2 /0.03M ------> moles OH- = (1.03 x 10^ -4 )

pOH= -log (1.03 x 10^-4)-------> pH=10.01

Solution 2 + StrongBase: 0.1 mL of 1.25M NaOH added to 10 mL of 30 mM WB ( HEPES,Na+ pKa= 7.55)

(1.03 x 10 ^-4 )+ (0.0001 L x1.25 M)= 1.2603 x 10^-4 moles OH-

pOH= -log(1.2603 x10^ -4)-------> pH=10.358

---------------------------

Solution3: 10 mL of 30 mM WA HEPES, pKa=7.55

Ka= 2.818 x 10^-7 = [H+]^2/ 0.03M -------> moles H+ = 8.455 x10^ -10

pH= -log (8.455 x 10^ -10) -------> pH=9.07

Solution 3 + StrongBase: 0.1 mL of 1.25M NaOH added to 10 mL of 30 mM WA ( HEPES pKa= 7.55)

0.01L x 0.03M= 0.0003 moles WA

0.0001L x 1.25 M= 0.000125 moles OH-

pH= 7.55 + log(0.000125/ (0.0003- 0.000125)) -------> pH=7.40

CHEM 2364 : Study Study Study For Anayltcal Exam 2

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