# lab4 - NAME and ID Number Quiz Section LAB PARTNER...

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NAME and ID Number: LAB PARTNER: Quiz Section: Total Points = 60 (template plus notebook pages) Notebok pages (5 pts) - Are yours organized and legible so that your TA can read them and find any information he/she might be looking for? PURPOSE AND METHOD DATA AND CALCULATIONS A. Standardization of NaOH Run 1 Run 2 Run 3** Mass of KHP 0.13 0.13 moles KHP 0.000641 0.000651 Initial buret reading, mL 0.05 0.01 Final buret reading, mL 6.57 6.61 mL NaOH titrated 6.52 6.60 L NaOH titrated 0.00652 0.00660 [NaOH], moles/L 0.0983 0.0986 Average, [NaOH] 0.0985 M 4 pts Table 1: NaOH Standardization How are the solubility products (K sp ) of KHP at various temperatures determined? ( 3 pts ) Titrate saturated KHP solutions at 0 C, 20 C, 35 C, 45 C, 55 C and 65 C to determine HP- concentrations. Use Ksp = [K+][HP-] = [HP-]^2 to calculate Ksp values as a function of temperature (K). How is K sp in a saturated solution of KHP in 0.50 M KCl determined? (3 pts ) Make K+ emissions standards using concentrations from the prelab. Measure emission intensity and plot calibration equation. Measure intensity of the KHP in 0.50 M KCl to determine the K+ concentration. Tritate 3ml of .50M KCI solution to find [HP-]. Use Ksp = [K+][HP-] to calculate Ksp values as a function of temperature (K). *Note: some of this information has been taken from the prelab. EXPERIMENT 4: THERMODYNAMICS II THE TEMPERATURE DEPENDENCE OF THE SOLUBILITY PRODUCT, HEAT OF SOLUTION, AND ENTROPY OF SOLUTION OF POTASSIUM HYDROGEN PHTHALATE By signing below, you certify that you have not falsified data and that you have not plagiarized any part of this lab report. Signature: (minus 5 pts if missing) Show the calculation for determining [NaOH]for one of the runs. May be typed or NEATLY handwritten. Moles NaOH / L NaOH = [NaOH] (Mass KHP / Molar mass KHP) / [(Final burret reading - Initial burret reading)/1000] = [NaOH] (.131g/204.22g) / [(6.57-.053)/1000] = [NaOH] .0983M = [NaOH] Run1 - Run2 / Run1 = Percent Error .0986-.0983 / .0983 = .3%

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B. 1.00 mL
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## This note was uploaded on 07/08/2008 for the course CHEM 162 taught by Professor N. during the Spring '08 term at University of Washington.

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lab4 - NAME and ID Number Quiz Section LAB PARTNER...

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