A student set out to analyze an impure sample of potash( potassium hydroxide) fo
ID: 1005289 • Letter: A
Question
A student set out to analyze an impure sample of potash( potassium hydroxide) for percent purity. The student will use an HCl solution that is standardized with Na2CO3. If the student weighs out a 0.5615g (_+ 0.0002g) sample of potash, calculate the answer for three questions 1) calculate the concentration and uncertainty (absolute and %relative) of the standardized HCl solution2) calcute the grams of KOH determined to be in the potash along with the uncertainty (absolute and % relative)
3) calcute the percent purity of the potash sample along with the uncertainty ( absolute and %relative)
MW KOH= 65.1056 g/mol MW Na2CO3= 105.98844 g/mol Standardized of HCl Grams Na2CO3= 0.2201 g (+_ 0.0002 g) Volume used mL HCl= 35.22 ( +_ 0.03 ml) Titration of potash sample vs HCl Volume used mL HCl= 40.36( 0.03mL) A student set out to analyze an impure sample of potash( potassium hydroxide) for percent purity. The student will use an HCl solution that is standardized with Na2CO3. If the student weighs out a 0.5615g (_+ 0.0002g) sample of potash, calculate the answer for three questions 1) calculate the concentration and uncertainty (absolute and %relative) of the standardized HCl solution
2) calcute the grams of KOH determined to be in the potash along with the uncertainty (absolute and % relative)
3) calcute the percent purity of the potash sample along with the uncertainty ( absolute and %relative)
MW KOH= 65.1056 g/mol MW Na2CO3= 105.98844 g/mol Standardized of HCl Grams Na2CO3= 0.2201 g (+_ 0.0002 g) Volume used mL HCl= 35.22 ( +_ 0.03 ml) Titration of potash sample vs HCl Volume used mL HCl= 40.36( 0.03mL) 1) calculate the concentration and uncertainty (absolute and %relative) of the standardized HCl solution
2) calcute the grams of KOH determined to be in the potash along with the uncertainty (absolute and % relative)
3) calcute the percent purity of the potash sample along with the uncertainty ( absolute and %relative)
MW KOH= 65.1056 g/mol MW Na2CO3= 105.98844 g/mol Standardized of HCl Grams Na2CO3= 0.2201 g (+_ 0.0002 g) Volume used mL HCl= 35.22 ( +_ 0.03 ml) Titration of potash sample vs HCl Volume used mL HCl= 40.36( 0.03mL)
Explanation / Answer
Q.1: Na2CO3(aq) + 2HCl(aq) -------- > H2CO3(aq) + 2NaCl(aq)
moles of Na2CO3 used = mass / molar mass = (0.2201 g +/- 0.0002 g) / 105.98844 g/mol
= (0.002077 +/- 2x10-6) mol
Let the concentration of HCl be 'M'
Volume of HCl used = (35.22 +/- 0.03) mL = (0.03522 +/- 0.00003) L
From the above balanced chemical reaction
2 x moles of Na2CO3 = moles of HCl
=> 2 x (0.002077 +/- 2x10-6) mol = M x (0.03522 +/- 0.00003) L
=> M = 2 x (0.002077 +/- 2x10-6) mol / (0.03522 +/- 0.00003) L =
=> M = (0.1179 +/- 0.0002) mol/L (absolute uncertainity) (answer)
or M = (0.1179 +/- 0.1697 %) mol/L (relative uncertainity) (answer)
Q.2: moles of KOH = moles of HCl = M x V = (0.1179 +/- 0.0002) mol/L x (0.04036 +/- 0.00003) L
= (0.004758 +/- 9x10-6) mol
Hence mass of KOH = (0.004758 +/- 9x10-6) mol x 65.1056 g/mol = (0.3098 +/- 0.0006) g (answer)
or mass of KOH = (0.3098 +/- 0.1937%) (answer)
Q.3: Percent purity = [(0.3098 +/- 0.0006) / (0.5615 +/- 0.0002) g] x 100 = (55.2 +/- 0.1) % (answer)
or Percent purity = (55.2 +/- 0.2%) % (answer)
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