A student set out to analyze the impure sample of potash (potassium hydroxide) f

Question

A student set out to analyze the impure sample of potash (potassium hydroxide) for its percent purity. The student will use an HCl solution that is standardized with Na2CO3. If the student ways out 0.5615g ((+/-)0.0002 g) sample of potash, calculate the answer for the three questions based on the following information. Remember to show your work airport your answers with your purpose significance figures

3) Calculate the % purity of the potash sample along with the uncertainty (absolute and % relative ).

PLEASE NOTE

MW of KOH = 65.1056 g/mol (negligible uncertainty)

MW of Na2CO3 = 105.98844 g/mol (negligible uncertainty)

Standardization of HCl

Mass of Na2CO3 = 0.2201g ((+/-)0.0002 g)

Volume used of (mL) HCl = 35.22 ((+/-)0.03 mL)

Titration of potash sample vs HCl

Volume used of (mL) HCl = 40.36 ((+/-)0.03 mL)

Explanation / Answer

Amount of Na2CO3 = [0.2201g   (+/-)0.0002g] /105.98844 g.mol-1

= 0.00208mol (+/-)1.88*10-6 mol

= 0.00208 mol (+/-)0.1814%

As 1 mol of Na2CO3 is neutralized by 2 mol of HCl, so amount of HCl in 35.22 (+/-)0.03 mL

= 2(0.00208 mol (+/-)0.1814%)

= 0.00415 mol (+/-)0.1814%

35.22 (+/-)0.03 mL = 35.22 mL (+/-)0.085%

Molarity of HCl

= [0.00415 mol (+/-)0.1814% / 35.22 mL (+/-)0.085%] * 1000

= 0.1179 M (+/-)0.266%

40.36mL (+/-) 0.03 mL = 40.36 mL (+/-) 0.074%

40.36 mL (+/-) 0.074% mL of HCl contains

= [0.1179 M (+/-)0.266% / 1000] * [40.36 mL (+/-) 0.074% mL] moles of HCl

= [1.179*10-4 (+/-)0.266%]* [40.36 mL (+/-) 0.074% mL] moles of HCl

= 4.758*10-3 mol   (+/-) 0.34% moles of HCl

As 1 mol of KOH is hydrolyzed by 1 mol of HCl, so amount of KOH = 4.758*10-3 mol   (+/-) 0.34% moles

= [4.758*10-3 mol   (+/-) 0.34% ] * 65.1056 g/mol

= 0.3098 g (+/-) 0.34%

= 0.3098 g (+/-) 0.001 g

0.5615g (+/-)0.0002 g = 0.5615 g (+/-) 0.035%

% purity

= [0.3098 g (+/-) 0.34%]/[ 0.5615g (+/-)0.035% g] * 100

= [0.5517 (+/-) 0.375%] * 100

= 55.17% (+/-) 0.375%                   ....(relative uncertainty)

Purity = 0.3098 g (+/-) 0.001 g     .....(absolute uncertainty)

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