where is this exercise take from?I mean, what is the name of the book? A 1.217g

Question

where is this exercise take from?I mean, what is the name of the book? A 1.217g sample of commercial KOH (MM=56.11 g/mol) contaminated by K2CO3 (MM=138.12 g/mol) and moisture (H20) was dissolved in water, and the resulting solution was diluted to 500.0 mL. *A 50.00mL aliquot of this solution was treated with 40.00mL of 0.05304M HCl and boiled to remove CO2. THe excess acid consumed 4.74 mL of 0.04983M NaOH. *An excess of neutral solution was added to another 50.00mL aliquot to precopitate the carbonate as BaCO3. THe solution was then titrated with 28.56mL of 0.05304M HC. Calculate the percentage of KOH,K2CO3 and H20 in the sample, assuming that these are the only compounds present.

Explanation / Answer

moles HCl used to titrate BaCO3 = 0.05304 M x 28.56 ml = 1.5148 mmol

moles K2CO3 present = 1.5148/2 = 0.7574 mmol in 50 ml aliquot

mass K2CO3 = 0.7574 mmol x 138.12 g x 10/mol/100 = 1.0 g

Percent K2CO3 in sample = 1.0 x 100/1.217 = 82.17%

Total HCl added to first 50 ml aliquot = 0.05304 M x 40 ml = 2.1216 mmol (KOH + K2CO3)

Excess unreacted HCl = 0.04983 M x 4.74 ml = 0.2362 mmol

HCl reacted (KOH + K2CO3) = 2.1216 - 0.2362 = 1.8854 mmol

Amount of KOH present = 1.8854 - 1.5148 = 0.3706 mmol

mass of KOH present = 0.3706 x 10 x 56/1000 = 0.207 g

percent KOH in sample = 0.207 x 100/1.217 = 17.0%

Percent water present = 100 - (percent K2CO3 + percent KOH) = 0.83%

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