Consider a mixture that may contain NaOH (Mw = 40.0), Na_2 CO_3 (Mw = 106) and N

Question

Consider a mixture that may contain NaOH (Mw = 40.0), Na_2 CO_3 (Mw = 106) and NaHCO_3 (MW = 84.0) along with inert material. A 2.3450 g sample containing this mixture was dissolved in water and diluted to 100.0 mL in a volumetric flask. Titration of a 50.0 mL portion to a phenolphthalein end point requires 22.46 mL of 0.1342 M HCl. Another 50.0 mL portion from the volumetric flask requires 46.32 mL of the 0.1342 M HCl when titrated to a cresol green end point. Which species are present? Calculate the percent of each species present in the original sample.

Explanation / Answer

Phenolphthalein will show the end point Na2CO3-----> NaHCO3. whereas bromocresol green shows color change in NaHCO3 ----> CO2 + H2O end point

Volume of HCl required to convert Na2CO3 ----> NaHCO3 = 22.46 mL

volume of HCl required to convert Na2CO3 ---> CO2 + H2O = 2 * 22.46 mL = 44.92 mL

to, neutralization of only NaHCO3 to co2 + h2O = 46.32mL - 44.92 mL = 1.4 mL

moles of NaHCO3 present in 50mL = (1.4/1000)L * 0.1342 M =0.19*10^-3 moles

moles of naHCO3 present in 100mL = 2* 0.19*10^-3 moles= 0.38*10^-3 moles

Mass of NaHCO3 present in 100mL = 0.38*10^-3 moles *84 g/mol = 0.032 gm

% of NaHCO3 =(0.032 /2.345)*100% = 1.36 %

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moles of Na2Co3 present = (44.92/1000) L * 0.1342 M = 6.03*10^-3 moles

moles of Na2CO3 present in 100mL = 2*6.03*10^-3 moles =0.0121 mol

Mass of Na2CO3 in 100mL = 0.0121 mol * 106g/mol = 1.28 gm

Mass% = (1.28gm/2.3450) *100 =54.58 %

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Mass % of NaOH = 100- (54.58 +1.36) =44.06 %

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Species present : OH-, CO3^2-, HCO3^-

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