I need help with an analytical question. 1. A weight of 1.300 g of sodium carbon

Question

I need help with an analytical question.

1. A weight of 1.300 g of sodium carbonate (MW = 105.989 g/mol) is dissolved in 250.0 mL water. If a 25.00 mL aliquot of the sodium carbonate solution is placed into an E-flask and titrated with 0.1034 M HCl, what should be the volume of HCl to reach the second endpoint (bromocresol green indicator)?

2. 3.    A 0.2500 g soda ash sample is dissolved in 75 mL of double distilled water. This solution was titrated with 0.1034 M HCl and a bromocresol green endpoint gave a volume of 21.16 mL. What is the percent soda ash (as %Na2CO3) that was in the original sample?

Explanation / Answer

(1) Na2CO3 + 2 HCl => 2 NaCl + CO2 + H2O


Initial moles of Na2CO3 = mass/molar mass of Na2CO3

= 1.300/105.989 = 0.0122654 mol


Moles of Na2CO3 used in titration = 25.00/250.0 x initial moles of Na2CO3

= 25.00/250.0 x 0.0122654 = 0.00122654 mol


Moles of HCl = 2 x moles of Na2CO3

= 2 x 0.00122654 = 0.00245308 mol


Volume of HCl = moles/concentration of HCl

= 0.00245308/0.1034

= 0.02372 L = 23.72 mL


(2) Moles of HCl = volume x concentration of HCl

= 21.16/1000 x 0.1034 = 0.00218794 mol


Moles of Na2CO3 = 1/2 x moles of HCl

= 1/2 x 0.00218794 = 0.00109397 mol


Mass of Na2CO3 = moles x molar mass of Na2CO3

= 0.00109397 x 105.989 = 0.11595 g


Percent Na2CO3 = mass of Na2CO3/mass of sample x 100%

= 0.11595/0.2500 x 100%

= 46.38%

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