(a)A 2.0 g sample of juice from a fresh lime was diluted with water and filleted

ID: 509438 • Letter: #

Question

(a)A 2.0 g sample of juice from a fresh lime was diluted with water and filleted to remove suspended matter. The clear liquid required 40.39mL of 0.04022M NaOH for titration to the phenolphthalein end point. Calculate the acidity as % citric acid.
(b) the carbon dioxide evolved on heating a 2.407 g sample of steel in a closed system was swept into a flask containing 50 mL of 0.05081 M barium hydroxide. After filtering the solution to remove the precipitated barium carbonate, the remaining base consumed 14.87 mL of 0.1125 M hydrochloric acid for titration to the phenolphthalein end point. Calculate the percentagw of carbon in the steel.

Explanation / Answer

3 NaOH (aq) + H3C6H5O7 (aq) Na3C6H5O7 (aq) + 3 H2O(l)

Step 1- Calculate the moles of NaOH-

40.39 ml of NaOH x 1L/1000 ml x 0.04022moles/1L = 0.0016245moles

Step 2- Calculate the moles of citric acid –

Moles of citric acid = 3 x moles of NaOH

= 3 x 0.0016245 = 0.0048735

Step 3- Calculate the mass of citric acid-

Molar mass of citric acid = 192.124 g/mol

Mass of citric acid = moles of citric acid x molar mass of citric acid

= 0.0048735mol x 192.124 g/mol

= 0.936g

Step 4- Calculate % of citric acid-

% of citric acid = (mass of citric acid/mass of sample) x 100

= (0.936g/2g) x 100

= 46.82%

Ba (OH)2 + CO2 BaCO3 + H2O

Ba(OH)2 + 2 HCl BaCl2 + 2 H2OS

Step-1 Calculate the moles of HCl-

14.87 ml x 1L/1000ml x 0.1125moles/1L = 0.001673 moles

Step 2- calculate the moles of barium hydroxide-

50 ml x 1L/1000ml x 0.05081mol/1L = 0.00254

Step -3 Calculate the moles of Ba (OH)2 neutralized by HCl

2 moles HCl neutralized 1 moles of Ba(OH)2

0.001673 moles of HCl neutralized 0.001673/2 = 0.0008365 moles Ba(OH)2

Moles of Ba(OH)2 remained = 0.00254 – 0.0008365 = 0.001704 moles

Step 4 Calculate the moles of CO2

Ba(OH)2 and CO2 reacts in 1:1 molar ratio

So, moles of CO2 = 0.001704

Step -5 calculate mass of CO2

Molar mass of CO2 = 44g/mol

Mass of CO2= 0.001704g x44g/mol = 0.07498 g CO2

Mass of C = 0.07498 x (12/44) = 0.02044g

Step6- % of C in steel-

(0.02044 /2.407) x 100 = 0.84%