Neutralization Titration of Soda Ash Three samples of Na2C03(purity - 99.93%), w
ID: 1062898 • Letter: N
Question
Neutralization Titration of Soda Ash Three samples of Na2C03(purity - 99.93%), weighing 0.2019g, 0.1988g, and 0.2011g, are individually dissolved into separate 50mL aliquots of deionized H_2O. These samples are then titrated with HC1. If the volumes of HC1 used to titrate each sample are 39.7mL, 35.3mL, and 36.2mL, respectively, what is the average molarity, to four significant figures, of HC1? (MW Na_2CO_3 = (105.9888g -mol^-1) A 10mL aliquot is obtained from a 100mL solution containing Na_2CO_3. The aliquot is treated and diluted to 50mL. 43.7mL of the HC1 used in question 1 is used to titrate the sample. What is the molarity, to four significant figures, of the original sample of Na_2CO_3 containing solution? What theoretical mass of Na_2CO_3 is present in the original sample? Assuming this sample was prepared using the same Na_2CO_3 used in question 1, what was the actual mass obtained to prepare the sample? Electrolysis of copper Two samples, weighing 0.1024g and 0.9991g, are obtained from an unknown copper ore, treated with acid, dissolved, and each individually diluted to 100mL. The solutions are each subjected to electrolysis to remove the copper from solution. The cathodes used in the electrolysis of each solution, respectively, had initial masses of 13.057g, and 13.084g. The final masses of the cathodes after the electrolysis were 13.141g and 13.935g, respectively. What the average % by mass of copper in the ore?Explanation / Answer
Answer : 1) Molecular weight of Na2CO3 = 105.9888 g
Means , 1 Mole of Na2CO3 = 105.9888 g in 1000 ml H2O
Therefore, X1 mole Na2CO3 = 0.2019 g in 50 ml H2O
X1 = [ 0.2019 * 50 ] / [ 105.9888 * 1000 ] = 9.5245 * 10 -5 M Na2CO3
X2 = [ 0.1988 * 50 ] / [ 105.9888 * 1000 ] = 9.3783 * 10 -5 M Na2CO3
X3 = [ 0.2011 * 50 ] / [ 105.9888 * 1000 ] = 9.4868 * 10 -5 M Na2CO3
Now, 50 ml of X1 M Na2CO3 requires 39.7 ml Y1 M HCl
Therefore, Y1 = [ 50 * X1 ] / 39.7 = 1.1989 * 10 -4 M HCl
Y2 = [ 50 * X2 ] / 35.3 = 1.3283 * 10 -4 M HCl
Y3 = [ 50 * X3 ] / 39.7 = 1.3103 * 10 -4 M HCl
Average Molarity of HCl = [ Y1 + Y2 + Y3 ] / 3 = 1.2791 * 10 -4 M HCl.
Answer 2 ):-
Now, 43.7 ml of 1.2791 * 10 -4 M HCl. required for 50 ml of X M Na2CO3
X = [ ( 1.2791 * 10 -4 ) * ( 43.7 ) ] / ( 50) = 1.1179 * 10 -4 M Na2CO3
Now, 10 ml of aliquot of Na2CO3 = 1.1179 * 10 -4 M Na2CO3
Therefore , 100 ml Sample Na2CO3 = 1.1179 * 10 -3 M Na2CO3
Answer (3) :- Therotical mass of Na2CO3
Molecular weight of Na2CO3 = 105.9888 g
Means , 1 Mole of Na2CO3 = 105.9888 g
therfore, 1.1179 * 10 -3 M Na2CO3 = [ 105.9888 * ( 1.1179 * 10 -3) ] /1 = 0.1184 g of Na2CO3
Answer (4) :- Actual mass of Na2CO3
Therotical mass = 0.1184 g of Na2CO3
Given % purity Na2CO3 = 99.938 %
Actual mass of Na2CO3 = [ 0.1184 * 99.938 ] /100 = 0.1183 g of Na2CO3