Hello I need help in calculating. My Lab was Analyis of Carbonate in Soda Ash CO

Question

Hello I need help in calculating.

My Lab was Analyis of Carbonate in Soda Ash

CO32- + 2H+ ------> H2CO3

I first weighed and dried out 2.1383g of Unknown Soda ash. Placed it into a 250mL volumetric flask, and filled with CO2 free distilled H20. I pipeted 25.0 mL of the unknown into a Flask and titrated with standard 0.10 M HCl.

For my trials I got the following:

Trial 1
took 18.2 ml of HCl to neutralize the solution

Trial 2
took 18.1 ml of HCl to neutralize the solution

Trial 3
took 18.2 ml of HCl to neutralize the solution

So here comes my question im drawing a huge roadblock with some easy things.

From the results calcuate the total carbonate ion. Calculate the concentration of carbonate in the solution. Then calculate the percent carbonate in the unkown sample. Express the composition of the solid unknown in a form such as 63.4 wt% Na2CO3

Any help would be much appreciated, thank you for anybody who takes the time to help.

Explanation / Answer

Given :

CO32- + 2H+ ------> H2CO3

Mass of soda ash= 2.1383 g

Volume of solution = 250.0 mL = 0.250 L

Volume taken for titration = 25.0 mL = 0.025 L

[HCl] =0.10 M

Volume of HCl for titration = (18.2 +18.1 + 18.2)/3 = 18.17 mL = 0.01817 L

Calculation of moles of CO32-

Mol ratio of CO­3­2- : H+ is 1 :2

So the moles of CO­3­2- = Moles of HCl x 1 mol CO­3­2- / 2 mol HCl

We calculated moles of HCl by using following equation

n HCl = Volume of HCl in L x molarity = .01817 L x .10 M = 0.001817 mol HCl

n CO­3­2- = 0.001817 mol HCl x 1 mol CO­3­2- / 2 mol HCl

= 0.000908 mol CO­3­2- = 9.08 E -4 mol CO­3­2-

Calculation of ions of CO­3­2-

Number of CO­3­2- ion = moles of CO­3­2- X      Avogadro number

= 9.08E-4 mol x 6.02 E23 CO­3­2- ion / 1 mol CO­3­2-

= 5.47 E20 CO­3­2- ion.

Calculation of concentration of CO­3­2-

Volume of solution of CO­3­2- taken= 25.0 mL = 0.025 L

So we now got moles and volume in L

Therefore

[CO­3­2- ]= mol / volume of solution in L

= 9.108 E-4 mol / 0.025 L

= 0.0363 M

Calculation of moles of sodium carbonate

Moles of sodium carbonate = moles of carbonate ions since mole ration of Na2CO3 is 1 : 1

Mol Na2CO3 = moles of carbonate ion = 9.108 E-4 mol

Calculation of mass of sodium carbonate = Moles x molarity

= 9.108 E-4 mol x 105.998 g /mol

= 0.0963 g

Calculation of mass percent = Mass of sodium carbonate / mass of sample ) x 100

=(0.0963/2.1383)

= 4.4 %

So the percent is 4.4 %

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