You need to prepare a solution with a specific concentration of Na+Na + ions; however, someone used the end of the stock solution of NaClNaCl, and there isn't any NaClNaCl to be found in the lab. You do, however, have some Na2SO4Na2SO4. Can you substitute the same number of grams of Na2SO4Na2SO4 for the NaClNaCl in a solution? Why or why not?

Question

Chemistry

Saul Huerta

12 April, 08:10

You need to prepare a solution with a specific concentration of Na+Na + ions; however, someone used the end of the stock solution of NaClNaCl, and there isn't any NaClNaCl to be found in the lab. You do, however, have some Na2SO4Na2SO4. Can you substitute the same number of grams of Na2SO4Na2SO4 for the NaClNaCl in a solution? Why or why not?

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Answers (1)

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Eileen Moody · Answer 1

Ionic equation

NaCl (aq) - -> Na + (aq) + Cl - (aq)

Na2SO4 (aq) - -> 2Na + (aq) + SO4^2 - (aq)

In NaCl solution, 1 mole of Na + is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na + is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

= 142 g/mol

Molecular weight of NaCl = 23 + 35.5

= 58.5 g/mol

Masses

% Mass of NA + in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100

= 46/142 * 100

= 32.4%

% Mass of NA + in NaCl = mass of Na+/total mass of NaCl * 100

= 23/58.5 * 100

= 39.3%

Therefore, the % mass of Na + in NaCl and Na2SO4 are different so it cannot be used.