A 500 mL solution is prepared in which 26.5 g of NaCl, 1.59 g of MgCl2, and 0.315 g of Na2CO3 is mixed and..

then dissolved in water.

a) determine the concentration (in M) of Cl- in solution.

b) determine the total concentration of all ions (collectively) in solution.

My final is tomorrow. I need some help on how to do this because i am very LOST!

Hi Bri,


I am french (Boulogne sur mer 62200 - FRANCE)



1°/ Determine the concentration (in M) of Cl(-) in solution.
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M(NaCl) = M(Na) + M(Cl)
M(NaCl) = 23 + 35,5
M(NaCl) = 58,5 g/mole

M(MgCl2) = M(Mg) + 2M(Cl)
M(MgCl2) = 24,3 + (2 x 35,5)
M(MgCl2) = 95,3 g/mole

M(Na2CO3) = 2M(Na) + M(C) + 3M(O)
M(Na2CO3) = (2 x 23) + 12 + (3 x 16)
M(Na2CO3) = 106 g/mole



n(NaCl) = m(NaCl) / M(NaCl)
n(NaCl) = 26,5 / 58,5
n(NaCl) = 0,453 mole

n(MgCl2) = m(MgCl2) / M(MgCl2)
n(MgCl2) = 1,59 / 95,3
n(MgCl2) = 1,668 • 10-2 mole

n(Na2CO3) = m(Na2CO3) / M(Na2CO3)
n(Na2CO3) = 0,315 / 106
n(Na2CO3) = 2,972 • 10-3 mole


n(Cl(-)) = n(NaCl) + 2n(MgCl2)
n(Cl(-)) = 0,453 + (2 x 1,668 • 10-2)
n(Cl(-)) = 0,4864 mole

[Cl(-)] = n(Cl(-)) / V
[Cl(-)] = 0,4864 / 0,500
[Cl(-)] = 0,9727 mole/Liter


2°/ Determine the total concentration of all ions (collectively) in solution.
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[ ions ] = ( 2n(NaCl) + 3n(MgCl2) + 3n(Na2CO3) ) / V
[ ions ] = ( (2 x 0,453) + (3 x 1,668 • 10-2) + (3 x 2,972 • 10-3) ) / 0,500
[ ions ] = 1,930 mole/Liter


I hope to have answered your question.
.