Question

a solution has 5.42 g of sodium carbonate in 0.367 l. what is its sodium ion concentration in mol/l? enter your answer in decimal format with three decimal places and no units.

Answer 1

Concentration of sodium ions, Na+ in mol/L = 0.278 mol/L

Sodium carbonate:

formula = Na₂ C03

Molar mass= 106.0 g/mol

first calculate moles:

Moles of Na₂CO3 = mass/Molar mass

                              =5.42/106·0g/m

                              = 0.0511 mol

Calculate concentration of Na₂CO3.

Concentration= moles/volume in L

                       =0.0511/0.367L

                       =0.1392 mol/L

Na₂C03 is a strong electrolyte that completely dissociates in aqueous solution: Na₂CO3 (aq) → 2 Na + (aq) + cost (aq)

Here, one mol Na₂CO3 produces 2 mols Na+ ions.

So, concentration of sodium ions, Na+ = 2(concentration of Na₂CO3)

                                      =2(0.1392 mol/L)

                                      = 0.278 mol/L

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