Question

In a waste-water treatment plant a 1.00x10³ L tank with 3.00x10^-2 M Na₂CO₃(aq) is mixed with a 2.00x10³ L tank with of 6.00x10^-3 M Na₂CO₃(aq). Assume sodium carbonate behaves as a strong electrolyte.

a) How many moles of sodium ions are in the product?

b) What is the concentration (in M) of sodium ions in the product?

Answer 1

A) Number of moles of sodium carbonate after mixing the two tanks can be calculated as below:

Total concentration of Na2CO3 = 3.00x10^-2 M + 6.00x10^-3 M

                                              = 0.036M

Total volume of water is: (1.00x10³ + 2.00x10³) L

                                  = 3000L

in 1 mole of Na2CO3 ------------ 2 moles of Na+ ions are there.

no. of moles = molarity * volume

                   = 0.036M * 3000L

                   = 108 moles of Na2CO3

Hence there will be 108* 2 = 216 moles of Na+ ions.

B) Concentration of sodium ions in the product:

Molarity = no.of moles/ volume in L

            = 216 moles/ 3000 L

           = 0.072M