In a waste-water treatment plant a 1.00x103 L tank with 3.00x10-2 M Na2CO3(aq) is mixed with a 2.00x103...

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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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.036M

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3000L

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

no. of moles = molarity * volume

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.036M * 3000L

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 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

Â Â Â Â Â Â Â Â Â Â Â = 216 moles/ 3000 L

Â Â Â Â Â Â Â Â Â Â = 0.072M

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