If 100.0 mL of 0.150 M sodium nitrate is added to 50.0 mL of 0.150 M sodium carbonate,what is the resulting molar concentration of the sodium ion?

Question

Quizplus · Accepted Answer

The total volume of the solution after mixing is 100.0 mL + 50.0 mL = 150.0 mL. Both sodium nitrate and sodium carbonate dissociate in water to give two moles of sodium ions per mole of sodium carbonate and one mole of sodium ion per mole of sodium nitrate. Therefore, the total moles of sodium ions are (0.150 mol/L * 0.100 L) from sodium nitrate + (0.150 mol/L * 0.050 L * 2) from sodium carbonate = 0.015 mol + 0.015 mol = 0.030 mol. The concentration of sodium ions is then 0.030 mol / 0.150 L = 0.200 M.

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