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Na2S (aq) is added to 13.3 mL of a solution of Ni2+ ions until NiS ceases...

Na2S (aq) is added to 13.3 mL of a solution of Ni2+ ions until NiS ceases to precipitate. The precipitate is dried and determined to be of mass 0.0543 g. Assuming 100% yield, what was the initial concentration of the Ni2+ solution.

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Answer #1

Molar mass of NiS = 58.7 + 32 = 90.7 gm/mol

Number of moles = mass/molar mass = 5.9867 * 10^(-4) moles

Molarity = Number of moles/Volume in(L)

=> 5.9867 * 10^(-4)/(13.3/1000)

=> 0.045M

Hence the initial concentration of Ni(2+) is 0.045M

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