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Suppose 2.57g of nickel(II) chloride is dissolved in 350.mL of a 38.0mM aqueous solution of potassium carbonate.

Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it.

Be sure your answer has the correct number of significant digits.

Respuesta :

Answer:

Molarity Ni^2+ = 0.0566 M

Explanation:

Step 1: Data given

Mass of nickel(II) chloride = 2.57 grams

volume of an aqueous 38.0 mM solution of potassium carbonate = 350 mL =0.350 L

The molar mass of nickel (II) chloride = 129.6 g/mol

Step 2: The balanced equation

NiCl2(aq) + K2CO3(aq) → 2KCl(aq) + NiCO3(s)

The dissociation of nickel (II) chloride:

NiCl2 →  Ni^2+ + 2Cl-

Step 3: Calculate moles NiCl2

Moles NiCl2 = mass NiCl2 / molar mass NiCl2

Moles NiCl2 = 2.57 grams / 129.6 g/mol

Moles NiCl2 = 0.0198 moles

For 1 mol NiCl2 we have 1 mol Ni^2+

For 0.0198 moles NiCl2 we have 0.0198 moles Nickel(II) cation

Step 4: Calculate molarity Ni^2+

Molarity Ni^2+ = moles / volume

Molarity Ni^2+ = 0.0198 moles / 0.350 L

Molarity Ni^2+ = 0.0566 M

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