Question

Suppose

50.mL

of a

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aqueous solution of potassium carbonate.

Calculate the final molarity of zinc cation in the solution. You can assume the volume of the solution doesn't change when the zinc iodide is dissolved in it.

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

Answer 1

Chemical Equation

ZnI2 + K2CO3 --------> ZnCO3 + 2KI

Given

Volume of K2CO3 = 50 ml = 0.050 L

Molarity of K2CO3 = 0.50 M

Number of moles of K2CO3 = Volume x Molarity

= 0.05 x 0.5 = 25 x 10^-3 moles

1 mole of K2CO3 produces 1 mole of Zinc Carbonate (ZnCO3)

So, 25 x 10^-3 moles produces 25x10^-3 moles of ZnCO3

There fore Number of Moles of ZnCO3 = 25x10^-3 moles

Now, ZnCO3 after dissolves in water and Produces Zinc Caption.

But Molarity of Zinc Cation only depends on solubility of ZnCO3.

From the source 4.692 x 10^-5 g of ZnCO3 dissolves in 100 ml of H2O at 20 C.

Molar mass of Zinc Carbonate = 125.388 (g/mol)

Number of moles of ZnCO3 =4.692 x 10^-5 (g) / 125.388 (g/mol) = 3.74 x 10^-7 moles

1 mole of ZnCO3 produces 1 mole of Zinc Cation (Zn^2+)

So, number of moles of Zn^2+ = 3.74 x 10^-7 moles

Volume = 100 ml = 0.1 L

Molarity of Zinc Cation = Number of Moles / Volume

= 3.74 x10^-7 mol / 0.1 L

= 3.74 x 10^-6 ( mol/L )

= 3.74 x 10^-6 M