Question

Suppose 9.07 g of copper(II) nitrate is dissolved in 100. mL of a 0.70 M aqueous solution of sodium chromate Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

 The molarity of the nitrate ion is1.0342 M

  Mass of copper(II) nitrate = 9.70grams

Volume of the 0.70 M aqueous solution of sodium chromate = 100 mL   =0.10 L

• The balanced equation is :-

  Cu(NO3)2 + Na2CrO4 → CuCrO4 + 2NaNO3 → 2Na+ + 2NO3-

For 1 mole Cu(NO3)2 consumed, we produce 2 moles of NO3-

• Calculate number of moles of Cu(NO3)2

           Number of moles = Mass of Cu(NO3)2/ Molar mass of Cu(NO3)2

            Moles Cu(NO3)2 = 9.70 grams /187.56 g/mol

             Moles Cu(NO3)2 =0.05171 moles

• Calculate moles of nitrate ion

         For 1 mole Cu(NO3)2 consumed, we produce 2 moles of NO3-

   For 0.05171mole of Cu(NO3)2,

        we have 2* 0.05171 = 0.10342 moles of NO3-

• Calculate molarity of the nitrate ion

         Molarity : Moles / volume

         Molarity NO3- =0.10342 moles / 0.10 L

          Molarity NO3- = 1.0342 M

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