Question

Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

Explanation:

Let us assume that the given data is as follows.

   mass of barium acetate = 2.19 g

   volume = 150 ml = 0.150 L    (as 1 L = 1000 ml)

   concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

        $Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}$

Hence, moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times Ba(C_{2}H_{3}O_{2})_{2}$  .......... (1)

As,   No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of $Ba(C_{2}H_{3}O_{2})_{2}$ will be calculated as follows.                          

     No. of moles = $\frac{mass}{\text{molar mass}}$  

                          =  $\frac{2.19 g}{255.415 g/mol}$   (molar mass of $Ba(C_{2}H_{3}O_{2})_{2}$ is 255.415 g/mol)            

                       = $8.57 \times 10^{-3}$

    Moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times 8.57 \times 10^{-3}$

                          = 0.01715 mol

Hence, final molarity will be as follows.

              Molarity = $\frac{\text{no. of moles}}{volume}$

                             = $\frac{0.01715 mol}{0.150 L}$

                             = 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.