Question

Suppose 10.3 g of lead (II) acetate is dissolved in 150. mL of a 0.40 M aqueous solution of ammonium sulfate.Calculate the final molarity of lead(II) cation in the solution. You can assume the volume of the solution doesn't change whenthe lead(II) acetate is dissolved in it.Round your answer to 3 significant digits.Ом3

Answer 1

We have a solution of ammonium sulfate and we add another solute, lead (II) acetate (Pb(C2H3O2)2), so the molarity of the solution will increase. Molarity is defined as moles of solute per volume of solution and can be described by the following equation:

`Molarity=(MolesSolute)/(Lsolution)`

In this case, the moles of solute will be the sum of the moles of both compounds. The volume will be the same, according to the statement, 150.mL or 0.150L.

So to find the final molarity of the solution we must find the moles of each compound and add them.

Moles of ammonium sulfate, (NH4)2SO4

We use the molarity equation:

`\begin{gathered} MolesSolute=Molarity* Lsolution \\ \\ MolesSolute=0.40M*0.150L=0.06mol(NH_4)_2SO_4 \end{gathered}`

Moles of lead (II) acetate, Pb(C2H3O2)2

We use the molar mass of the compound. The molar mass of Pb(C2H3O2)2 is 325.29g/mol. So, the moles will be:

`molPb\left(C_2H_3O_2\right)_2=givenMass,gPb\left(C_2H_3O_2\right)_2*(1molPb\left(C_2H_3O_2\right)_2)/(MolarMass,gPb\left(C_2H_3O_2\right)_2)`
`\begin{gathered} molPb\left(C_2H_3O_2\right)_2=10.3gPb\left(C_2H_3O_2\right)_2*(1molPb\left(C_2H_3O_2\right)_2)/(325.29gPb\left(C_2H_3O_2\right)_2) \\ \\ molPb\left(C_2H_3O_2\right)_2=0.0317Pb\left(C_2H_3O_2\right)_2 \end{gathered}`

Now, we calculate the final molarity:

`\begin{gathered} Molarity=(MolesSolute)/(Lsolution) \\ \\ Molarity=(0.06mol(NH_4)_2SO_4+0.0317molPb\left(C_2H_3O_2\right)_2)/(0.150L) \\ \\ Molarity=0.611M \end{gathered}`

Answer: The final molarity of the solution is 0.611M