Question

Suppose of lead(II) nitrate is dissolved in of a 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 when the lead(II) nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

The final molarity of lead(II) cation in the solution is 0 M.

Explanation:

Mass of lead(II) nitrate = 1.14 g

Molar mass of lead nitrate = 331 g/mol

Moles of lead (II) nitrate = $\frac{1.14 g}{331 g/mol}=0.003444 mol$

Moles of ammonium sulfate = n

$Moles(n)=Molarity(M)\times Volume (L)$

Volume of the solution 43.0 mM  ammonium sulfate = 200 ml = 0.200 L

Molarity of the solution = 43.0mM = 0.043 M

$n=0.043 M\times 0.2 L = 0.0086 mol$

$Pb(NO_3)_2+(NH_4)_2SO_4\rightarrow PbSO_4+2NH_4NO_3$

According to reaction, 1 mol of lead nitrate reacts with 1 moles of ammonium sulfate. Then 0.003444 mol will react with:

$\frac{1}{1}\times 0.003444 mol=0.003444 mol$ of ammonium sulfate

As we can see that lead (II) nitrate is present in limiting amount and it will get completely consumed by ammonium sulfate to produce lead sulfate due to which no moles of lead(II) ions will remain in the solution.

The concentration of lead(II) ions = $[Pb^{2+}]$

$[Pb^{2+}]=\frac{0.00 mol}{0.200 L}=0 M$

The final molarity of lead(II) cation in the solution is 0 M.