Question

When $35$ mL of $0.15\mathrm{M}$ lead nitrate solution is mixed with $20\mathrm{mL}$ of $0.12\mathrm{M}$ chromic sulfate solution, _________ $\times {10}^{-5}$ moles of lead sulfate precipitate out. (Round off to the nearest integer).

Answer 1

Precipitate

• When you pour a liquid solution into it, you get a precipitate, which is an insoluble solid.
• When the solubility of solid declines as a result of a change in the solution, other than a chemical reaction or a change in temperature, a solid is synthesized. In meteorology, a precipitate is either liquid or solid water.

Explanation

$$\begin{gathered} 3\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3⟶3{\mathrm{PbSO}}_4+2\mathrm{Cr}{\left({\mathrm{NO}}_3\right)}_3 \\ 35\mathrm{mL}20\mathrm{mL} \\ 0.15\mathrm{M}0.12\mathrm{M} \\ M\times V35\times 0.15=5.2520\times 0.12=2.4 \end{gathered}$$

For $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ 3 moles, we need 1 mole of $C{r}_2{\left({\mathrm{SO}}_4\right)}_3$

For $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ 5.25 moles, we require 1/3×5.25 mole of ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$ , i.e., 1.75 moles

Meanwhile, we have 2.4 moles. As a consequence, ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$ is an excess reagent, while $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ is a limiting reagent (LR)

$3\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3⟶3{\mathrm{PbSO}}_4+2\mathrm{Cr}{\left({\mathrm{NO}}_3\right)}_3$

Moles of ${\mathrm{PbSO}}_4$ formed = moles of $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ consumed
= 5.25 m mol = 525 × 10^-5 moles

Conclusion

When 35 mL of 0.15 M lead nitrate solution is treated with 20 mL of 0.12 M chromic sulfate solution, $\mathbf{525}\times {10}^{-5}$moles of lead sulfate precipitate.