Answer to Question #96609 in General Chemistry for Brianna

Question

In the laboratory you dissolve 21.9 g of lead(II) acetate in a volumetric flask and add water to a total volume of 500 mL.

What is the molarity of the solution? M.

What is the concentration of the lead cation? M.

What is the concentration of the acetate anion? M.

Answer 1

Moles of lead (II) acetate

$$n(Pb(CH_3COO)2) =\frac{m}{M} = \frac{21.9 g}{325.29 \frac{g}{mol}}=0.0673 mol$$

1) Molarity of the solution

$$c=\frac{n}{V} = \frac{0.0673 mol}{0.5 L}=0.135 M$$

2) When this salt gets into water it dissociates

$$Pb(CH_3COO)_2 \rightarrow Pb^{2+}+2CH_3COO^-$$

As 1 mole of $Pb(CH_3COO)_2$ gives 1 mole of $Pb^{2+}$ , then moles of $Pb^{2+}$

$$n(Pb^{2+})= n(Pb(CH_3COO)_2)=0.0673 mol$$

$$c(Pb^{2+})=\frac{n}{V}=\frac{0.0673 mol}{0.5 L} = 0.135 M$$

3) as 1 mole of $Pb(CH_3COO)_2$ gives 2 moles of $CH_3COO^-$ ,then moles of $CH_3COO^-$

$$n(CH_3COO^-)=2\times n(Pb(CH_3COO)_2)=2\times 0.0673 =0.135 mol$$

$$c(CH_3COO^-) =\frac{n}{V} = \frac{0.135 mol}{0.5 L} =0.270 M$$