Question

precipitate all of the Ca+ and Pb+ as CaF2() and PbF26). The precipitate was dried and wei

Answer 1

The number of mmol of NaF are $\text { 0.575 M} \times \text { 62.5 mL} = \text { 35.94 mmol }$

35.94 mmol of NaF will give $= \frac{\text { 35.94 mmol }}{2}=\text { 17.97 mmol }$ of the precipitate.

Let x mmol of precipitate are CaF2. 17.97-x mmol of precipitate will be PbF2

The molar mass of CaF2 is 78 g/mol. The mass of $CaF_2 = \text { 78 g/mol } \times \text { x mmol } \times \text {0.001 mol/mmol } = \text { 0.078x g}$

The molar mass of PbF2 is 245.2 g/mol.

$PbF_2 = \text { 245.2 g/mol } \times \left ( 17.97-x \right )\text { mmol } \times \text {0.001 mol/mmol } = \text { 4.41}-0.2452x \: g$

Total mass of the precipitate

$= \text { 0.078x g} +\left ( \text { 4.41}-0.2452x \: g \right )$

$= \text { 4.41}-0.1672x \: g$

But this is equal to 3.993 g

$3.993 \: g = \text { 4.41}-0.1672x \: g$

$0.1672x=0.413$

$x=2.47$ mmol

$[Ca^{2+}]=[CaF_2]=\frac{2.47 \: mmol}{45.0 \: mL} = 0.0549 \: M$