Question

300 mL of 0.200 M Pb(NO3)2 is added to 200 mL of 0.0500 M NaCl(aq) at 25°C. Will a precipitate of PbCl2 form at 25 °C? Tip: You have to calculate appropriate Qsp for this system. Ks- 1.7x10 for PbCl2 in water at 25 °C. (You MUST show your work to justify your answer.)

Answer 1

Step-by-step explanation:

It is given that volume of
`Pb(NO_(3))_(2)` is 300 mL and molarity is 0.200 M.

Volume of NaCl is 200 mL and molarity is 0.050 M.

The chemical reaction will be as follows.

`PbCl_(2)(s) \rightleftharpoons Pb^(2+)(aq) + 2Cl^(-)(aq)`

`K_(sp)` for
`PbCl_(2)` is given as
`1.7 * 10^(-5)`.

As, molarity is number of moles present in liter of solution.

Hence, moles of
`Pb^(2+)`(aq) will be calculated as follows.

moles of
`Pb^(2+)`(aq) =
`0.300 L * 0.200 M`

= 0.06 mol

`[Pb^(2+)]` =
`(0.06 mol)/(0.5 L)`

= 0.120 M

Mole of
`Cl^(-)(aq)` =
`0.2 L * 0.05 M`

= 0.010 M

Now, Q =
`[Pb^(2+)][Cl^(-)]^(2)`

=
`0.120 * (0.010)^(2)`

=
`1.2 * 10^(-5)`

As, Q <
`K_(sp)` hence, there will be no formation of
`PbCl_(2)` precipitate.