Upon mixing `50.0 mL` of `0.1 M` lead nitrate solution with `50.0 mL` of `0.05 M` chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed and what is the molar concertration of chromic suplhate left in the solution?

Options

A. 0.0005,0.0084
B. 0.0084,0.005
C. 0.005,0.00084
D. 0.05,0.00084

Answer

A

Explanation

`underset({:(50 xx 0.1),(= 5 "mmol"):})(overset(3 "mmol")(3Pb(NO_(3))_(2))) + overset(1 "mmol")Cr_(2)(SO_(4))_(3) rarr underset({:(50 xx 0.05),(= 2.5 "mmo'"):})overset(3 "mmol")(3PbSO_(4) darr) + overset(2 "mmol")(2Cr(NO_(3))_(3))`
First find the limiting reagent.
3 mmol of `Pb (NO_(3))_(2) implies` 1 mmol of `Cr_(2) (SO_(4))_(3)`
5 mmol of `Pb (NO_(3))_(2) rarr (1)/(3) xx 5`
`implies (5)/(3) = 1.66 "mmol"`
So `Pb(NO_(3))_(2)` is the limiting reagent.
i. 3 mmol of `Pb(NO_(3))_(2) implies 3` mmol of `PbSO_(4)`
5 mmol of `Pb(NO_(3))_(2) implies 5` mmol
`implies (5)/(1000) "mol" implies 0.005 "mol"`
Species left in the solution are `Cr_(2)(SO_(4))_(3)` and `Cr(NO_(3))_(3)`.
to calculate the concentration of `Cr_(2)(SO_(4))_(3)`:
Initial mmol = 2.5
Reacted mmol = 1.65
Left mmoles `= 2.5 - 1.6 = 0.84 "mmol"`
Total volume `= 50 + 50 = 100 mL`
Concentration `= (0.84)/(100) = 0.0084 M`
iii. To calculate the concentration of `Cr(NO_(3))_(3)`.
3 mmol of `Pb(NO_(3))_(2) implies 2` mmol of `Cr(NO_(3))_(3)`.
5 mmol of `Pb(NO_(3))_(2) implies (2)/(3) xx 5`
`implies (10)/(3) = 3.33 "mmol"`
Concentration `= (3.33)/(100) = 0.33 M`