Upon mixing 50.0 mL of 0.1 M lead
nitrate solution with 50.0 mL of 0.05 M
chromic sulphate (Cr2(SO4)3)
solution, precipitation of lead sulphate
takes place. Calculate the molar
concentration of the species left behind
in the final solution.
Pb(NO3)2 + Cr2(SO4)3
→ PbSO4 + +Cr(NO3)3
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Answered by
1
Answer:
=5mmol
50×0.1
3Pb(NO
3
)
2
)
3mmol
+
Cr
2
)SO
4
)
3
1mmol
→
=2.5mmol
50×0.05
3PbSO
4
)
3mmol
↓+
2mmol
2Cr(NO
3
)
2
First find the limiting reagent.
3 mmol of Pb(NO
3
)
2
⇒ 1 mmol of Cr
2
(SO
4
)
3
5 mmol of Pb(NO
3
)
2
⇒
3
1
×5
⇒
3
5
=1.66mmol
So, Pb(NO
3
)
2
is the limiting reagent.
(i) 3 mmol of Pb(NO
3
)
2
⇒3 mmol of PbSO
4
5 mmol of Pb(NO
3
)
2
⇒5mmol
⇒
1000
5
mol⇒0.005mol
(ii) 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 =2.5−1.66=0.84mmol
Total Volume =50+50=100mL
Concentration=
100
0.84
=0.0084M
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