Question icon
Grade upto college level Physical Chemistry

Upon mixing 45.0 ml. of 0.25 M lead nitrate solution with 25.0 ml of 0.10 M chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble.

Profile image of Amit Saxena
12 Years agoGrade upto college level
Answers icon

1 Answer

Profile image of Navjyot Kalra
12 Years ago
Calculation of number of moles in 45 ml. of 0.025 M Pb(NO3)2
Moles of Pb(NO3)2 = 0.25 * 45/1000 = 0.01125
∴ Initital moles of Pb2+ = 0.01125
Moles of NO_{3}^{-} = 0.01125 * 2 = 0.02250 [1 mole Pb(NO3)2 ≡ 2 moles of NO3]
Calculation of number of moles in 25 ml. of 0.1 M chromic sulphate
Moles of chromic sulphate (Cr2(SO4)3 = 0.1 * 25/1000 = 0.0025 moles.
Moles of SO_{4}^{-} = 0.0025 * 3 = 0.0075 [1 Mole of chromic sulphate ≡ 3 moles of SO42-]
Moles of PbSO4 formed = 0.0075 [SO42- is totally consumed]
Moles of Pb2+ left = 0.01125 – 0.0075 = 0.00375
Moles of NO_{3}^{-}left = 0.02250 [NO3- remain untreated]
Moles of chromium ions = 0.0025 * 2 = 0.005
Total volume of the solution = 45 + 25 = 70 ml.
∴ Molar concentration of the species left
(i) Pb2+ = 0.00375/70 * 1000 = 0.05357 M
(ii) NO_{3}^{-}= 0.0225/70 * 1000 = 0.3214 M
(iii) Cr3+ = 0.005/70 * 1000 = 0.0714 M