IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

sulekha_hi

Upon mixing 45.0 mL OF lead nitrate soilution with 25 mL OF 0.10M chromic sulphate,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 final solution.Assume that lead sulphate is completely insoluble.

By using the formula concentration= Milli equivalent / total vol* valency i got mole of PbSO4 precipitated as 0.0075

Then how to find concentrations of ions.

amrita

concentration of Pb2+ ions=(no. of moles of pb2+) x 1000/vol.[molarity]
no.of.moles of Pb2+ ions= no.of. moles of PbSo4 x 1
concentration of So4 2- ions= (no.of moles of so4 2-)x 1000/vol.
no.of moles of So4 2- = no. of moles of pbso2 x 1
for example,
no.of moles of No3 - in Pb(No3)2=no. of moles of Pb(No3)2 x 2

plz dont mind if it's wrong......

arun-rashi

in this question you cannot give concentration of lead nitrate .in these question you apply conservation of milliequivalents for example 2 millieq of BaCl2 reacts with 6 millieq of Al2(SO4)2 to give AlCl3 and BaSO4 ,HERE limiting fector is barium choloride so 2 millieq of barium sulphate solid formed we easily cal wt of this and also cal concentration of AlCl3 it is soluble=M*V/fector *total volume .we can say that aftyer mixing volume increase and also remember that moles =equivalents/velency or molerity =normality/fector.

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