You are given a sample containing NaCl ( aq ) and NaBr ( aq ), both with concent

Question

You are given a sample containing NaCl (aq) and NaBr (aq), both with concentrations of 0.020 M. Some of the figures below represent a series of snapshots of what molecular-level views of the sample would show as a 0.200 M Pb(NO3)2 (aq) solution is slowly added. In these figures, Br? is red-brown and Cl? is green. Also, spectator ions are not shown, nor are ions whose concentrations are much less than the other ions. The Ksp constants for PbCl2 and PbBr2 are 1.7 10-5 and 6.6 10-6, respectively.

Order the figures in a time sequence to show what happens as the lead(II) nitrate solution is added, excluding any that do not "make sense".

Which terms complete the statement below from the italicized words?

Will all, some, or none of the figures were excluded because PbBr2 will precipitate before, after, or at the same time as PbCl2

Explanation / Answer

Nacl-> Na+ + Cl-

NaBr->Na+ +Br-

since NaCl and NaBr are salts they dissociate in solution completely.. thus [Br-]=0.200 and [Cl-]=0.200

Now lead nitrate Pb(N03)2 is added to the solution.. thus forming compounds PbCl2 and PbBr2..

Among the above the one which requires lesser cooncentration of [Pb]+2 will precipitate first

therefore [Pb+2] required for PbCl2 to precipitate =

[Pb+2][Cl-]=Ksp(PbCl2), therefore [Pb+2]= 1.7*10^-5/0.2 =     8.7*10^-5 M

similarly [Pb+2] required for PbBr2 to precipitate is Ksp(PbBr2)/[Br-]=6.6*10^-6/0.2=     3.3*10^-5   M

clearly the concentration of lead(in the form of lead nitrate) required to precipitate PbBr2 is lesser .. therefore PbBr2 will get precipitated first.. when the concentration of lead reaches 8.7*10^-5 M ,both the compounds precipitate..

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