19 March, 15:36

A precipitation reaction is caused by mixing 100. mL of 0.25 M K2Cr2O7 solution with 100. mL of 0.25 M Pb (NO3) 2 solution. When the precipitate forms, it is filtered from the mixture. Which describes the changes in concentration of the spectator ions K + and NO3 - in the reaction mixture as the reaction occurs? K + is doubled, NO3 - halvedNeither is affectedBoth are doubledK + is halved, NO3 - halved

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1. 19 March, 15:56
0
Both K + and NO3 - concentrations are halved

Explanation:

Since the reaction is

K₂Cr₂O₇ + Pb (NO₃) ₂ 2K (NO₃) + PbCr₂O₇

knowing that K + and NO3 - are spectator ions, their concentration do not change due to reaction but it will change due to the mixing of volumes of the reactants since we are mixing volumes of different concentrations.

for K + initially we have:

c K + initial = 0.25 M

c K + final = n / V = C i * Vin / Vfinal = 0.25 M * 100 ml / 200 ml = 0.125 M

the 200 ml comes from mixing the 100 ml of Pb (NO₃) ₂ and 100 ml of K₂Cr₂O₇ (assuming that the volume of the precipitate is insignificant relative to the solution, it does not occlude a significant portion of the saturated solution and assuming that any volume contraction/expansion due to non ideal behaviour of the solution is negligible)

the same happens with NO3-, therefore its concentration is also halved