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if a titration using these solutions started with 25.00 mL of potassium chromate solution (0.5056 M),...

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Chemistry

if a titration using these solutions started with25.00 mL of potassium chromate solution (0.5056 M), what volume of0.2456 M lead (II) nitrate solution would be required to reach theend point? what MASS of the products would be created in thatreaction?

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4.0 Ratings (545 Votes)

Pb(NO3)2 (aq) + K2CrO4 (aq) ======> PbCrO4(s) + 2 KNO3 (aq)

1 mole of Pb(NO3)2 consume 1 mole of K2CrO4

number of mole of K2CrO4 is = molarity x volume in L.

                                                        = 0.5056 x 0.025

                                                         = 0.1264 mole.

so,0.1264 mole of lead (II) nitrate solution will needed to consue all K2CrO4.

volume of Pb(NO3)2 = no.of mole/ molarity

                                     =0.1264 / 0.2456 =0.05146 L

froduct is formed = 1 mole of PbCrO4 and 2 mole of KNO3

          mass of product = 1x 323.2 + 2x 101.1

                                        = 525.4 g
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