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

if a titration using these solutions started with 25.00 mL of potassium chromate solution (0.5056 M), what volume of 0.2456 M lead (II) nitrate solution would be required to reach the end point? what MASS of the products would be created in that reaction?

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Expert Solution

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

queen_honey_blossom answered 9 months ago

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