What is the solubility of Ag2CrO4 in 0.10 M K2CrO4?

2 Answers

Relevance
  • ?
    Lv 7
    2 months ago

    Ag2CrO4(s) = 2 Ag+(aq) + CrO4 2- (aq) Ksp = 1.2 x 10-12 

    We have for a solution of potassium chromate 0.1M

    Ksp=[Ag+]^2*[CrO24−]

    if s moles of the salt dissolves we get

    (2s)2*(s+0.1) = 1.2×10−12

    since s is small compared with 0,1 you can write:

    Ksp = 4s^2 *.1 =0.4s^2 = 1,2×10−12

    =SQRT( 1.2×10−12 / 0.4) 

    = 1.73*10^-6 moles/ liter

  • 2 months ago

    It depends on the Ksp of Ag2CrO4.

Still have questions? Get your answers by asking now.