anonymous
  • anonymous
Calculate the solubility of silver chloride in 0.17M AgNO3. (Ksp=1.8E-10)
OCW Scholar - Introduction to Solid State Chemistry
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
AgNO3 dissociates into Ag+ and NO3-, therefore our Ksp will be: Ksp = [Ag+][NO3-] because AgNO3 is a solid and solids & liquids are never counted in equilibrium problems because their concentrations are always constant. X moles of the silver chloride will dissociate which means X moles of both Ag+ and NO3- will be in the solution. To do this, I always set up a chart. AgNO3(s) -> Ag+ NO3- .17 0 0 Initial concentration -x +x +x Change in concentration .17-x x x Equilibrium concentration Now let's take the equilibrium values and then plug them into our Ksp. Ksp = 1x10-8 Ksp = [Ag+][NO3-] 1x10-8 = [Ag+][NO3-] *Now recall our equilibrium values. 1x10-8 = (x)(x) 1x10-8 =x^2 x = The square root of 1x10-8 which is 1x10-4

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