Multiple question as per guideline only first should be answered, but for convenience of student both questions are answered :
1. We have, Ksp (solubility product) = [Ag+]2 [CrO42-]
We will take your calculated value as per question :
Ksp (solubility product) = [Ag+]2 [CrO42-] = 1.77*10-14
How mang grams of Ag2CrO4 will dissolve per litre of solution ?
or we have to calculate molar solubility :
Ag2CrO4 2 Ag+ + CrO42-
Initial concentration (M) 1 0 0
Change (M) Negligible +2s + s ( s = solubilty of Ag2CrO4)
At Equilibrium ~1 2s s
At Equilibrium Ksp = [Ag+]2 [CrO42-]
or Ksp = [Ag+]2 [CrO42-] = (2s)2 (s) = 4s3
or 1.77*10-14 = 4s3
Solubility (s) = 1.64*10-5 M
Molar mass of Ag2CrO4 = 331.73 g/mol
So, amount of Ag2CrO4 dissolved = 1.64*10-5 mol/L *331.73 g/mol = 5.45*10-3 g/L
2. When we mix solutions of AgNO3 and K2CrO4 following reaction takes place :
2 AgNO3 (aq) + K2CrO4 (aq) Ag2CrO4(s) + 2 K+(aq) + 2 NO3- (aq)
Net ionic reaction : 2 Ag+ (aq) + CrO42-(aq) Ag2CrO4(s)
So, it is evident that 2 moles of 2 Ag+ reacts with one mole of CrO42-(aq), so we had [AgNO3] larger than .[K2CrO4].
As per Ksp we have ions in solution :
[Ag+] = 3.28* 10-5 and [CrO42-] = 1.64*10-5
After formation of Ag2CrO4 we have following equilibrium ;
Ag2CrO4(s) 2 Ag+ (aq) + CrO42-(aq) ; Ksp = [Ag+]2 [CrO42-]
If, we add equal amount of both than Ag+ will act as limiting reactant and almost all of Ag+ will be in form of Ag2CrO4(s) and Ksp ( product of of ionic concentrations in solution) will product of higher concentration of [CrO42-] and very lower concentration of [Ag+] ( product of of ionic concentrations in solution) ; as Ksp is a constant value irrespective of amount of Ag2CrO4. Ag2CrO4(s) starts forming only after reaction Quotient exceeds solubility product.
need help with these two questions QUESTIONS 1. Based on the value of the Kap that...