Question: How many grams of AgI can be dissolved in 150.0 mL of a solution that initially contains NaCn at a concentratoin of .440 M?
Would the equation be Ag(Cn)2 -> Ag+ 2CN? What numbers go into what place in the ice table. does the .440 M of CN get squared?
I think the equations you want are as follows:
AgI(s) ==> Ag^+ + I^-
Ksp = (Ag^+)(I^-) = look up.
Ag^+ + 2CN^- ==> Ag(CN)2^-
Kf = [Ag(CN)2^-]/(Ag^+)(CN^-)^2 = look up.
Please check my work? so the ksp value is 8.3E-17 and the Kf value is 5.6E18. i times those two together to get a K value of 464.8. which when i add the two equations together gives me AgI + 2CN ==> Ag(CN)2 + I. I'm kind of confused on were i go from there. does the .440M get pluged into the ice table and the e would be .440 for the CN and the Ag(Cn)2 would be nothing and the AgI -X/.1500L. Than you would solve for X. Is that how it is done?
There is no AgI in your final equation(and I'm not suggesting there should be); also, note that AgI as a solid has a molarity of 1 (by definition so the AgI-x has no meaning). (Technically it's the activity that is 1.)
Also note that the problem states 150.0 mL and not 1500 L.
To answer this question, we need to determine the maximum amount of AgI that can be dissolved in the given solution.
First, let's write the correct equation for the dissociation of NaCN:
NaCN -> Na+ + CN-
Since the compound AgI is not mentioned in the question, we cannot assume its presence. Therefore, we don't need to write an equation involving Ag(CN)2.
Now, to determine the amount of AgI that can dissolve, we must consider the reaction between AgI and CN-. This reaction can be represented as:
AgI + CN- -> Ag(CN) + I-
To find the maximum amount of AgI that can dissolve, we need to compare the initial concentration of CN- (given as 0.440 M) with the concentration of AgI that completely reacts with it.
A 1:1 stoichiometry ratio exists between the reactants AgI and CN-. Therefore, the maximum concentration of AgI that can dissolve is also 0.440 M.
Now, to calculate the number of grams of AgI that corresponds to a concentration of 0.440 M, we need the molar mass of AgI, which is approximately 234.772 grams/mole.
We can use the following equation to find the mass:
Mass = Concentration x Volume x Molar mass
Given that the concentration is 0.440 M and the volume is 150.0 mL (which can be converted to 0.150 L), we can substitute these values into the equation:
Mass = 0.440 M x 0.150 L x 234.772 g/mol
Solving this, we get:
Mass ≈ 15.72 grams
Therefore, approximately 15.72 grams of AgI can be dissolved in 150.0 mL of the given solution.