Silver chloride has Ksp=1.6x10^-10 and silver cheomate has Ksp=9.0x10^-12. We have 1.00 liter of a solution contains both nacl(0.10M) and Na2cro4(0.10M) solid AgNo3 is added slowly and the solution is stirred well.

a) which precipitate first Agcl or AgCro4?
B)what are the concentration of [Ag^+],[Cl-] and [CrO4^2-] at the point when the second salt just begins to precipitate?
C) is the method of selective precipitatation a good one for separating the anions in the solution?[hint calculate the percent of the fist ion (to precipitate) remaining in solution at the point the second anion begins to precipitate,in part (b) 99.5+% removed indicates good seperation!]

I divided ksp of each one /M and then found which precipitate first is that right can you pls explain parts b and c thanks!

First you need to be sure part a is right. Did you get AgCl pptng first? Ksp/0.1 is Ag^+ needed for pptn AgCl but sqrt (Ksp/0.1) is Ag^+ needed for pptn of Ag2CrO4.

Adding AgNO3 drop wise to the mixed solution will ppt AgCl first.

B. AgCl will continue to ppt (with no Ag2CrO4) until the Ksp for Ag2CrO4 is reached. What is the (Ag^+) at that point. (Ag^+)(CrO4^2-) = 9E-12
Since CrO4 is 0.1, then Ag^+ (from above) is sqrt(9E-12/0.1) = 9.48E-6
Knowing that is Ag^+ you can calculate Cl^- from Ksp AgCl. Of course the CrO4^2- is 0.1 (ok, it's 1 ion less than that).

C.
You have the initial Ag^+ from part A. You know Ag^+ when Ag2CrO4 just begins to ppt (from part B).
Calculate the percentage Ag still in solution and if it is 0.5% or less then the two can be separated this way.

To determine which precipitate, AgCl or AgCrO4, forms first, we need to compare the solubility product constants (Ksp) of both compounds.

a) AgCl has a Ksp of 1.6x10^-10, while AgCrO4 has a Ksp of 9.0x10^-12. The compound with the lower Ksp value will have a lower solubility, and therefore, precipitate first.

In this case, AgCl has the lower Ksp value (1.6x10^-10 < 9.0x10^-12), so AgCl will precipitate first.

b) When the second salt begins to precipitate, the solution is at the point of equilibrium, where the concentrations of both the dissolved ions and the precipitated ions reach a stable state.

Let's first write out the balanced chemical equations for the reactions involved:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)

To find the concentrations of [Ag+], [Cl-], and [CrO4^2-] at the equilibrium point, we need to consider their initial concentrations and how they change during the precipitation process.

We start with 1.00 liter of a solution containing both NaCl (0.10M) and Na2CrO4 (0.10M). Since NaCl is a strong electrolyte, it will dissociate completely into Na+ and Cl- ions. Similarly, Na2CrO4 will dissociate completely into 2Na+ and CrO4^2- ions.

Initially, the concentration of [Ag+] is zero, and the concentrations of [Cl-] and [CrO4^2-] are 0.10M.

When AgNO3 is added slowly, it reacts with Cl- ions according to the balanced chemical equation AgCl(s) ⇌ Ag+(aq) + Cl-(aq). Ag+ ions from AgNO3 combine with Cl- ions to form insoluble AgCl precipitate.

As AgCl precipitates, Cl- ions are consumed, reducing their concentration in the solution. At the same time, Ag+ ions are released, increasing their concentration.

At the equilibrium point, the concentrations of [Ag+], [Cl-], and [CrO4^2-] will depend on their solubility product expressions and the stoichiometry of the precipitate reactions.

For Ag+ and Cl- ions:
Ksp (AgCl) = [Ag+][Cl-] = 1.6x10^-10

For Ag+ and CrO4^2- ions:
Ksp (Ag2CrO4) = [Ag+]^2[CrO4^2-] = 9.0x10^-12

At the point when the second salt just begins to precipitate, the product of the concentrations of the ions involved in the respective Ksp expressions will be equal to their Ksp values.

Solving these equations together, we can find the concentrations:

1.6x10^-10 = [Ag+][Cl-]
9.0x10^-12 = [Ag+]^2[CrO4^2-]

Since [Ag+] = [Cl-] at the equilibrium point due to their 1:1 stoichiometry, we can substitute [Ag+] for [Cl-] in the second equation:

9.0x10^-12 = ([Cl-])^2[CrO4^2-]

Since we know the initial concentration of [Cl-] is 0.10M, we can substitute it into the equation:

9.0x10^-12 = (0.10)^2[CrO4^2-]

Solving for [CrO4^2-]:

[CrO4^2-] = 9.0x10^-12 / (0.10)^2

c) To determine if the method of selective precipitation is good for separating the anions in the solution, we need to calculate the percent of the first ion remaining in solution at the point the second anion begins to precipitate.

Let's calculate the percentage of Cl- remaining in solution:

Percent remaining = ([Cl-] remaining / [Cl-] initial) x 100

At the point when Ag2CrO4 starts to precipitate, [Cl-] = 0.10 - [Cl-] precipitated (since [Cl-] precipitated = [Ag+] at the equilibrium point)

Substituting the values:

Percent remaining = ([Cl-] - [Ag+]) / [Cl-] x 100

where [Ag+] is determined from the previous calculation.

If the percentage is close to 100% (e.g., 99.5% or higher), it indicates good separation.