[Karela]

Okay so I am fine until I get to 'G'. I was using the method my professor used in class but then while looking online I saw someone posted this quiz a few years back and the person who answered had a different method and a different solution so now I am unsure as to what method to use. I will post both and hopefully someone will tell me which is the right one. Thank you.

When a solution of silver nitrate and potassium chromate are mixed, a precipitate forms.  Given 855cm^3 of a 0.0033M silver nitrate is mixed with 434 cm^3 of a 0.0110M potassium chromate, find:

a. Write the balanced equation for each substance when placed in water.

AgNO₃(s) Ag¹⁺(aq) + NO₃^1-(aq)

K₂CrO₄(s) 2K¹⁺(aq) + CrO₄^2-(aq)

b. Write the equation for the Net Ionic reaction

Ag₂CrO₄(s) 2Ag¹⁺(aq) + CrO₄^2-(aq)

c. Write the Ksp expression

K_sp = [Ag¹⁺(aq)]² × [CrO₄^2-(aq)]

d. List and calculate the concentration of all ions in solution after mixing but BEFORE any reaction takes place.

For AgNO₃ the moles is (0.855L)(0.00330M) = 0.00282 mol, and it's a 1:1 ratio throughout so Ag¹⁺ and NO₃^1- both have 0.00282 mols.
For K₂CrO₄ the moles is (0.434L)(0.0110M) = 0.00477 mol and it's a 1:2:1 ratio so K¹⁺ has 0.00954 mol and CrO₄^2- has 0.00477 mols.

[Ag¹⁺] = (0.00282mol)/(1.29L) = [0.00219]
[NO₃^1-] = (0.00282mol)/(1.29L) = [0.00219]
[K¹⁺] = (0.00954mol)/(1.29L) = [0.00740]
[CrO₄^2-] = (0.00477mol)/(1.29L) = [0.00370]

e. Calculate a trial Ksp, also called Q.

Q = [Ag¹⁺(aq)]² × [CrO₄^2-(aq)]

Q = [0.00219]² × [0.00370]

Q = [1.77×10^-8]

f. Compare the trial Q with the Ksp and explain how you could tell if a precipitate forms.  Does a precipitate form?

The trial K_sp or Q is larger than K_sp which is 1.12×10^-12
Because Q > K, so a precipitate does form.

*g. List and calculate the concentration of all ions in solution after mixing and AFTER and reaction has taken place.

This was what I thought to do:

2Ag¹⁺(aq) + CrO₄^2-(aq)  Ag₂CrO₄(s)
You have 0.00282 mol of Ag and 0.00477 mol of CrO₄. For every one mole of chromate ion, you need two moles of silver ion. I divided the chromate ion by two (0.00477)/(2) = 0.00239 moles. So I thought there would be 0.00 moles of silver ion left, 0.00239 moles of chromate ion left, and 0.00239 moles of silver chromate formed. Then take the moles of silver and chromate and divide by the volume of the solution (1.29L). I obtained:
[Ag¹⁺] = [0.00] 
[NO₃^1-] = [0.00219]
[K¹⁺] = [0.00740]
[CrO₄^2-] = [0.00185]

This is what the person online said to do:

Have the silver ion be 2x and the chromate ion to be just x, and then plug those into the k_sp equation
1.12×10^-12 = [2x]²×

1.12×10^-12 = [4x]³
x = 6.54×10^-5

Therefore Ag⁺ would be 2(6.54×10^-5) = [1.31×10^-4]
and CrO₄ is [6.54×10^-4]

I just do not know which method is correct. I know how to solve 'H' through 'K', but the values are rather dependent on my answer to 'G' so I am really hoping someone can tell me which method is correct. Also I am unsure, if it is the second method, would the concentrations of NO₃^2- and K¹⁺ be the same as they were originally?

h. Calculate the mass in grams of any precipitate found.
i. Using the unit of moles, count the number of positive and negative charges of the ions remaining in solution after reaction.
j. Are the number of positive (+) and negative (-) charges equal to Zero?
k. Explain your answer to j above.

[Borek]

Have the silver ion be 2x and the chromate ion to be just x

That's incorrect - that WOULD be correct for a solution prepared by putting solid silver chromate in water and letting it dissolve till the solution becomes saturated. That's completely different kind of a problem.

I have not checked every detail of the first solution, but looks to me like your calculation of the final concentrations is wrong. Basically it is a limiting reagent type problem, although final concentration of neither ion is zero - you have a saturated solution of chromate. You should calculate what was left in excess after precipitation, calculate concentration of this ion, then use this concentration and K_sp to calculate concentration of the other ion.

It still doesn't have to be the correct result, there are some simplifying assumptions behind this approach.

[Karela]

THANK YOU!! It didn't even occur to me to think about limiting reactants. I think I got the answer now, thank you so much.