net ionic equations

When a solution of silver nitrate and potassium chromate are mixed, a precipitate forms.?
Given 855 cm^3 of a 0.00330 silver nitrate is mixed with 434 cm^3 of a 0.0110 M potassium chromate, find: (please show all work)

a. Write the balanced equation for each substance when placed in water.
b. Write the equation for the NET IONIC REACTION
c. Write the Ksp expression
d. List and calculate the concentration of ALL IONS in solution after mixing but BEFORE any reaction takes place.
e. Calculate a trial Ksp, also called Q
f. Compare the trial Q with the Ksp and explaine how you can tell if a precipitate forms. Does a precipitate form?
g. List and calculate the concentration of ALL IONS in solution after mixing and AFTER ANY REACTION has taken place
h. Calculate the mass in grams of any precipitate formed
i. Using the unit of moles, COUNT the number of POSITIVE and NEGATIVE CHARGES of the ions remaining in solution after the reaction
j. Are the number of positive and negative charges equal to zero?
k. Explain your answer to J above

0.855L of 0.0033M AgNO3 is mixed with 0.434L of 0.011M K2CrO4

a)
Dissociation of silver nitrate:

AgNO3 < Ag(+) + NO3(-)

Dissociation of potassium chromate:
K2CrO4 < 2K(+) + CrO4 (2-)

I've included the charges on the ions, but not the subscripts.

b) General solubility rules state that nitrates are soluble and alkali metal salts are soluble, so the likely precipitate is silver chromate, Ag2CrO4

Net ionic equation:

2Ag(+) + CrO4(2-) < Ag2CrO4

c) For the dissociation of sparing soluble salts, the solid is placed as the reactant, so:

Ag2CrO4 < 2Ag(+) + CrO4(2-)

K=[P]/[R]
Ksp = [P]
= [Ag(+)]^2[CrO4{2-)]

d) At the moment of mixing:

[Ag(+)] = (0.0033)(0.855) / (1.289)
=0.00219M
[NO3(-)] = (0.0033)(0.855) / (1.289)
=0.00219M
[K(+)] = (2*0.011)(0.434) / (1.289)
=0.0074M

[CrO4(2-)] = (0.01)(.434) / (1.289)
=0.0037M

e) TIP = [Ag(+)]^2[CrO4(2-)]
= (0.00219)^2(0.0037)
=1.77x10^-8

f) The Trial Product uses the current scenario to predict whether precipitation will occur. If the TIP is greater than the Ksp of the salt, a precipitate will form. The Ksp of silver chromate is 1.12x10^-12, so yes a precipitate will form.

g) Assume completion reactions for all salts other than silver chromate because of the solubility of the nitrate salts and the alkali metal salts.

Ag2CrO4 < 2Ag(+) + CrO4(2-)

Ksp = [Ag(+)]^2 [CrO4(2-)]
Let the concentration of CrO4(2-) be x and the concentration of Ag(+) be 2x
1.12x10^-12 = (2x)^2 (x)
1.12x10^-12 = 4x^3
x=6.54x10^-5

Therefore,

[Ag(+)] = 2(6.54x10^-5)
=1.308x10^-4M
[CrO4(2-)] = 6.54x10^-5M

h) To solve for the mass of precipitate formed, find the number of moles of CrO4(2-) ions before any reaction has occured and subtract the number of moles of CrO4(2-) ions after the reaction. This is because in the dissociation reaction of silver chromate, there is a 1:1 mole ratio between silver chromate and the chromate ions.

n (before) = CV
= (0.0219)(1.289)
= 0.0028 mol

n (after) = CV
= (1.308x10^-4)(1.289)
= 1.686x10^-4 mol

n (precipitate) = n (before) - n (after)
= 0.0028 - 1.686x10^-4
=0.0026 mol

m (precipitate) = n*MM
=(0.0026)(331.73)
= 0.8625g

i) In the solution, there are 1.686x10^-4 moles of chromate ions, and because of the mole relationship, there are 3.372x10^-4 moles of silver ions. Each chromate ion has a charge of 2- and each silver ion has a charge of 1-.

+ve charge = (1)(3.372x10^-4)
=3.372x10^-4
-ve charge = (2)(1.686x10^-4)
=3.372x10^-4

j) The net charge equals zero.

k) The net charge equals zero because the solution ions in solution have been formed by the dissociation of the neutral salt Ag2CrO4.