Need help with equilibrium questions

Question

Need a solution to this question with working out if possible:

230 mL of 0.0015 mol/L silver nitrate is added to 130 mL of 0.010 mol/L calcium acetate. Does a precipitate form? Ksp for silver acetate is 0.0020. (7 marks) 

Answer 1

Does a precipitate form presumably of AgC2H3O2 with a Ksp of 2 x 10-3.

You are given a solution where 0.23 L of of a 0.0015 M solution of AgNO3 is added to 0.130 L of a 0.01M solution of Ca(C2H3O2)2.

Balanced equation:

2AgNO3 + Ca(C2H3O2)2 -------> 2AgC2H3O2 + Ca(NO3)2

Acetates are usually soluble with Group 1A metals and is minimally soluble with other metals,

Nitrates are essentially a soluble ion. So again presumably we are to determine if a precipitate will form of AgC2H3O2

when adding two solutions together you are diluting the Concentration of each compound which must be accounted for in your calculations,

Step 1) Convert the M given of AgNO3 and Ca(C2H3O2) to moles of each.

For AgNO3 0.23L x 0.0015mol/L = 3.45 x 10-4 mol which equals 3.45 x 10-4 mol of Ag+

(one Mole of AgNO3 = 1 mole of Ag+ and 1 mole of NO3-}

For Ca(C2H3O2)2 0.130L X 0.01 mol/L = 1.3 x 10-3 mol

(Note that for every 1 mole of Ca(C2H3O2)2 there are 1 mole of Ca+2 and 2 moles of C2H3O2-)

to account that there is twice the number of moles of C2H3O2- we multiply the moles found above by 2)

Moles of C2H3O2- = 2 x 1.3 x 10-3 = 2.6 x 10-3 mol

Step 2) Accounting for the fact we are adding solutions and therefore diluting its concentrations.

0.230 L of AgNO3 is added to 0.130 L of Ca(C2H3O2)2 we must use the total volume of both solutions.

0.230L + 0.130L = 0.36 L total Volume

Now take the Moles of each ion in solution we calculated above, Ag+ and C2H302- and divide by the total volume to obtain the new concentrations

For Ag+ 3.45 x 10-4mol/0.36L = 9.58 x 10-4 mol/L

For C2H3O2- 2.6 x 10-3 mol/0.36L = 7.22 x 10-3 mol/L

Step 3) We now need to use the equilibrium equation for AgC2H3O2 <----> Ag+ + C2H3O2-

and calculate just as if we were using the Ksp, but instead use the reaction quotient (ion product) given the symbol Q which is the initial concentrations before adjusting to equilibrium concentrations and compare that value to the Ksp.

a) if Q is greater than the Ksp than there will be a precipitate as the concentration is greater than the equilibrium concentration and the reaction will shift away from solubility of both ions but to the left to the solid compound.

b) if Q is less than the Ksp than there will be no precipitate forming as the initial concentrations are less than the equilibrium concentrations.

c) if Q is = to the Ksp the solution is at equilibrium.

4) Calculating Q and comparing it to the Ksp of AgC2H3O2

using the same formulation as if using the Ksp

AgC2H3O2 <-----> Ag+ + C2H3O2-

Q = {Ag+}{C2H3O2-}

Q = (9.58 x 10-4)(7.22 x 10-3 = 6.91 x 10-6

Ksp = 2 x 10-3

Q is less than the given Ksp so no precipitate will form.