Ksp for Ag(C2H3O2)2 is 1.9E-3. I don't know how advanced this class is but the bottom line is the silver acetate will ppt BUT it is fairly soluble so I don't know if we calculate the solid then add back the solubility. First up we'll just calculate the solid that forms and see how that agrees with the choices in the problem for answers.
millimoles Ca(CH3COO) = mL x M = 15.5 mL x 0.33 M = 5.12
millimoles 20.5 mL x 0.45 M = 9.22
Ca(CH3COO)2(aq) + 2 AgClO3(aq) --> 2 AgCH3COO(s) + Ca(ClO3)2(aq)
I............5.12....................9.22.........................0..........................0
Limiting reagent is AgClO3 and it will use 5.12/2 = 2.56mmoles Ca(Ac)2.
C.......... -4.61....................-9.22................+9.22.......................+4.61
E.........0.51..........................0.....................9.22...........................4.61
So you have 0.51 millimoles Ca(CH3COO)2 left unreacted. Assuming that the AgCH3COO(s) is so insoluble that no acetate is added from the solid, the millimoles/mL of Ca(CH3COO)2 is 0.51mmoles/36 mL = 0.014 M. Since there are two mols CH3COO^- for every mole Ca(CH3COO)2 the acetate concentration is 0.014 x 2 = ?
15.5 mL of 0.33 mol/L calcium acetate reacts with 20.5mL of 0.45 mol/L silver chlorate. Determine the concentration of the acetate ion in the final solution.
Ca(CH3COO)2(aq) + 2 AgClO3(aq) --> 2 AgCH3COO(s) + Ca(ClO3)2(aq)
A) 0.014 mol/L
B) 0.028 mol/L
C) 0.0 mol/L
D) 0.37 mol/L
Which one is the right answer?
1 answer
8 answers