Answer to Question #279811 in Chemistry for Koala009

Question

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 is the right answer?

Answer 1

According to the equation, n (AgClO_3(aq)) = 2 x n (Ca(CH₃COO)_2(aq)).

C_M = n / V

n = C_M x V

n (Ca(CH₃COO)₂)= 0.33 x 0.0155 = 0.005115 mol

n (AgClO_3(aq))_available = 0.45 x 0.0205 = 0.009225 mol

n (AgClO_3(aq))_required = 2 x n (Ca(CH₃COO)_2(aq)) = 2 x 0.005115 = 0.01023 mol

The required amount of Ca(CH₃COO)_2(aq) is greater than its available quantity. Therefore, its remaining concentration is C) 0.0 mol/L.