Question
Ba2+(aq) + EDTA4−(aq) ⇔ Ba(EDTA)2−(aq) K = 7.7 × 107
The polyatomic ion C10H12N2O84− is commonly abbreviated as EDTA4−. The ion can form complexes with metal ions in aqueous solutions. A complex of EDTA4− with Ba2+ ion forms according to the equation above. A 50.0 mL volume of a solution that has an EDTA4−(aq) concentration of 0.30 M is mixed with 50.0 mL of 0.20 M Ba(NO3) 2 to produce 100.0 mL of solution.
(a) Considering the value of K for the reaction, determine the concentration of Ba(EDTA)2−(aq) in the 100.0 mL of solution. Justify your answer.
(b) The solution is diluted with distilled water to a total volume of Ba2+ (aq) 1.00 L. After equilibrium has been reestablished, is the number of moles of present in the solution greater than, less than, or equal to the number of moles of Ba2+(aq) present in the original solution before it was diluted? Justify your answer.
▶️Answer/Explanation
Ans:
(a)
| Based on the K value, the reaction goes essentially to completion. Ba2+(aq) is the limiting reactant. The concentration of Ba2+ when the solutions are first mixed but before any reaction takes place is 0.20 M/2 = 0.10 M. Thus the equilibrium concentration of Ba(EDTA)2- (aq) is 0.10 M. |
(b)
| The number of moles of Ba2+(aq) increases because the percent dissociation of Ba(EDTA)2- (aq) increases as the solution is diluted. OR A mathematical justification such as the following: The dilution from 100.0 mL to 1.00 L reduces the concentrations of all species to one tenth of their original values. Immediately after the dilution, the reaction quotient, Q, can be determined as shown below. \(Q = \frac{\frac{1}{10}[Ba(EDTA)^{2-}]}{\frac{1}{10}[Ba^{2+}]\times \frac{1}{10}[EDTA^{4-}]}=10K\) Because Q > K, the net reaction will produce more reactants to move toward equilibrium, so the number of moles of Ba2+(aq) will be greater than the number in the original solution. |