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2016 FRQ q6-(a) 이 문제를 이런 식으로 답변 해도 되나요?

Because of the large K value, the reaction will essentially go to completion (mostly products).

In the new 100mL solution the the concentration of Ba2+ is 0.1M, EDTA4- is 0.15M, but Ba2+ and EDTA4- react in 1:1 ratio to form BA(EDTA)2-, so Ba2+ is the limiting reactant.

Thus only 0.1M BA(EDTA)2- will be formed.

[BA(EDTA)2-] = 0.1M