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Unformatted text preview: Then (0.06 x) M Ag+ reacts with CrO 4 2-. CrO 4 2-used is (0.06 x) / 2 = (0.03 0.5x ) M 2Ag + (aq) + CrO 4 2-(aq) Ag 2 CrO 4 (s) Initial 0.06 0.165 Change-(0.06 x)-(0.03 0.5x) Equilibrium x 0.165-(0.03-0.5x) = 0.162 + 0.5x When , no precipitate will form Solve the equation gives: So at the equilibrium, the concentration of silver ion is . It is easy to find that CrO 4 2-is surplus. So silver ion is very dilute at the equilibrium....
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- Spring '11
- Analytical Chemistry