# precipitation answer - Then(0.06 – x M Ag reacts with CrO...

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precipitation Will a precipitate form when 250 ml of 0.33 M Na2CrO4 are added to 250 ml 0.12 M AgNO3 ? (Ksp Ag2CrO4= 1.1 e-12) What is the concentration of the silver ion remaining in solution? My answers to this question is Yes a precipitate will form and [Ag+]=0.060M Could you check my work? Thank you 2Ag + (aq) + CrO 4 2- (aq)  Ag 2 CrO 4 (s) After the solutions are mixed together, [CrO42-] = 0.33 * 250/ (250 + 250) = 0.165 M [Ag+] = 0.12 / 2 = 0.06 M Therefore, a precipitate forms. Assume x M Ag + remains in the solution

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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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precipitation answer - Then(0.06 – x M Ag reacts with CrO...

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