Map a A solution prepared by mixing 50.0 mL of 0.400 M AgNO, and 50.0 ml of 0.40

Question

Map a A solution prepared by mixing 50.0 mL of 0.400 M AgNO, and 50.0 ml of 0.400 M TINO, was titrated with 0.800 M NaBr in a cell containing 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIBr is Kup 3.6 x10-s and the solubility constant of AgBr is Ksp 5.0x10-13 a silver indicator electrode and a reference electrode of constant potential (a) Which precipitate forms first? O TIBr O AgBr (b) Which of the following expressions shows how the cell potential, E, depends on [Ag'1? (1) E=0.799-00591610g([Ag+])]-0.175 (II) E=0.175[0.799-0.05916|og[Ag+]] (III) E=10.799-0.05916logl 11-0.175 (IV) E= 0.175-10.799-0.05916!og! 0.05916104 e Ag Scroll down to view the rest of the question. Ag (c) Calculate the first and second equivalence points of the titration.

Explanation / Answer

1.The solubility product for AgBr is low comparing to TiBr so AgBr will precipitate first

2.As per the nernst equation , the equation for this system is E= 0.175 - [ 0.799 - 0.05916log[ A + ]]

3.first equalence point is 25 ml and second equalence point is 50 ml

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