a solution prepared by mixing 60ml of .1m agno3 and 60ml 0f .1m tino3 was titrated...

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Chemistry

a solution prepared by mixing 60ml of .1m agno3 and 60ml 0f .1mtino3 was titrated with .2M nabr in a cell containing a silverindicator electrode and a reference electrode of constant potential.175 v. the reference electrode is attached to the positiveterminal of the potentiometer, and the silver electrode is attachedto the negative terminal. the solubility constant of tibr isksp=3.6x10^-6 and the solubility constant of agbr isksp=5x10^-13.
AgBr precipitates first
the following expression shows how the cell potential E dependson {ag+]Â Â Â Â Â E=.175-[.799-.05916log(1/[ag+])]
1. Calculate the first and second equivalence points oftitration.
2. what is the cell potential when the following volumes of .2Mnabr have been added: a)1ml, b)16.8ml, c)29ml, d)29.9ml, e)30.3ml,f)46.8ml, g)60ml, h)63.6ml

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4.4 Ratings (860 Votes)

1 For first question the first equivalence point at 30 ml of NaBr an second equivalance point at 60 ml of NaBr Mixture of solution contains 60 ml of 01 M AgNO3 an 60 ml of 01 M TiNO3 an See Answer

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