Answered: A 25.00 mL solution of 0.08830 M Nal is…

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter15: Equilibria Of Other Reaction Classes

Section: Chapter Questions

Problem 99E: A 0.125-M solution of 0.075 Mn(NO3)2 is saturated with H2S([H2S]=0.10M). At what pH does Mns begin...

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A 25.0025.00 mL solution of 0.088300.08830 M NaINaI is titrated with 0.051500.05150 M AgNO3AgNO3. Calculate pAg+pAg+ following the addition of the given volumes of AgNO3AgNO3. The KspKsp of AgIAgI is 8.3×10−178.3×10−17.

37.40 mLpAg+=37.40 mLpAg+=

VepAg+=VepAg+=

47.30 mLpAg+=47.30 mLpAg+=

A 25.00 mL solution of 0.08830 M Nal is titrated with 0.05150 M AgNO.. Calculate pAg* following the addition of the
given volumes of AgNO3. The Kp of AgI is 8.3 x 10-17.
37.40 mL pAg*
V. pAg+
47.30 mL pAg*

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