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A 100.00mL solution of 0.100 M NaCl was titrated with 0.100 M AgNO3. A SCE indicator electrode (E+=0.241V) was the anode...

A 100.00mL solution of 0.100 M NaCl was titrated with 0.100 M AgNO3. A SCE indicator electrode (E+=0.241V) was the anode and an Ag wire was the cathode for this precipitation titration. Calculate the voltage after the addition of 65 mL of AgNO3 given that Ksp(AgCl)=1.8x10^(-10). Answer=0.081 V. Please show steps, thank you! :)

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SC & indicator electrode here SLE means saturated Colomel Electrode. The titrating Reaction Ag+ + cl is : - Agl (s). At 65 meTo find cell voltage of given equation, Agt lua-] Kap & Age ] = ks 18 x 10-10 Geven 2 Ksp = [Ag+ ] = 1.8810-10 0.0212 (Ag + JI on 1. substititing E= 0.558+ value of Agt in this 0.0591 log 18.5810-9) - 0.558 + 0.0591 (-8.0705) -0.550 a 0.47 69 = 0.081

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