Calculate E for the titration of 50.00 mL of 0.100 M Ag⁺ when 50.00 mL of 0.150 M Cl⁻ is added. Silver wire is the indicator electrode and the saturated Ag-AgCl electrode is the reference electrode. E = 0.197 V for the saturated Ag-AgCl electrode, E°_Ag⁺= 0.7993 V and K_sp = 1.8 × 10⁻⁸ for AgCl.

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Solution millimales of d Ag c millimales of Agt cl is in excess ct ... View the full answer