Question

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! :)

Answer 1

SC & indicator electrode here SLE means saturated Colomel Electrode. The titrating Reaction Ag+ + cl is : - Agl (s). At 65 me 65% of ct remains addition, has precipitated in solution. and 35% of ce of solution .. Dilution factor - Initial volume Total volume Now total solume=165ml Initial volume of t = 100mL .. Dilution factor = 100 165 Tal - Fraction x Remaining original o concentration x Délution factor fraction Original Remaining = 0.350 concentration of cl = 0.1oom. la) : (0'350) (0.100 M) /100 1 165 Ta] = 0.0212 m Scanned with CamScanner

To 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 + J 3 8.5 x 10-9 m 1 the reference potential is fined at 0.241 v because reference cell is saturated. The Nernst Equation for entrie cell is E = E - E 10.799 -0.0591 log t 4 -20.2419 Potential indicator of Ag /Agt electrode Potential of SLE reference electrode The TE = cell voltage is therefore 0.558 +0.059 1 log [Ag+] /

I 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 V. I The cell voltage after addition of 65 ml 210.0818.. ra Scanned CamScan