A silver wire dipped in 0.1M HCl solution saturated with AgCl develops oxidation potential of -0.25V. If Eo Ag+/Ag=-0.799V then Ksp of AgCl
Answer (1)
27th May, 2019
Hello Aastha,
Eel = E0el – 0.0592/ n log[Ag+]
Substitute the values in the above expression
-0.25V = -0.799V – 0.0592/1 log[Ag+]
log[Ag+] = -9.2736
[Ag+]= 5.3* 10-10
Hence the solubility product will be
Ksp = [Ag+][Cl-] = 5.3* 10-10 * 0.1
= 5.3 * 10-11
Hope this helps you
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