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Calculate the solubility product of AgCl...

Calculate the solubility product of AgCl from the two half reactions and standard electrode potentials at `25^@C`
`Ag^(+) + e to Ag(s) E^@ = 0.799V`
` AgCl + e to Ag(s) + Cl^(-) E^@ = 0.222V`

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`AgCl to Ag^(+) Cl^(-) , E^@ = 0.222-0.799`
` E^@ = 0.0591 log [Ag^+] Cl^-] = 0.0591 log K_(sp)`
`1.66 xx 10^(-10)`
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