12. Solving a chemistry problem
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12. Solving a chemistry problem
12. Calculate the AgBr solubility product at 298 K if the EMF of the chain is (-)Ag,AgBr|KBraq||KClaq|AgCl,Ag( )
C=0.1M C=0.1M
at 298 K is 0.151 V. The solubility product of AgCl is 1.56 * 10-10, and the equivalent electrical conductivities of KCl and KBr are 128.9 and 131.8, respectively (equivalent electrical conductivities are given to take into account the diffusion potential).
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