Calculate the molar solubility of Ag2CrO4 Ksp=1.1x10^-12, including net ionic equations. a)0.045 M K2CrO4 b)0.0085 M...

Calculate the molar solubility of Ag2CrO4 Ksp=1.1x10^-12, including net ionic equations. a)0.045 M K2CrO4 b)0.0085 M KOH

Expert Solution

a) 0.045M K2CrO4

At equilibrium 2x (0.045+x)

assuming (0.045+x) = 0.045

Molar solubility = 2 * 2.47 * 10^(-6) = 4.94 * 10^(-6) M

b) Since KOH doesn't have any ion in the common, hence the equation will become

At equilibrium 2x x

Hence the molar solubility of Ag2CrO4 is equal to 6.50 * 10^(-5)

queen_honey_blossom answered 3 years ago

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