The K(SP) of Ag(2)C(2)O(4) at 25^(@)C is...

The `K_(SP)` of `Ag_(2)C_(2)O_(4)` at `25^(@)C` is `1.29xx10^(-11)mol^(3)L^(-3)`. A solution of `K_(2)C_(2)O_(4)` containing `0.152` mole in 500 mL water is shaken at `25^(@)C` with excess of `Ag_(2)CO_(3)` till the equilbirium is reached.
`Ag_(2)CO_(3)+K_(2)C_(2)O_(4)hArrAg_(2)C_(2)O_(4)+K_(2)CO_(3)`
Ar equilibrium the solution contains `0.0358` mole of `K_(2)CO_(3)`. Assuming degree of dissociation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)` to be same, calculate `K_(SP)` of `Ag_(2)CO_(3)`.

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The solubility of Ag_(2)C_(2)O_(4) at 25^(@)C is 1.20 xx 10^(-11) . A solution of K_(2)C_(2)O_(4) containing 0.15mol in 500mL water is mixed with excess of Ag_(2)CO_(3) till the following equilibrium is established: Ag_(2)CO_(3) + K_(2)C_(2)O_(4) hArr Ag_(2)C_(2)O_(4) + K_(2)CO_(3) At equilibrium, the solution constains 0.03 mol of K_(2)CO_(3) . Assuming that the degree of dissociation of K_(2)C_(2)O_(4) and K_(2)CO_(3) to be equal, calculate the solubility product of Ag_(2)CO_(3) . [Take 100% ionisation of K_(2)C_(2)O_(4) and K_(2)CO_(3)]

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The solubility product ( K_(sp) ) of Ca(OH)_(2) at 25^@ is 4.42xx10^(-5) . A 500 mL of saturated solution of Ca(OH)_(2) is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)_(2) in milligrams is precipitated? At equilibrium, the solution contains 0.0358 mole of K_(2)CO_(3) . Assuming the degree of dissociation of K_(2)C_(2)O_(4) and K_(2)CO_(3) to be equal, calculate the solubility product of Ag_(2)CO_(3) .

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