Question

Concentration of Ag^+ ions in saturated solution of Ag2CrO4 at 20°C is 1.5×10^-4 mol/ L At 20°C the solubility product of Ag2CrO4 is

Answer 1

Answer: Solubility product of $Ag_2CrO_4\text{ is }13.5\times 10^{-12}mol^3/L^3$

Explanation: Solubility product is defined as the product of the concentrations of ions each raised to the power their stoichiometric coefficients. It is represented by $K_{sp}$

For the given chemical reaction:

$Ag_2CrO_4\rightleftharpoons 2Ag^++CrO_4^{2-}$

$K_{sp}=[Ag^+]^2[CrO_4^{2-}]$

We are given, $[Ag^+]=1.5\times 10^{-4}mol/L$

And in the product 2 moles of Silver ions are produced, so the concentration of silver ions will be = $2\times (1.5\times 10^{-4})=3.0\times 10^{-4}mol/L$

And concentration of $CrO_4=1.5\times 10^{-4}M$

Putting the above values in solubility product equation, we get

$K_{sp}=(3.0\times 10^{-4}mol/L)^2(1.5\times 10^{-4}mol/L)=13.5\times 10^{-12}mol^3/L^3$

Answer 2

Answer:

1.68 into 10 to the power minus 12 will be the right answer of this question

Explanation:

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