Assuming the maximum concentration of each solution is y mol/L

On mixing the solutions the volume of the concentration of each solution is reduced to half. After mixing the maximum concentration of each solution is

\( \dfrac{y}{2}\) mol/L.

Thus, [FeSO_{4}]

= [Na_{2}S] = \( \dfrac{y}{2}\)M

So, [Fe^{2+}]

= [FeSO_{4}] = \( \dfrac{y}{2}\) M

\( FeS(s) \leftrightarrow Fe^{2+}_{(aq)} + S^{2-}_{(aq)}\)

Ksp = \( [Fe^{2+}][S^{2-}] \)

\( =6.3×10^{−18}\)

\( = (\dfrac{y}{2})(\dfrac{y}{2})\dfrac{y^{2}}{4} \)

\( = 6.3\times 10^{-18}\)

Thus, y = \( 5.02\times 10^{-9}\)

Thus, if the concentration of FeSO_{4} and Na_{2}SO_{4 }are equal to or less than that of

\( 5.02\times 10^{-9}M\), then there won’t be precipitation of FeS.

Answered by Abhisek | 1 year agoThe concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^{–19} M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO_{4}, MnCl_{2}, ZnCl_{2} and CdCl_{2} . in which of these solutions precipitation will take place?

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_{sp} is 9.1 × 10^{–6}).

The ionization constant of benzoic acid is 6.46 × 10^{–5} and Ksp for silver benzoate is 2.5 × 10^{–13}. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_{sp} = 7.4 × 10^{–8} ).

The solubility product constant of Ag_{2}CrO_{4} and AgBr are 1.1 × 10^{–12} and 5.0 × 10^{–13}, respectively. Calculate the ratio of the molarities of their saturated solutions.