What is the maximum concentration of equimola | vedclass

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(A) Let the maximum concentration of each solution be $x \ mol/L$. After mixing equal volumes,the concentration of each ion in the mixture becomes hal

Vedclass · Accepted Answer

$5.02 	imes 10^{-9} \ M$

Vedclass · Answer

(A) Let the maximum concentration of each solution be $x \ mol/L$. After mixing equal volumes,the concentration of each ion in the mixture becomes half of the initial concentration,i.e.,$x/2 \ M$.$	herefore [Fe^{2+}] = \frac{x}{2} \ M$ and $[S^{2-}] = \frac{x}{2} \ M$.For no precipitation,the ionic product must be less than or equal to the solubility product $(K_{sp})$.$Q_{sp} = [Fe^{2+}][S^{2-}] \leq K_{sp}$$(\frac{x}{2})(\frac{x}{2}) \leq 6.3 	imes 10^{-18}$$\frac{x^2}{4} \leq 6.3 	imes 10^{-18}$$x^2 \leq 25.2 	imes 10^{-18}$$x \leq \sqrt{25.2} 	imes 10^{-9}$$x \leq 5.02 	imes 10^{-9} \ M$.Thus,the maximum concentration is $5.02 	imes 10^{-9} \ M$.

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes,there is no precipitation of iron sulphide? (For iron sulphide,$K_{sp} = 6.3 \times 10^{-18}$).

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