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 2 added 27 characters in body edited Feb 20 at 15:37 Soumik Das 4,42911 gold badge55 silver badges2323 bronze badges Yes, It should be the other way around.(i.e. $$\ce{[Fe^3+] << [H3O+]}$$) You are absolutely right in the logic that $$K_{sp} << K_w$$. Thus due to the common ion effect, the dissociation of $$\ce{Fe(OH)3}$$ should be very less. So, the solubility of Ferric hydroxide in water should be extremely less. By mathematical treatment also, this is justifiable. Consider the solubility of Ferric hydroxide in water to be $$S$$ $$mol.L^{-1}$$$$\mathrm{mol.L^{-1}}$$. and consider the concentration of $$\ce{H3O+}$$ is $$x \ mol.L^{-1}$$$$x \ \mathrm{mol.L^{-1}}$$. Thus the total concentration of $$\ce{OH-}$$ will be $$(3S +x) \ mol.L^{-1}$$$$(3S +x) \ \mathrm{mol.L^{-1}}$$ . So, for finding $$S$$ and $$x$$ we need to solve two equations which are as follows, $$x(3S +x)^3 = 2 \times10^{-39}$$ and, $$x(3S +x) = 10^{-14}$$ Exact solution of these two equations is difficult to find, but if we substitute $$(3S+x)$$ by $$\frac{10^{-14}}{x}$$ in first equation we will find that $$S = 2\times10^3 \times x^3$$. Now $$x$$ is between $$0$$ and $$1$$ ( roughly of the order of $$10^{-7}$$) and thus if it is cubed, it will become even smaller. So, it is seen that $$S <. So, thus it is justified that $$\ce{[H3O+] >> [Fe^3+]}$$. In fact it will be found that $$\ce{[H3O+] \approx 10 ^{-7} M}$$ and $$\ce{[Fe^3+] \approx 2 \times 10^{-18} M}$$ . So, your justification is absolutely valid. Yes, It should be the other way around.(i.e. $$\ce{[Fe^3+] << [H3O+]}$$) You are absolutely right in the logic that $$K_{sp} << K_w$$. Thus due to the common ion effect, the dissociation of $$\ce{Fe(OH)3}$$ should be very less. So, the solubility of Ferric hydroxide in water should be extremely less. By mathematical treatment also, this is justifiable. Consider the solubility of Ferric hydroxide in water to be $$S$$ $$mol.L^{-1}$$. and consider the concentration of $$\ce{H3O+}$$ is $$x \ mol.L^{-1}$$. Thus the total concentration of $$\ce{OH-}$$ will be $$(3S +x) \ mol.L^{-1}$$ . So, for finding $$S$$ and $$x$$ we need to solve two equations which are as follows, $$x(3S +x)^3 = 2 \times10^{-39}$$ and, $$x(3S +x) = 10^{-14}$$ Exact solution of these two equations is difficult to find, but if we substitute $$(3S+x)$$ by $$\frac{10^{-14}}{x}$$ in first equation we will find that $$S = 2\times10^3 \times x^3$$. Now $$x$$ is between $$0$$ and $$1$$ ( roughly of the order of $$10^{-7}$$) and thus if it is cubed, it will become even smaller. So, it is seen that $$S <. So, thus it is justified that $$\ce{[H3O+] >> [Fe^3+]}$$. In fact it will be found that $$\ce{[H3O+] \approx 10 ^{-7} M}$$ and $$\ce{[Fe^3+] \approx 2 \times 10^{-18} M}$$ . So, your justification is absolutely valid. Yes, It should be the other way around.(i.e. $$\ce{[Fe^3+] << [H3O+]}$$) You are absolutely right in the logic that $$K_{sp} << K_w$$. Thus due to the common ion effect, the dissociation of $$\ce{Fe(OH)3}$$ should be very less. So, the solubility of Ferric hydroxide in water should be extremely less. By mathematical treatment also, this is justifiable. Consider the solubility of Ferric hydroxide in water to be $$S$$ $$\mathrm{mol.L^{-1}}$$. and consider the concentration of $$\ce{H3O+}$$ is $$x \ \mathrm{mol.L^{-1}}$$. Thus the total concentration of $$\ce{OH-}$$ will be $$(3S +x) \ \mathrm{mol.L^{-1}}$$ . So, for finding $$S$$ and $$x$$ we need to solve two equations which are as follows, $$x(3S +x)^3 = 2 \times10^{-39}$$ and, $$x(3S +x) = 10^{-14}$$ Exact solution of these two equations is difficult to find, but if we substitute $$(3S+x)$$ by $$\frac{10^{-14}}{x}$$ in first equation we will find that $$S = 2\times10^3 \times x^3$$. Now $$x$$ is between $$0$$ and $$1$$ ( roughly of the order of $$10^{-7}$$) and thus if it is cubed, it will become even smaller. So, it is seen that $$S <. So, thus it is justified that $$\ce{[H3O+] >> [Fe^3+]}$$. In fact it will be found that $$\ce{[H3O+] \approx 10 ^{-7} M}$$ and $$\ce{[Fe^3+] \approx 2 \times 10^{-18} M}$$ . So, your justification is absolutely valid. 1 answered Feb 20 at 11:15 Soumik Das 4,42911 gold badge55 silver badges2323 bronze badges Yes, It should be the other way around.(i.e. $$\ce{[Fe^3+] << [H3O+]}$$) You are absolutely right in the logic that $$K_{sp} << K_w$$. Thus due to the common ion effect, the dissociation of $$\ce{Fe(OH)3}$$ should be very less. So, the solubility of Ferric hydroxide in water should be extremely less. By mathematical treatment also, this is justifiable. Consider the solubility of Ferric hydroxide in water to be $$S$$ $$mol.L^{-1}$$. and consider the concentration of $$\ce{H3O+}$$ is $$x \ mol.L^{-1}$$. Thus the total concentration of $$\ce{OH-}$$ will be $$(3S +x) \ mol.L^{-1}$$ . So, for finding $$S$$ and $$x$$ we need to solve two equations which are as follows, $$x(3S +x)^3 = 2 \times10^{-39}$$ and, $$x(3S +x) = 10^{-14}$$ Exact solution of these two equations is difficult to find, but if we substitute $$(3S+x)$$ by $$\frac{10^{-14}}{x}$$ in first equation we will find that $$S = 2\times10^3 \times x^3$$. Now $$x$$ is between $$0$$ and $$1$$ ( roughly of the order of $$10^{-7}$$) and thus if it is cubed, it will become even smaller. So, it is seen that $$S <. So, thus it is justified that $$\ce{[H3O+] >> [Fe^3+]}$$. In fact it will be found that $$\ce{[H3O+] \approx 10 ^{-7} M}$$ and $$\ce{[Fe^3+] \approx 2 \times 10^{-18} M}$$ . So, your justification is absolutely valid.