Calculate the concentration of aqueous ammonia necessary to initiate the precipitation of iron(II) hydroxide from a 0.0030 M solution of FeCl2. Strategy For iron(II) hydroxide to precipitate from solution, the product [Fe^2+][OH^−]^2 must be greater than its Ksp. First, we calculate [OH^−] from the

Question

Calculate the concentration of aqueous ammonia necessary to initiate the precipitation of iron([latex]II[/latex]) hydroxide from a 0.0030 M solution of [latex]FeCl_{2}[/latex].

Strategy For iron([latex]II[/latex]) hydroxide to precipitate from solution, the product [latex][Fe^{2+}][OH^{-}]^{2}[/latex] must be greater than its [latex]K_{sp}[/latex] . First, we calculate [latex][OH^{-}][/latex] from the known [latex][Fe^{2+}][/latex] and the [latex]K_{sp}[/latex] value listed in Table 16.2. This is the concentration of [latex]OH^{-}[/latex] in a saturated solution of [latex]Fe(OH)_{2}[/latex]. Next, we calculate the concentration of [latex]NH_{3}[/latex] that will supply this concentration of [latex]OH^{-}[/latex] ions. Finally, any [latex]NH_{3}[/latex] concentration greater than the calculated value will initiate the precipitation of [latex]Fe(OH)_{2}[/latex] because the solution will become supersaturated.

Accepted Answer

Ammonia reacts with water to produce [latex]OH^{-}[/latex] ions, which then react with [latex]Fe^{2+}[/latex] to form [latex]Fe(OH)_{2}[/latex]. The equilibria of interest are

[latex]NH_{3}(aq)+H_{2}O(l) \xrightleftharpoons[]{} NH^{+}_{4}(aq)+OH^{-}(aq) [/latex]

[latex]Fe^{2+}(aq)+2OH^{-}(aq) \xrightleftharpoons[]{} Fe(OH)_{2}(s) [/latex]

First we find the [latex]OH^{-}[/latex] concentration above which [latex]Fe(OH)_{2}[/latex] begins to precipitate. We write

[latex]K_{sp}=[Fe^{2+} ][OH^{-} ] ^{2}=1.6 imes 10^{-14} [/latex]

Because [latex]Fe(OH)_{2}[/latex] is a strong electrolyte, [latex][Fe^{2+}][/latex] = 0.0030 M and

[latex][OH^{-} ]^{2}=\frac{1.6 imes 10^{-14} }{0.0030}=5.3 imes 10^{-12} [/latex]

[latex][OH^{-}]=2.3 imes 10^{-6} M[/latex]

Next, we calculate the concentration of [latex]NH_{3}[/latex] that will supply [latex]2.3 × 10^{-6}[/latex] M [latex]OH^{-}[/latex] ions. Let x be the initial concentration of [latex]NH_{3}[/latex] in mol/L. We summarize the changes in concentrations resulting from the ionization of [latex]NH_{3}[/latex] as follows.

[latex]\begin{matrix} & NH_{3}(aq) & & + & H_{2}O(l) & \xrightleftharpoons[]{} &NH^{+}_{4} (aq) &+&OH^{-}(aq) \ Initial (M): &x & && &&0.00 &&0.00&&& \ Change (M): & -2.3 imes 10^{-6} & &&&&+2.3 imes 10^{-6} &&+2.3 imes 10^{-6}& \ \hline Equilibrium (M):&(x-2.3 imes 10^{-6} )&&&&&2.3 imes 10^{-6} &&2.3 imes 10^{-6}& \end{matrix}[/latex]

Substituting the equilibrium concentrations in the expression for the ionization constant (see Table 15.4),

[latex]K_{b}=\frac{[NH^{+}_{4} [OH^{-} ] }{[NH_{3} ]} [/latex]

[latex]1.8 imes 10^{-5}=\frac{(2.3 imes 10^{-6})(2.3 imes 10^{-6}) }{(x-2.3 imes 10^{-6})} [/latex]

Solving for x, we obtain

[latex]x = 2.6 × 10^{-6} M[/latex]

Therefore, the concentration of [latex]NH_{3}[/latex] must be slightly greater than [latex]2.6 × 10^{-6} M[/latex] to initiate the precipitation of [latex]Fe(OH)_{2}[/latex].

Compound	$K_{sp}$	Compound	$K_{sp}$
Aluminum hydroxide $[Al(OH)_{3} ]$	$1.8\times 10^{-33}$	Lead $(II)$ chromate $(PbCrO_{4})$	$2.0\times 10^{-14}$
Barium carbonate $(BaCO_{3})$	$8.1\times 10^{-9}$	Lead $(II)$ fluoride $(PbF_{2})$	$4.1\times 10^{-8}$
Barium fluoride $(BaF_{2})$	$1.7\times 10^{-6}$	Lead $(II)$ iodide $(PbI_{2})$	$1.4\times 10^{-8}$
Barium sulfate $(BaSO_{4})$	$1.1\times 10^{-10}$	Lead $(II)$ sulfide (PbS)	$3.4\times 10^{-28}$
Bismuth sulfide $(BiSO_{3})$	$1.6\times 10^{-72}$	Magnesium carbonate $(MgCO_{3})$	$4.0\times 10^{-5}$
Calcium sulfide (CdS)	$8.0\times 10^{-28}$	Magnesium hydroxide $[Mg(OH)_{2}]$	$1.2\times 10^{-11}$
Calcium carbonate $(CaCO_{3})$	$8.7\times 10^{-9}$	Manganese $(II)$ sulfide (MnS)	$3.0\times 10^{-14}$
Calcium fluoride $(CaF_{2})$	$4.0\times 10^{-11}$	Mercury $(I)$ sulfide $(Hg_2 Cl_2)$	$3.5\times 10^{-18}$
Calcium hydroxide $[Ca(OH)_{2}]$	$8.0\times 10^{-6}$	Mercury $(I)$ sulfide (HgS)	$4.0\times 10^{-54}$
Calcium phosphate $[Ca_{3}(PO_{4} )_{2}]$	$1.2\times 10^{-26}$	Nickel $(II)$ sulfide (NiS)	$1.4\times 10^{-24}$
Chromium $(II)$ hydroxide $Cr(oh)_3$	$3.0\times 10^{-29}$	Silver bromide (AgBr)	$7.7\times 10^{-13}$
Cobalt $(II)$ sulfide (CoS)	$4.0\times 10^{-21}$	Silver chloride $(Ag_{2}CO_{3})$	$8.1\times 10^{-12}$
Copper $(I)$ bromide (CuBr)	$4.2\times 10^{-8}$	Silver chloride (AgCl)	$1.6\times 10^{-10}$
Copper $(I)$ iodide (CuI)	$5.1\times 10^{-12}$	Silver iodide (AgI)	$8.3\times 10^{-17}$
Copper $(II)$ hydroxide $[Cu(OH)_{2}]$	$2.2\times 10^{-20}$	Silver sulfate $(Ag_{2}SO_{4})$	$1.4\times 10^{-5}$
Copper $(II)$ sulfide (CuS)	$6.0\times 10^{-37}$	Silver sulfide $(Ag_{2}S)$	$6.0\times 10^{-51}$
Iron $(II)$ hydroxide $[Fe(OH)_{2}]$	$1.6\times 10^{-14}$	Strontium carbonate $(SrCO_{3})$	$1.6\times 10^{-9}$
Iron $(III)$ hydroxide $[Fe(OH)_{3}]$	$1.1\times 10^{-36}$	Strontium sulfate $(SrSO_{4})$	$3.8\times 10^{-7}$
Iron $(II)$ sulfide (FeS)	$6.0\times 10^{-19}$	Tin $(II)$ sulfide (SnS)	$1.0\times 10^{-26}$
Lead $(II)$ carbonate $(PbCO_{3})$	$3.3\times 10^{-14}$	Zinc hydroxide $[Zn(OH)_{2}]$	$1.8\times 10^{-14}$
Lead $(II)$ chloride $(PbCl_{2})$	$2.4\times 10^{-4}$	Zinc sulfide (ZnS)	$3.0\times 10^{-23}$

Question 16.14: Calculate the concentration of aqueous ammonia necessary to ...

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