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Chapter 15: Q42 E (page 874)

Iron concentrations greater than 5.4脳10-6M in water used for laundrypurposes can cause staining. What [OH-] is required to reduce [Fe2+] to this level by precipitation of Fe(OH)2?

Short Answer

Expert verified

When the concentration of OH- ions in water is 1.82.10-5 iron would precipitate asFe(OH)2 and no stains would left on laundry.

Step by step solution

01

Define iron concentrations

In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical descriptions can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration.

02

Finding the value of the dissociation

Equation for Ksp

K鈭唒= [Fe2+] [OH-]2

Concentrations ofFe2+bigger than[Fe2+]=5.4.10-6 cause stains on laundry if present in water

For

Fe(OH)2 ,Ksp=1.8.10-15

03

Calculating required OH-1

The solubility of ions, or the molarity of ions needed to start precipitating

Ksp== [Fe2+] [OH-]2

\begin{aligned}{{K_{sp}}=\left[{{\rm{F}}{{\rm{e}}^{2+}}}\right]{{\left[{{\rm{O}}{{\rm{H}}^-}}\right]}^2}}\\{1.8\cdot{{10}^{-15}}=5.4\cdot{{10}^{-6}}{{\left[{{\rm{O}}{{\rm{H}}^-}}\right]}^2}}\\{\left[{{\rm{O}}{{\rm{H}}^-}}\right]=\sqrt{\frac{{1.8\cdot{{10}^{-15}}}}{{5.4\cdot{{10}^{-6}}}}}}\\{\left[{{\rm{O}}{{\rm{H}}^-}}\right]=1.82\cdot{{10}^{-5}}}\end{aligned}

When the concentration ofOH- ions in water is 1.82.10-15 iron would precipitate as Fe(OH)2 and no stains would left on laundry.

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Most popular questions from this chapter

Question: Calculate the equilibrium concentration of Ni2+ in a 1.0 M solution\(\left( {Ni{{\left( {N{H_3}} \right)}_6}} \right){\left( {N{O_3}} \right)_2}\).

Question: 28. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate \({K_{sp}}\)for each of the slightly soluble solids indicated:

(a) \(AgBr:\left( {A{g^ + }} \right) = 5.7 \times 1{0^{ - 7}}M,\left( {B{r^ - }} \right) = 5.7 \times 1{0^{ - 7}}M\)

(b) \(CaC{O_3}:\left( {C{a^{2 + }}} \right) = 5.3 \times 1{0^{ - 3}}M,\left( {C{O_3}^{2 - }} \right) = 9.0 \times 1{0^{ - 7}}M\)

(c) \(Pb{F_2}:\left( {P{b^{2 + }}} \right) = 2.1 \times 1{0^{ - 3}}M,\left( {{F^ - }} \right) = 4.2 \times 1{0^{ - 3}}M\)

(d) \(A{g_2}Cr{O_4}:\left( {A{g^ + }} \right) = 5.3 \times 1{0^{ - 5}}M,3.2 \times 1{0^{ - 3}}M\)

(e) \(In{F_3}:\left( {I{n^{3 + }}} \right) = 2.3 \times 1{0^{ - 3}}M,\left( {{F^ - }} \right) = 7.0 \times 1{0^{ - 3}}M\)

Precipitation and Dissolution

1. Complete the changes in concentrations for each of the following reactions:

\(\begin{array}{l}(a)AgI(s) \to nA{g^ + }(aq) + {I^ - }(aq)\\ x \_ \\(b)CaC{O_3}(s) \to C{a^{2 + }}(aq) + C{O_3}^{2 - }(aq)\\ \_\quad x\\(c)Mg{(OH)_2}(s) \to nM{g^{2 + }}(aq) + 2O{H^ - }(aq)\\ x \quad \_\_\\(d)M{g_3}{\left( {P{O_4}} \right)_2}(s) \to n3M{g^{2 + }}(aq) + 2P{O_4}^{3 - }(aq)\\ x\_\\(e)C{a_5}{\left( {P{O_4}} \right)_3}OH(s) \to n5C{a^{2 + }}(aq) + 3P{O_4}^{3 - }(aq) + O{H^ - }(aq)\\ \_ \_ x\end{array}\)

Calculate the molar solubility of \({\bf{Ba}}{{\bf{F}}_{\bf{2}}}\) in a buffer solution containing\({\bf{0}}.{\bf{20}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{HF}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{0}}.{\bf{20}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{NaF}}\) .

Question: Explain why addition of \(N{H_3}\;or\;HN{O_3}\) to a saturated solution of \(Cu{(OH)_2}\)in contact with solid \(Cu{(OH)_2}\) increases the solubility of the solid.
See all solutions

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