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17.4: Solubility of Salts

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    60764
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    Introduction

    In general chemistry 1 students memorized a series of solubility rules (section 3.4.3) to predict if an ionic compound was soluble or not. Using these rules we would predict that insoluble salts formed precipitates and soluble salts dissolved. In this section we will apply chemical equilibria to the concept of solubility and introduce a type of equilibrium constant, the solubility constant, to allow us to calculate how soluble a salt really is.

    Solubility Equilibrium defines the dynamic equilibria between a precipitate and its dissolved ions when the rate of dissolution equals the rate of crystallization and the resulting solution is a saturated solution, that contains the maximum concentration of dissolved ions that coexist with the undissolved solute (precipitate).

    What do we mean by solubility of a salt? If I say one mole of sodium sulfate dissolves in water I am really meaning for every mole of sodium sulfate that dissolves, two moles of sodium are formed and one mole of sulfate. We will use the concept of solubility to describe the moles the salt that dissolves to form a saturated solution, and the actual ion concentration depends on the formula.

    Solubility Product

    This is the equilibrium constant for the process of dissolution. Consider the generic salt

    \[M_xA_y(s) \leftrightharpoons xM^{+m}(aq) + yA^{-n}(aq)\]
    Noting from charge neutrality that x(+m)+y(-n)=0, then the equilibrium constant expression is:
    \[K_{sp}=[M^{+m}]^x[A^{-n}]^y\]

    The solid reactant is not part of the equilibrium, and it is called the solubility product because it is the product of the concentration of all the ions a salt breaks into. It is important to realize this is a heterogeneous equilibrium that defines a saturated solution and the solid is part of the process, although it does not influence the equilibrium concentration. That is, if all the solid is dissolved, you may have an unsaturated solution, and the solubility product defines the concentration of a saturated solution. At the end of this chapter is a solubility product table for ionic compounds at 25oC.

    Lets look at the solubility product for Calcium phosphate

    \[Ca_3(PO_4)_2(s) \leftrightharpoons 3Ca^{+2} + 2PO_4^{-3}\]
    \[K_{sp}(Ca_3(PO_4)_2)=[Ca^{+2}]^3[PO_4^{-3}]^2\]

    Solubility Calculations

    As in most equilibrium calculations, there are two types of problems. Knowing K, what are the equilibrium concentrations, and knowing the equilibrium concentration of one thing, what is K

    Solubility of a Salt

    The solubility of a salt is "x" (from the ICE table), but we will not be using ICE tables in solubility product calculations, and are just using it now to explain what is going on. From the following ICE table you can see that for every x moles of the calcium phosphate that dissolves per liter (its solubility), the solution gains 3 times that many moles of calcium and twice that many of phosphate, and there is no calcium phosphate floating around, just calcium and phosphate.

    ICE Table \(Ca_{3}(PO_4)_{2}(s)\) \( \leftrightharpoons \) \(Ca^{+2}\) \(PO_4^{-3}\)
    Initial [solid]   0 0
    Change still solid   +3x +2x
    Equilibrium still solid   3x 2x

    That is, x the extent of reaction is a ICE diagram is the solubility of a salt. So you do not use a rice diagram, you say that for every x moles of solid that dissolves, the ions are that times the number of the ions in the formula So for Calcium phosphate

    \[[Ca^{+2}]= 3x \; and \; [PO_4^{-3}] =2x]\]

    \[K_{sp}(Ca_{3}(PO_4)_{2})=[3x]^3[2x]^2 = 108x^5\]

    Solving for the solubility (x) and noting that Ksp(calcium phosphate) = 2.07 × 10−33

    \[x=\sqrt[5]{\frac{K_{sp}}{108}}=\left ( \frac{2.07x10^{-33}}{108} \right )^{\frac{1}{5}}=1.14x10^{-7} = 114nM\]

    So there are three questions that can be answered here

    1. The solubility of calcium phosphate is 114nM in water (x)
    2. The solubility of calcium is 341 nM in water (3x)
    3. The solubility of phosphate is 228 nM in water (2x)

    Exercise \(\PageIndex{1}\)

    Which is more soluble, silver thiocyanide (Ksp = 1.1 x 10-12) or silver sulfite (Ksp = 1.50 x 10-14)

    Answer

    Silver sulfite, you can only compare Ksp values if the number of ions are the same, see video below.

    For silver thiocyanide,

    \[ K_{sp}=[Ag^+][SCN^-]=x^2 \\ x=\sqrt{k_{sp}} =\sqrt{1.1x10^{-12}} = 1.0x10^{-6} \nonumber\]

    For silver sulfite,

    \[ K_{sp}=[Ag^+][SO_3^{-2}]^2=x(2x)^2 = 4x^3 \\ x= \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{1.50 x 10^{-14}}{4}} \\ = 1.56x10^{-5}\]

    Video \(\PageIndex{1}\): Solution to exercise \(\PageIndex{1}\).

    Determining Ksp

    If a saturated solution of La(IO3)3 has an iodate concentration of 0.006M, what is Ksp?

    \[La(IO_3)_3 \leftrightharpoons La^{+3}+3IO_3^-\]

    The lanthanum ion concentration = x and the iodate ion concentration = 3x=0.0060. So x=0.0020

    \[K_{sp}=[La^{+3}][IO_3^{-3}]^3=[0.0020]0.0060]^3=4.2x10^{-10}\]

    Solubility and Common ion Effect

    In section 17.1.3 solubility was introduced as an example of the common ion effect, and this problem was explained using ICE table and Le Chatelier's Principle.

    What is the solubility of Calcium phosphate in a 0.100M sodium phosphate solution?

    This is the same problem as above except that there is a common ion as the soluble sodium phosphate introduces phosphate that shifts the calcium phosphate to the left, reducing its solubility.

    \[Ca_3PO_4(s) \leftrightharpoons 3Ca^{+2} + 2PO_4^{-3}\]
    \[K_{sp}(Ca_3PO_4)=[Ca^{+2}]^3[PO_4^{-3}]^2\]

    noting that \([PO_4^{-3}] = 0.100M+3x\) and ignoring the x (see section 17.1.3)

    \[K_{sp}(Ca_3PO_4)=[3x]^3[.1]^2 = 0.27x^3\]

    \[x=\left ( \frac{K_{sp}}{0.27} \right )^{\frac{1}{3}}=\left ( \frac{2.07x10^{-33}}{0.27} \right )^{\frac{1}{3}}=1.97x10^{-11}\]

    Exercise \(\PageIndex{2}\)

    What is the lead concentration for a saturated solution of lead(II)bromide, and then mathematically demonstrate Le Chatlier's principle by determining the lead concentration when in the presence of a common ion by making the solution 1M in sodium bromide.

    Answer

    As the following video shows, the lead is 0.0118 in the absence of sodium bromide and is reduced to 6.6x10-6 in the presence of 1 M NaBr. This demonstrates Le Chatlier's principle as adding the bromide with the soluble sodium bromide salt pushed the equilbria of the insoluble lead(II)bromide salt to the left (consuming the bromide added), and thus removing the lead.

    Video \(\PageIndex{2}\): Solution to exercise \(\PageIndex{2}\).

    Solubility Product Table

    This values relate to 25oC.

    table 17.4: Solubility Product Constants
    Compound Name Compound Formula Ksp
    Aluminum phosphate AlPO4 9.84 × 10−21
    Barium bromate Ba(BrO3)2 2.43 × 10−4
    Barium carbonate BaCO3 2.58 × 10−9
    Barium chromate BaCrO4 1.17 × 10−10
    Barium fluoride BaF2 1.84 × 10−7
    Barium iodate Ba(IO3)2 4.01 × 10−9
    Barium nitrate Ba(NO3)2 4.64 × 10−3
    Barium sulfate BaSO4 1.08 × 10−10
    Barium sulfite BaSO3 5.0 × 10−10
    Beryllium hydroxide Be(OH)2 6.92 × 10−22
    Bismuth arsenate BiAsO4 4.43 × 10−10
    Bismuth iodide BiI3 7.71 × 10−19
    Cadmium carbonate CdCO3 1.0 × 10−12
    Cadmium fluoride CdF2 6.44 × 10−3
    Cadmium hydroxide Cd(OH)2 7.2 × 10−15
    Cadmium iodate Cd(IO3)2 2.5 × 10−8
    Cadmium phosphate Cd3(PO4)2 2.53 × 10−33
    Cadmium sulfide CdS 8.0 × 10−27
    Calcium carbonate CaCO3 3.36 × 10−9
    Calcium fluoride CaF2 3.45 × 10−11
    Calcium hydroxide Ca(OH)2 5.02 × 10−6
    Calcium iodate Ca(IO3)2 6.47 × 10−6
    Calcium phosphate Ca3(PO4)2 2.07 × 10−33
    Calcium sulfate CaSO4 4.93 × 10−5
    Cesium perchlorate CsClO4 3.95 × 10−3
    Cesium periodate CsIO4 5.16 × 10−6
    Cobalt(II) arsenate Co3(AsO4)2 6.80 × 10−29
    Cobalt(II) hydroxide Co(OH)2 5.92 × 10−15
    Cobalt(II) phosphate Co3(PO4)2 2.05 × 10−35
    Copper(I) bromide CuBr 6.27 × 10−9
    Copper(I) chloride CuCl 1.72 × 10−7
    Copper(I) cyanide CuCN 3.47 × 10−20
    Copper(I) iodide CuI 1.27 × 10−12
    Copper(I) thiocyanate CuSCN 1.77 × 10−13
    Copper(II) arsenate Cu3(AsO4)2 7.95 × 10−36
    Copper(II) oxalate CuC2O4 4.43 × 10−10
    Copper(II) phosphate Cu3(PO4)2 1.40 × 10−37
    Copper(II) sulfide CuS 6.3 × 10−36
    Europium(III) hydroxide Eu(OH)3 9.38 × 10−27
    Gallium(III) hydroxide Ga(OH)3 7.28 × 10−36
    Iron(II) carbonate FeCO3 3.13 × 10−11
    Iron(II) fluoride FeF2 2.36 × 10−6
    Iron(II) hydroxide Fe(OH)2 4.87 × 10−17
    Iron(III) hydroxide Fe(OH)3 2.79 × 10−39
    Iron(III) sulfide FeS 6.3 × 10−18
    Lanthanum iodate La(IO3)3 7.50 × 10−12
    Lead(II) bromide PbBr2 6.60 × 10−6
    Lead(II) carbonate PbCO3 7.40 × 10−14
    Lead(II) chloride PbCl2 1.70 × 10−5
    Lead(II) fluoride PbF2 3.3 × 10−8
    Lead(II) hydroxide Pb(OH)2 1.43 × 10−20
    Lead(II) iodate Pb(IO3)2 3.69 × 10−13
    Lead(II) iodide PbI2 9.8 × 10−9
    Lead(II)selenite PbSeO4 1.37 × 10−7
    Lead(II) sulfate PbSO4 2.53 × 10−8
    Lead(II) sulfide PbS 8.0 × 10−28
    Lithium carbonate Li2CO3 8.15 × 10−4
    Lithium fluoride LiF 1.84 × 10−3
    Lithium phosphate Li3PO4 2.37 × 10−11
    Magnesium carbonate MgCO3 6.82 × 10−6
    Magnesium fluoride MgF2 5.16 × 10−11
    Magnesium hydroxide Mg(OH)2 5.61 × 10−12
    Magnesium phosphate Mg3(PO4)2 1.04 × 10−24
    Manganese(II) carbonate MnCO3 2.24 × 10−11
    Manganese(II) iodate Mn(IO3)2 4.37 × 10−7
    Mercury(I) bromide Hg2Br2 6.40 × 10−23
    Mercury(I) carbonate Hg2CO3 3.6 × 10−17
    Mercury(I) chloride Hg2Cl2 1.43 × 10−18
    Mercury(I) fluoride Hg2F2 3.10 × 10−6
    Mercury(I) iodide Hg2I2 5.2 × 10−29
    Mercury(I) oxalate Hg2C2O4 1.75 × 10−13
    Mercury(I) sulfate Hg2SO4 6.5 × 10−7
    Mercury(I) thiocyanate Hg2(SCN)2 3.2 × 10−20
    Mercury(II) bromide HgBr2 6.2 × 10−20
    Mercury (II) iodide HgI2 2.9 × 10−29
    Mercury(II) sulfide (red) HgS 4 × 10−53
    Mercury(II) sulfide (black) HgS 1.6 × 10−52
    Neodymium carbonate Nd2(CO3)3 1.08 × 10−33
    Nickel(II) carbonate NiCO3 1.42 × 10−7
    Nickel(II) hydroxide Ni(OH)2 5.48 × 10−16
    Nickel(II) iodate Ni(IO3)2 4.71 × 10−5
    Nickel(II) phosphate Ni3(PO4)2 4.74 × 10−32
    Palladium(II) thiocyanate Pd(SCN)2 4.39 × 10−23
    Potassium hexachloroplatinate K2PtCl6 7.48 × 10−6
    Potassium perchlorate KClO4 1.05 × 10−2
    Potassium periodate KIO4 3.71 × 10−4
    Praseodymium hydroxide Pr(OH)3 3.39 × 10−24
    Rubidium perchlorate RbClO4 3.00 × 10−3
    Scandium fluoride ScF3 5.81 × 10−24
    Scandium hydroxide Sc(OH)3 2.22 × 10−31
    Silver(I) acetate AgCH3CO2 1.94 × 10−3
    Silver(I) arsenate Ag3AsO4 1.03 × 10−22
    Silver(I) bromate AgBrO3 5.38 × 10−5
    Silver(I) bromide AgBr 5.35 × 10−13
    Silver(I) carbonate Ag2CO3 8.46 × 10−12
    Silver(I) chloride AgCl 1.77 × 10−10
    Silver(I) chromate Ag2CrO4 1.12 × 10−12
    Silver(I) cyanide AgCN 5.97 × 10−17
    Silver(I) iodate AgIO3 3.17 × 10−8
    Silver(I) iodide AgI 8.52 × 10−17
    Silver(I) oxalate Ag2C2O4 5.40 × 10−12
    Silver(I) phosphate Ag3PO4 8.89 × 10−17
    Silver(I) sulfate Ag2SO4 1.20 × 10−5
    Silver(I) sulfide Ag2S 6.3 × 10−50
    Silver(I) sulfite Ag2SO3 1.50 × 10−14
    Silver(I) thiocyanate AgSCN 1.03 × 10−12
    Strontium arsenate Sr3(AsO4)2 4.29 × 10−19
    Strontium carbonate SrCO3 5.60 × 10−10
    Strontium fluoride SrF2 4.33 × 10−9
    Strontium iodate Sr(IO3)2 1.14 × 10−7
    Strontium sulfate SrSO4 3.44 × 10−7
    Thallium(I) bromate TlBrO3 1.10 × 10−4
    Thallium(I) bromide TlBr 3.71 × 10−6
    Thallium(I) chloride TlCl 1.86 × 10−4
    Thallium(I) chromate Tl2CrO4 8.67 × 10−13
    Thallium(I) iodate TlIO3 3.12 × 10−6
    Thallium(I) iodide TlI 5.54 × 10−8
    Thallium(I) thiocyanate TlSCN 1.57 × 10−4
    Thallium(III) hydroxide Tl(OH)3 1.68 × 10−44
    Tin(II) hydroxide Sn(OH)2 5.45 × 10−27
    Tin(II) sulfide SnS 1.0 × 10−25
    Yttrium carbonate Y2(CO3)3 1.03 × 10−31
    Yttrium fluoride YF3 8.62 × 10−21
    Yttrium hydroxide Y(OH)3 1.00 × 10−22
    Yttrium iodate Y(IO3)3 1.12 × 10−10
    Zinc arsenate Zn3(AsO4)2 2.8 × 10−28
    Zinc carbonate ZnCO3 1.46 × 10−10
    Zinc fluoride ZnF2 3.04 × 10−2
    Zinc hydroxide Zn(OH)2 3 × 10−17
    Zinc selenide ZnSe 3.6 × 10−26
    Zinc sulfide (wurtzite) ZnS 1.6 × 10−24
    Zinc sulfide (sphalerite) ZnS 2.5 × 10−22

    Test Yourself

    Homework: Section 17.4

    Contributors and Attributions

    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:


    This page titled 17.4: Solubility of Salts is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Robert Belford.

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