{"id":7688,"date":"2021-06-08T21:57:18","date_gmt":"2021-06-08T21:57:18","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/some-special-types-of-equilibria\/"},"modified":"2021-10-07T20:42:17","modified_gmt":"2021-10-07T20:42:17","slug":"some-special-types-of-equilibria","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/some-special-types-of-equilibria\/","title":{"raw":"Some Special Types of Equilibria","rendered":"Some Special Types of Equilibria"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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    \r\n \t
  1. Identify several special chemical equilibria and construct their K<\/em>a<\/sub> expressions.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nIn one sense, all chemical equilibria are treated the same. However, there are several classes of reactions that are noteworthy because of either the identities of the reactants and products or the form of the K<\/em>eq<\/sub> expression.\r\n

    Weak Acids and Bases<\/h1>\r\nIn Chapter 12 \"Acids and Bases\"<\/a>, we noted how some acids and bases are strong and some are weak. If an acid or base is strong, it is ionized 100% in H2<\/sub>O. HCl(aq) is an example of a strong acid:\r\n

    HCl(aq)\u00a0\u2192 H+<\/sup>(aq)\u00a0\u00a0+\u00a0\u00a0Cl\u2212<\/sup>(aq) (100%)<\/p>\r\nHowever, if an acid or base is weak, it may dissolve in H2<\/sub>O but does not ionize completely. This means that there is an equilibrium between the unionized acid or base and the ionized form. HC2<\/sub>H3<\/sub>O2<\/sub> is an example of a weak acid:\r\n

    HC2<\/sub>H3<\/sub>O2<\/sub>\u00a0(aq) \u21c4 H+<\/sup>(aq)\u00a0+\u00a0C2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup>(aq)<\/p>\r\nHC2<\/sub>H3<\/sub>O2<\/sub> is soluble in H2<\/sub>O (in fact, it is the acid in vinegar), so the reactant concentration will appear in the equilibrium constant expression. But not all the molecules separate into ions. This is the case for all weak acids and bases.\r\n\r\nAn acid dissociation constant<\/strong>, K<\/em>a<\/sub>, is the equilibrium constant for the dissociation of a weak acid into ions. Note the a<\/em> subscript on the K<\/em>; it implies that the substance is acting as an acid. The larger K<\/em>a<\/sub> is, the stronger the acid is. Table 13.1 \"Acid Dissociation Constants for Some Weak Acids\" lists several acid dissociation constants. Keep in mind that they are just equilibrium constants.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Table 13.1 Acid Dissociation Constants for Some Weak Acids<\/caption>\r\n
    Acid<\/th>\r\nK<\/em>a<\/sub><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    HC2<\/sub>H3<\/sub>O2<\/sub><\/td>\r\n1.8 \u00d7 10\u22125<\/sup><\/td>\r\n<\/tr>\r\n
    HClO2<\/sub><\/td>\r\n1.1 \u00d7 10\u22122<\/sup><\/td>\r\n<\/tr>\r\n
    H2<\/sub>PO4<\/sub>\u2212<\/sup><\/td>\r\n6.2 \u00d7 10\u22128<\/sup><\/td>\r\n<\/tr>\r\n
    HCN<\/td>\r\n6.2 \u00d7 10\u221210<\/sup><\/td>\r\n<\/tr>\r\n
    HF<\/td>\r\n6.3 \u00d7 10\u22124<\/sup><\/td>\r\n<\/tr>\r\n
    HNO2<\/sub><\/td>\r\n5.6 \u00d7 10\u22124<\/sup><\/td>\r\n<\/tr>\r\n
    H3<\/sub>PO4<\/sub><\/td>\r\n7.5 \u00d7 10\u22123<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNote also that the acid dissociation constant refers to one<\/em> H+<\/sup> ion coming off the initial reactant. Thus the acid dissociation constant for H3<\/sub>PO4<\/sub> refers to this equilibrium:\r\n

    H3<\/sub>PO4<\/sub>(aq) \u21c4 H+<\/sup>(aq)\u00a0+ H2<\/sub>PO4<\/sub>\u2212<\/sup>(aq)\u00a0\u00a0\u00a0\u00a0K<\/em>a\u00a0<\/sub>= 7.5\u00d710\u22123<\/sup><\/p>\r\nThe H2<\/sub>PO4<\/sub>\u2212<\/sup> ion, called the dihydrogen phosphate ion, is also a weak acid with its own acid dissociation constant:\r\n

    H2<\/sub>PO4<\/sub>\u2212<\/sup>(aq) \u21c4 H+<\/sup>(aq)\u00a0+\u00a0HPO4<\/sub>2\u2212<\/sup>(aq)\u00a0\u00a0\u00a0\u00a0K<\/em>a\u00a0<\/sub>= 6.2\u00d710\u22128<\/sup><\/p>\r\nThus for so-called polyprotic<\/em> acids, each H+<\/sup> ion comes off in sequence, and each H+<\/sup> ion that ionizes does so with its own characteristic K<\/em>a<\/sub>.\r\n

    \r\n

    Example 13.10<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWrite the equilibrium equation and the K<\/em>a<\/sub> expression for HSO4<\/sub>\u2212<\/sup> acting as a weak acid.\r\n\r\nSolution<\/em>\r\nHSO4<\/sub>\u2212<\/sup> acts as a weak acid by separating into an H+<\/sup> ion and an SO4<\/sub>2\u2212<\/sup> ion:\r\n

    HSO4<\/sub>\u2212<\/sup>(aq) \u21c4 H+<\/sup>(aq) + SO4<\/sub>2\u2212<\/sup>(aq)<\/p>\r\nThe K<\/em>a<\/sub> is written just like any other equilibrium constant, in terms of the concentrations of products divided by concentrations of reactants:\r\n

    K<\/em>a<\/sub> = [H+<\/sup>][SO4<\/sub>2\u2212<\/sup>][HSO4<\/sub>\u2212<\/sup>]<\/p>\r\nTest Yourself<\/em>\r\nWrite the equilibrium equation and the K<\/em>a<\/sub> expression for HPO4<\/sub>2\u2212<\/sup> acting as a weak acid.\r\n\r\nAnswer<\/em>\r\nHPO4<\/sub>2\u2212<\/sup>(aq)\u2009\u21c4\u2009H+<\/sup>(aq) + PO4<\/sub>3\u2212<\/sup>(aq)\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009K<\/em>a<\/sub> = [H+<\/sup>][PO4<\/sub>3\u2212<\/sup>][HPO4<\/sub>2\u2212<\/sup>]\r\n\r\n<\/div>\r\n<\/div>\r\nThe K<\/em>a<\/sub> is used in equilibrium constant problems just like other equilibrium constants are. However, in some cases, we can simplify the mathematics if the numerical value of the K<\/em>a<\/sub> is small, much smaller than the concentration of the acid itself. Example 13.11 illustrates this.\r\n

    \r\n

    Example 13.11<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is the pH of a 1.00 M solution of HC2<\/sub>H3<\/sub>O2<\/sub>? The K<\/em>a<\/sub> of HC2<\/sub>H3<\/sub>O2<\/sub> is 1.8 \u00d7 10\u22125<\/sup>.\r\n\r\nSolution<\/em>\r\nThis is a two-part problem. We need to determine [H+<\/sup>] and then use the definition of pH to determine the pH of the solution. For the first part, we can use an ICE chart:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nHC2<\/sub>H3<\/sub>O2<\/sub>(aq)<\/th>\r\n\u21c4<\/th>\r\nH+<\/sup>(g)<\/th>\r\n+<\/th>\r\nC2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup>(g)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n1.00<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n\u2212x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n1.00 \u2212 x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe now construct the K<\/em>a<\/sub> expression, substituting the concentrations from the equilibrium row in the ICE chart:\r\n

    [latex]K_{\\text{a}}=\\dfrac{([\\ce{H+}][\\ce{C2H3O2-}])}{[\\ce{HC2H3O2}]}=\\dfrac{(x)(x)}{(1.00-x)}=1.8\\times 10^{-5}[\/latex]<\/p>\r\nHere is where a useful approximation comes in: at 1.8 \u00d7 10\u22125<\/sup>, HC2<\/sub>H3<\/sub>O2<\/sub> will not ionize very much, so we expect that the value of x<\/em> will be small. It should be so small that in the denominator of the fraction, the term (1.00 \u2212 x<\/em>) will likely be very close to 1.00. As such, we would introduce very little error if we simply neglect the x<\/em> in that term, making it equal to 1.00:\r\n

    [latex](1.00-x)\\approx 1.00\\text{ for small values of }x[\/latex]<\/p>\r\nThis simplifies the mathematical expression we need to solve:\r\n

    [latex]\\dfrac{(x)(x)}{1.00}=1.8\\times 10^{-5}[\/latex]<\/p>\r\nThis is much easier to solve than a more complete quadratic equation. The new equation to solve becomes:\r\n

    [latex]x^2=1.8\\times 10^{-5}[\/latex]<\/p>\r\nTaking the square root of both sides:\r\n

    [latex]x=4.2\\times 10^{-3}[\/latex]<\/p>\r\nBecause x<\/em> is the equilibrium concentrations of H+<\/sup> and C2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup>, we thus have:\r\n

    [latex][\\ce{H+}]=4.2\\times 10^{-3}\\text{ M}[\/latex]<\/p>\r\nNotice that we are justified by neglecting the x<\/em> in the denominator; it truly is small compared to 1.00. Now we can determine the pH of the solution:\r\n

    [latex]\\text{pH}=-\\log[\\ce{H+}]=-\\log(4.2\\times 10^{-3})=2.38[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nWhat is the pH of a 0.500 M solution of HCN? The K<\/em>a<\/sub> of HCN is 6.2 \u00d7 10\u221210<\/sup>.\r\n\r\nAnswer<\/em>\r\n4.75\r\n\r\n<\/div>\r\n<\/div>\r\nWeak bases also have dissociation constants, labelled K<\/em>b<\/sub> (the b<\/em> subscript stands for base). However, values of K<\/em>b<\/sub> are rarely tabulated because there is a simple relationship between the K<\/em>b<\/sub> of a base and the K<\/em>a<\/sub> of its conjugate acid:\r\n

    [latex]K_{\\text{a}}\\times K_{\\text{b}}=1.0\\times 10^{-14}[\/latex]<\/p>\r\nThus it is simple to calculate the K<\/em>b<\/sub> of a base from the K<\/em>a<\/sub> of its conjugate acid.\r\n

    \r\n

    Example 13.12<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is the value of K<\/em>b<\/sub> for C2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup>, which can accept a proton and act as a base?\r\n\r\nSolution<\/em>\r\nTo determine the K<\/em>b<\/sub> for C2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup>, we need to know the K<\/em>a<\/sub> of its conjugate acid. The conjugate acid of C2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup> is HC2<\/sub>H3<\/sub>O2<\/sub>. The K<\/em>a<\/sub> for HC2<\/sub>H3<\/sub>O2<\/sub> is in Table 13.1 \"Acid Dissociation Constants for Some Weak Acids\" and is 1.8 \u00d7 10\u22125<\/sup>. Using the mathematical relationship between K<\/em>a<\/sub> and K<\/em>b<\/sub>:\r\n

    [latex](1.8\\times 10^{-5})K_{\\text{b}}=1.0\\times 10^{-14}[\/latex]<\/p>\r\nSolving,\r\n

    [latex]K_{\\text{b}}=\\dfrac{1.0\\times 10^{-14}}{1.8\\times 10^{-5}}=5.6\\times 10^{-10}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nWhat is the value of K<\/em>b<\/sub> for PO4<\/sub>3\u2212<\/sup>, which can accept a proton and act as a base? The K<\/em>a<\/sub> for HPO4<\/sub>2\u2212<\/sup> is 2.2 \u00d7 10\u221213<\/sup>.\r\n\r\nAnswer<\/em>\r\n4.5 \u00d7 10\u22122<\/sup>\r\n\r\n<\/div>\r\n<\/div>\r\n

    Autoionization of Water<\/h1>\r\nIn Chapter 12 \"Acids and Bases\"<\/a>, we introduced the autoionization of water \u2014 the idea that water can act as a proton donor and proton acceptor simultaneously. Because water is not a strong acid (Table 12.1 \"Strong Acids and Bases\"<\/a>), it must be a weak acid, which means that its behaviour as an acid must be described as an equilibrium. That equilibrium is as follows:\r\n

    H2<\/sub>O(\u2113)\u00a0+\u00a0H2<\/sub>O(\u2113) \u21c4 H3<\/sub>O+<\/sup>(aq)\u00a0+\u00a0OH\u2013<\/sup>(aq)<\/p>\r\nThe equilibrium constant includes [H3<\/sub>O+<\/sup>] and [OH\u2212<\/sup>] but not [H2<\/sub>O(\u2113)] because it is a pure liquid. Hence the expression does not have any terms in its denominator<\/em>:\r\n

    K<\/em> = [H3<\/sub>O+<\/sup>][OH\u2212<\/sup>] \u2261 K<\/em>w<\/sub> = 1.0 \u00d7 10\u221214<\/sup><\/p>\r\nThis is the same K<\/em>w<\/sub> that was introduced in Chapter 12 and the same 1.0 \u00d7 10\u221214<\/sup> that appears in the relationship between the K<\/em>a<\/sub> and the K<\/em>b<\/sub> of a conjugate acid-base pair. In fact, we can rewrite this relationship as follows:\r\n

    [latex]K_{\\text{a}}\\times K_{\\text{b}}=K_{\\text{w}}[\/latex]<\/p>\r\n\r\n

    Insoluble Compounds<\/h1>\r\nIn Chapter 4 \"Chemical Reactions and Equations\"<\/a>, in section \"Types of Chemical Reactions: Single- and Double-Displacement Reactions\"<\/a>, on chemical reactions, the concept of soluble and insoluble compounds was introduced. Solubility rules were presented that allow a person to predict whether certain simple ionic compounds will or will not dissolve.\r\n\r\nDescribing a substance as soluble or insoluble is a bit misleading because virtually all substances are soluble; they are just soluble to different extents. In particular for ionic compounds, what we typically describe as an insoluble<\/em> compound can actually be ever so slightly soluble; an equilibrium is quickly established between the solid compound and the ions that do form in solution. Thus the hypothetical compound MX does in fact dissolve but only very slightly. That means we can write an equilibrium for it:\r\n

    MX(s) \u21c4 M+<\/sup>(aq)\u00a0+\u00a0X\u2013<\/sup>(aq)<\/p>\r\nThe equilibrium constant for a compound normally considered insoluble is called a solubility product constant<\/strong>. The equilibrium constant for a compound normally considered insoluble. and is labelled K<\/em>sp<\/sub> (with the subscript sp<\/em>, meaning \u201csolubility product\u201d). Because the reactant is a solid, its concentration does not appear in the K<\/em>sp<\/sub> expression, so like K<\/em>w<\/sub>, expressions for K<\/em>sp<\/sub> do not have denominators. For example, the chemical equation and the expression for the K<\/em>sp<\/sub> for AgCl, normally considered insoluble, are as follows:\r\n

    AgCl(s) \u21c4 Ag+<\/sup>(aq)\u00a0+\u00a0Cl\u2013<\/sup>(aq)\u00a0\u00a0\u00a0\u00a0Ksp<\/sub> = [Ag+<\/sup>][Cl\u2013<\/sup>]<\/p>\r\nTable 13.2 \"Solubility Product Constants for Slightly Soluble Ionic Compounds\" lists some values of the K<\/em>sp<\/sub> for slightly soluble ionic compounds.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Table 13.2 Solubility Product Constants for Slightly Soluble Ionic Compounds<\/caption>\r\n
    Compound<\/th>\r\nK<\/em>sp<\/sub><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    BaSO4<\/sub><\/td>\r\n1.1 \u00d7 10\u221210<\/sup><\/td>\r\n<\/tr>\r\n
    Ca(OH)2<\/sub><\/td>\r\n5.0 \u00d7 10\u22126<\/sup><\/td>\r\n<\/tr>\r\n
    Ca3<\/sub>(PO4<\/sub>)2<\/sub><\/td>\r\n2.1 \u00d7 10\u221233<\/sup><\/td>\r\n<\/tr>\r\n
    Mg(OH)2<\/sub><\/td>\r\n5.6 \u00d7 10\u221212<\/sup><\/td>\r\n<\/tr>\r\n
    HgI2<\/sub><\/td>\r\n2.9 \u00d7 10\u221229<\/sup><\/td>\r\n<\/tr>\r\n
    AgCl<\/td>\r\n1.8 \u00d7 10\u221210<\/sup><\/td>\r\n<\/tr>\r\n
    AgI<\/td>\r\n8.5 \u00d7 10\u221217<\/sup><\/td>\r\n<\/tr>\r\n
    Ag2<\/sub>SO4<\/sub><\/td>\r\n1.5 \u00d7 10\u22125<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
    \r\n

    Example 13.13<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWrite the K<\/em>sp<\/sub> expression for Ca3<\/sub>(PO4<\/sub>)2<\/sub>.\r\n\r\nSolution<\/em>\r\nRecall that when an ionic compound dissolves, it separates into its individual ions. For Ca3<\/sub>(PO4<\/sub>)2<\/sub>, the ionization reaction is as follows:\r\n

    Ca3<\/sub>(PO4<\/sub>)2<\/sub>(s) \u21c4 3Ca2+<\/sup>(aq) + 2PO4<\/sub>3\u2013<\/sup>(aq)<\/p>\r\nHence the K<\/em>sp<\/sub> expression is\r\n

    K<\/em>sp<\/sub> = [Ca2+<\/sup>]3<\/sup>[PO4<\/sub>3\u2212<\/sup>]2<\/sup><\/p>\r\nTest Yourself<\/em>\r\nWrite the K<\/em>sp<\/sub> expression Ag2<\/sub>SO4<\/sub>.\r\n\r\nAnswer<\/em>\r\nK<\/em>sp<\/sub> = [Ag+<\/sup>]2<\/sup>[SO4<\/sub>2\u2212<\/sup>]\r\n\r\n<\/div>\r\n<\/div>\r\nEquilibrium problems involving the K<\/em>sp<\/sub> can also be done, and they are usually more straightforward than other equilibrium problems because there is no denominator in the K<\/em>sp<\/sub> expression. Care must be taken, however, in completing the ICE chart and evaluating exponential expressions.\r\n

    \r\n

    Example 13.14<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat are [Ag+<\/sup>] and [Cl\u2212<\/sup>] in a saturated solution of AgCl? The K<\/em>sp<\/sub> of AgCl is 1.8 \u00d7 10\u221210<\/sup>.\r\n\r\nSolution<\/em>\r\nThe chemical equation for the dissolving of AgCl is\r\n

    AgCl(s) \u21c4 Ag+<\/sup>(aq) + Cl\u2013<\/sup>(aq)<\/p>\r\nThe K<\/em>sp<\/sub> expression is as follows:\r\n

    K<\/em>sp<\/sub> = [Ag+<\/sup>][Cl\u2212<\/sup>]<\/p>\r\nSo the ICE chart for the equilibrium is as follows:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAgCl(s)<\/th>\r\n\u21c4<\/th>\r\nAg+<\/sup>(aq)<\/th>\r\n+<\/th>\r\nCl\u2212<\/sup>(aq)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n\u2212x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n<\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that we have little in the column under AgCl except the stoichiometry of the change; we do not need to know its initial or equilibrium concentrations because its concentration does not appear in the K<\/em>sp<\/sub> expression. Substituting the equilibrium values into the expression:\r\n

    [latex](x)(x)=1.8\\times 10^{-10}[\/latex]<\/p>\r\nSolving,\r\n

    [latex]\\begin{align*}\r\nx^2&= 1.8\\times 10^{-10} \\\\\r\nx&=1.3\\times 10^{-5}\r\n\\end{align*}[\/latex]<\/p>\r\nThus [Ag+<\/sup>] and [Cl\u2212<\/sup>] are both 1.3 \u00d7 10\u22125<\/sup> M.\r\n\r\nTest Yourself<\/em>\r\nWhat are [Ba2+<\/sup>] and [SO4<\/sub>2\u2212<\/sup>] in a saturated solution of BaSO4<\/sub>? The K<\/em>sp<\/sub> of BaSO4<\/sub> is 1.1 \u00d7 10\u221210<\/sup>.\r\n\r\nAnswer<\/em>\r\n1.0 \u00d7 10\u22125<\/sup> M\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Example 13.15<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat are [Ca2+<\/sup>] and [PO4<\/sub>3\u2212<\/sup>] in a saturated solution of Ca3<\/sub>(PO4<\/sub>)2<\/sub>? The K<\/em>sp<\/sub> of Ca3<\/sub>(PO4<\/sub>)2<\/sub> is 2.1 \u00d7 10\u221233<\/sup>.\r\n\r\nSolution<\/em>\r\nThis is similar to Example 13.14, but the ICE chart is much different because of the number of ions formed.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nCa3<\/sub>(PO4<\/sub>)2<\/sub>(s)<\/th>\r\n\u21c4<\/th>\r\n3Ca2+<\/sup>(aq)<\/th>\r\n+<\/th>\r\n2PO4<\/sub>3\u2212<\/sup>(aq)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n\u2212x<\/em><\/td>\r\n+3x<\/em><\/td>\r\n+2x<\/em><\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n<\/td>\r\n+3x<\/em><\/td>\r\n+2x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor every unit of Ca3<\/sub>(PO4<\/sub>)2<\/sub> that dissolves, three Ca2+<\/sup> ions and two PO4<\/sub>3\u2212<\/sup> ions are formed. The expression for the K<\/em>sp<\/sub> is also different:\r\n

    K<\/em>sp<\/sub> = [Ca2+<\/sup>]3<\/sup>[PO4<\/sub>3\u2212<\/sup>]2<\/sup> = 2.1 \u00d7 10\u221233<\/sup><\/p>\r\nNow when we substitute the unknown concentrations into the expression, we get:\r\n

    [latex](3x)^3(2x)^2=2.1\\times 10^{-33}[\/latex]<\/p>\r\nWhen we raise each expression inside parentheses to the proper power, remember that the power affects everything inside the parentheses, including the number. So:\r\n

    [latex](27x^3)(4x^2)=2.1\\times 10^{-33}[\/latex]<\/p>\r\nSimplifying:\r\n

    [latex]108x^5=2.1\\times 10^{-33}[\/latex]<\/p>\r\nDividing both sides of the equation by 108, we get:\r\n

    [latex]x^5=1.9\\times 10^{-35}[\/latex]<\/p>\r\nNow we take the fifth root of both sides of the equation (be sure you know how to do this on your calculator):\r\n

    [latex]x=1.1\\times 10^{-7}[\/latex]<\/p>\r\nWe are not done yet. We still need to determine the concentrations of the ions. According to the ICE chart, [Ca2+<\/sup>] is 3x<\/em>, not x<\/em>. So\r\n

    [latex]\\begin{align*}\r\n[\\ce{Ca2+}]&=3x=3\\times 1.1\\times 10^{-7}=3.3\\times 10^{-7}\\text{ M} \\\\\r\n{[\\ce{PO4^{3-}}]}&=2x = 2\\times 1.1\\times 10^{-7}=2.2\\times 10^{-7}\\text{ M}\r\n\\end{align*}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nWhat are [Mg2+<\/sup>] and [OH\u2212<\/sup>] in a saturated solution of Mg(OH)2<\/sub>? The K<\/em>sp<\/sub> of Mg(OH)2<\/sub> is 5.6 \u00d7 10\u221212<\/sup>.\r\n\r\nAnswer<\/em>\r\n[Mg2+<\/sup>] = 1.1 \u00d7 10\u22124<\/sup> M; [OH\u2212<\/sup>] = 2.2 \u00d7 10\u22124<\/sup> M\r\n\r\n<\/div>\r\n<\/div>\r\n

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    Food and Drink App: Solids in Your Wine Bottle<\/h1>\r\nPeople who drink wine from bottles (as opposed to boxes) will occasionally notice some insoluble materials in the wine, either crusting the bottle, stuck to the cork, or suspended in the liquid wine itself. The accompanying figure shows a cork encrusted with coloured crystals. What are these crystals?\r\n\r\n[caption id=\"attachment_756\" align=\"aligncenter\" width=\"470\"]\"Cork\" The red crystals on the top of the wine cork are from insoluble compounds that are not soluble in the wine.[\/caption]\r\n\r\nOne of the acids in wine is tartaric acid (H2<\/sub>C4<\/sub>H4<\/sub>O6<\/sub>). Like the other acids in wine (citric and malic acids, among others), tartaric acid imparts a slight tartness to the wine. Tartaric acid is rather soluble in H2<\/sub>O, dissolving over 130 g of the acid in only 100 g of H2<\/sub>O. However, the potassium salt of singly ionized tartaric acid, potassium hydrogen tartrate (KHC4<\/sub>H4<\/sub>O6<\/sub>; also known as potassium bitartrate and better known in the kitchen as cream of tartar), has a solubility of only 6 g per 100 g of H2<\/sub>O. Thus, over time, wine stored at cool temperatures will slowly precipitate potassium hydrogen tartrate. The crystals precipitate in the wine or grow on the insides of the wine bottle and, if the bottle is stored on its side, on the bottom of the cork. The colour of the crystals comes from pigments in the wine; pure potassium hydrogen tartrate is clear in its crystalline form, but in powder form it is white.\r\n\r\nThe crystals are harmless to ingest; indeed, cream of tartar is used as an ingredient in cooking. However, most wine drinkers don\u2019t like to chew their wine, so if tartrate crystals are present in a wine, the wine is usually filtered or decanted to remove the crystals. Tartrate crystals are almost exclusively in red wines; white and rose wines do not have as much tartaric acid in them.\r\n\r\n<\/div>\r\n
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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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