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Solubility Equilibrium

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  Chemical Equilibrium – Solubility The solubility of a substance is dependent on the forces holding the crystal together (the lattice energy) and the solvent acting on these forces. For now, we will consider only water as the solvent. As the solid dissolves, water molecules surround the ions in the solution by a process called hydration. During hydration, energy is released. The extent to which the energy of hydration is greater than the lattice energy determines the solubility.The euilibrium involved in the solubility of a substance is! a #  b  (s)   a n$  (a) $ b # n%  (a) ( a general euation)where is the cation of change n$ and # is the anion of charge n%. The subscripts of the salt become the coefficients for the ions. The euilibrium mass action expression would be! &  sp  '  n$  a  # n%   b  *otice that only the aueous ions are part of the mass action expression. +olids are never included. Also notice that the coefficient for each ion becomes the exponent in the mass action expression.There are four common types of solubility euilibria problems! solubility in pure water (will a precipitate form or finding the molar solubility), solubility in the presence of a common ion and selective precipitation. e will loo- at each of these in turn. Solubility in pure water  *umerical problems involving solubility of a compound in pure water may fall into two categories! () determining whether a precipitate will form when two aueous solutions are mixed or (/) determining the actual concentration of ions in a saturated solution.0et1s examine the mixing of two solutions. The problem might read as follows!+uppose 22.2 m0 of a 2.222 3a3l /  solution and /22.2 m0 of a 2.2/22 *a 4 56 7  solution were mixed. ould a precipitate form8To tac-le this problem, you must do a stoichiometry problem first. +tart by writing the balanced chemical euation!4 3a3l /  (a) $ / *a 4 56 7  (a)   3a 4 (56 7 ) /  (s) $ 9 *a3l (a)  3onvert this euation into net ionic form by crossing out the spectator ions from the total ionic form of the euation!  4 3a /$  (a) $ 9 3l  :  (a) $ 9 *a $  (a) $ / 56 74%  (a)   3a 4 (56 7 ) /  (s) $ 9 *a $  (a) $ 9 3l  :  (a) 4 3a /$  (a) $ / 56 74%  (a)   3a 4 (56 7 ) /  (s) Find the initial concentration of 3a /$  ions and 56 74%  ions! +++ == ///// 3a 2.222  3a3lmol3amol x 03a3lmol2.222 )(3a %4774%47744%7 56 2.2/22  56 *amol 56mol x 056 *amol2.2/22  56  ==  *ext, determine the new, diluted concentration of calcium and phosphate ions resulting from the mixing of the two solutions! +++ == /// 3a 2.22444  02.422;0)3amol0)(2.222(2.22  3a %47%474%7 56 2.244  02.422)56mol0)(2.2/22(2./22  56  == 3alculate the solubility product uotient, <, based on the mass action expression of the solubility euilibrium!< ' 3a /$  4 56 74%  / < ' 2.22444  4 2.244  /  ' 9.=4 x 2 %/ 3ompare the value of < to the value of &  sp . >f the < ≤  &  sp , then the solubility has not  been exceeded and a precipitate will not form. >f the < ? &  sp , then the solubility has been exceeded and a precipitate will form. The &  sp  of 3a 4 (56 7 ) /  is .2 x 2 %/9 . +ince <, 9.=4 x 2 %/ , exceeds the &  sp  of .2 x 2 %/9 , a precipitate of 3a 4 (56 7 ) /  will form in this case.A second type of solubility problem would as- to find the solubility of a pure substance in water. +uppose the uestion as-s for the solubility of calcium phosphate in water. The euilibrium ta-ing place is!3a 4 (56 7 ) /  (s)   4 3a /$  (a) $ / 56 74%  (a)  The &  sp  of calcium phosphate is .2 x 2 %/9 , and is either given in the problem or can be found on a table of solubility product constants. @elow the euilibrium, show what is occurring initially, the change needed to establish euilibrium and the conditions at euilibrium!3a 4 (56 7 ) /  (s)   4 3a /$  (a) $ / 56 74%  (a) >nitial 2 23hange $ 4x$ /xuilibrium 4x /x>nitially, calcium phosphate has not dissociated, thus the concentrations of calcium ions and phosphate ions is considered to be 2. Thin- of it as a reference point. To establish euilibrium, some calcium phosphate must dissociate as a 4x amount of calcium ions and/x amount of phosphate ions. Thus, at euilibrium, the amount of calcium ions is 4x and the amount of phosphate ions is /x. These values are substituted into the mass action expression!&  sp  ' 3a /$  4 56 74%  /  ' .2 x 2 %/9  ' (4x) 4 (/x) /  ' .2 x 2 %/9 As a reminder, 3a 4 (56 7 ) /  is a solid, and is therefore not a part of the mass action expression.+olve the above algebraic expression for x!2B x =  ' .2 x 2 %/9  x =  ' .2 x 2 %/9 ;2B ' C.4 x 2 %/C   9%=/C% 2x /.= 2x C.4 x == The value of x is the molar solubility of calcium phosphate. The actual concentrations of 3a /$  ions and 56 74%  ions in solution is given by!3a /$  ' 4x ' 4(/.= x 2 %9 ) ' .= x 2 %9   3a /$ 56 74%  ' /x ' /(/.= x 2 %9 ) ' =.2 x 2 %9   56 74% Esing the molar solubility, one may calculate the actual mass of calcium phosphate dissolved in one liter of water! solution;0)(563ag2x D.B  )(563amol )(563ag42.B  x solution0)(563amol2x /.= /747%/74/74/74 9% =  Solubility in the presence of a common ion +uppose a salt of low solubility is found in the presence of a soluble salt in which the cation or anion of the soluble salt is common to the cation or anion of the insoluble salt. For instance, if a uantity of sodium phosphate is introduced to a saturated solution of calcium phosphate, what would happen to the solubility of calcium phosphate8 0et1s loo- at this problem ualitatively, first. The common ion is the phosphate ion. According to 0e3hatelier1s 5rinciple, if a stress is placed on an euilibrium, the euilibrium will shift in the direction which relieves the stress. The stress in this case is excess phosphate ion the euilibrium will shift to the left, and one would see additional  precipitate of calcium phosphate form.3a 4 (56 7 ) /  (s)   4 3a /$  (a) $ / 56 74%  (a) uilibrium shifts to the left in the presence of excess phosphate ion.The lowered solubility of calcium phosphate becomes evident when the problem is approached uantitatively. A typical problem would read!hat is the solubility of calcium phosphate in a 2.2 solution of sodium  phosphate8The euilibrium ta-ing place is!3a 4 (56 7 ) /  (s)   4 3a /$  (a) $ / 56 74%  (a) &eep in mind that while excess phosphate will be present from the sodium phosphate, sodium ion is Gust a spectator ion during the process.To start the problem, set up an euilibrium table!3a 4 (56 7 ) /  (s)   4 3a /$  (a) $ / 56 74%  (a) >nitial 2 2.2 3hange $ 4x $ /xuilibrium 4x 2.2 $ /x *otice how the initial amount of phosphate is not Hero. This initial amount of phosphate ion is contributed by the 2.2 sodium phosphate solution. For euilibrium to become established, a 4x amount of calcium ion is formed and an additional /x amount of  phosphate ion from the dissociation of calcium phosphate is introduced. Thus, at euilibrium, the amount of calcium ion in solution is 4x and the amount of phosphate ion in solution is 2.2 $ /x.
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