Aqueous Equilibria By: Chris Via

Common-Ion Effect C.I.E.- the dissociation of a weak electrolyte by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte. The addition of this strong electrolyte causes the equilibrium to shift as well as the pH to change.

Example The dissociation of HC 2 H 3 O 2 is: HC 2 H 3 O 2 H +1 + C 2 H 3 O 2 -1 but if NaC 2 H 3 O 2 is added to the solution, the addition of C 2 H 3 O from NaC 2 H 3 O 2 causes the equilibrium to shift to the left and also reducing the concentration of H +1, therefore raising the pH of the solution.

Practice Problem Calculate the fluoride -ion concentration and pH of a solution containing 0.1 mol of HCl and 0.2 mol of HF in 1.0L solution. (K a of HF = 6.8 x ) Now you want to set up an ice box of the dissociation of HF.

Continued HFH+F-F- K a =6.8*10 -4 = [H + ][F - ]/[HF]=(.1+x)x/(.2-x)

Solution In this case we can assume x is too small to make a difference so (0.1+ x) becomes just 0.1 So we are left with 0.1 = 6.8 x x =1.4 x M= [F -1 ] [H +1 ]= (0.1+ x) = 0.1M -log ([H + ]) = -log (0.1) = 1 The pH=1.00 and the [F -1 ]= 1.4 x M

Buffered Solutions Buffers are solutions that resist changes in pH upon the addition of small amounts of acids or bases. A key example of a buffer is human blood which will stay around a pH of 7.4 Buffers resist changes in pH because they contain both an acidic species to neutralize OH -1 ions and a basic species to neutralize H +1 ions.

Buffers However, it is key that the acid and base species of the buffer do not neutralize each other. To do this, buffers are often prepared by mixing a weak acid or base with a salt of that acid or base. Ex, the HC 2 H 3 O 2 and C 2 H 3 O 2 -1 buffer can be prepared by adding NaC 2 H 3 O 2 to a solution of HC 2 H 3 O 2.

Buffer Capacity Buffer Capacity is the amount of an acid or base the buffer can neutralize before the pH begins to change significantly. The greater the amounts of conjugate acid- base pair, the more resistant the solution is to change in pH. Henderson-Hasselbalch equation is used to calculate the pH of buffers: pH = pK a + log[base]/[acid]

Sample Exercise What is the pH of a buffer that is 0.12 M lactic acid and.1 M sodium lactate? (K a for lactic acid = 1.4 x ) You want to start out with an ice box of the dissociation of lactic acid lactic acid hydrogen lactate

Problem Continued K a = 1.4 x = x (0.1+x)/(0.12-x) Once again in this case x will be too small so it can be ignored. X=1.7 x = [H +1 ] pH= -log (1.7 x ) = 3.77 With the Henderson-Hasselbalch equation: pH= pK a + log([base]/[acid]) pH= (-0.08) =3.77

Titrations An acid-base titration is when an acid is added to a base or vice versa. Acid-Base indicators are usually used to determine the equivalence point which is the point at which stoichiometrically equivalent quantities of acid and base have been brought together.

Acid-Base Titrations When strong acids and bases are mixed the pH changes very dramatically near the equivalence point a single drop a this point could change the pH by a number of units.

Titrations

Polyprotic Acids When titrating with Polyprotic acids or bases the substance has multiple equivalence points. So for example, in a titration of Na 2 CO 3 with HCl. There are two distinct equivalence points on the titration curve.

Solubility-Product Constant K sp = Solubility Product Constant It expresses the degree to which the solid is soluble in water. The equation for K sp is K sp = [ion] a [ion] b An example is the expression of the K sp of BaSO 4 : K sp = [Ba 2+ ][SO 4 2- ]

Factors that Affect Solubility They are: –the presence of common ions –the pH of the solution –presence of complexing agents The presence of common ions in a solution will reduce the solubility and make the equilibrium shift left.

pH and Solubility The general rule of solubility and pH is: The solubility of slightly-soluble salts containing basic anions increases as the pH of the solution is lowered. This is because the OH - ion is insoluble in water while the H + ion is very soluble, therefore when a basic solution has a low concentration of OH - ions the salt will be easier to dissolve.

Continued The more basic the anion, the more the solubility is influenced by the pH of the solution. However, salts with anions of strong acids are unaffected by changes in pH.

Complex Ions A characteristic of most metal ions is their ability to act as a Lewis acid when interacting with water (Lewis base). However, when these metal ions interact with Lewis bases other than water, the solubility of that metal salt changes dramatically.

Complex Ion Complex Ion-when a metal ion (Lewis acid) is bonded together with a Lewis base other than water. –Ex. Ag(NH 3 ) 2 +, Fe(CN) 6 -4 The stability of a complex ion can be judged by the size of the equilibrium constant for its formation. –This is called the formation constant, K f –the larger K f the more stable the ion is.

Amphoterism Amphoteric-a metal hydroxide that is capable of being dissolved in strong acids or strong bases, but not in water because it can act like an acid or a base. –Ex. Al +3, Cr +3, Zn +2, and Sn +2 However, these metal ions are more accurately expressed as Al(H 2 O) This is a weak acid and as it is added to a strong base it loses protons and eventually forms the neutral and water-soluble Al(H 2 O) 3 (OH) 3

Precipitation of Ions The reaction quotient, Q, can with the solubility product constant to determine if a precipitation will occur. If Q > K sp -precipitation occurs until Q = K sp If Q = K sp -equilibrium exists (saturated solution) If Q < K sp -solid dissolves until Q = K sp Q is sometimes referred to as the ion product because there is no denominator.

Selective Precipitation Selective Precipitation-the separation of ions in an aqueous solution by using a reagent that forms a precipitate with one or more of the ions. The sulfide ion is widely used to separate metal ions because the solubilities of the sulfide salts span a wide range and are dependent on the pH of the solution.

Continued An example is a solution containing both Cu +2 and Zn +2 mixed with H 2 S gas. CuS ends up forming a precipitate, but ZnS does not. This is because CuS has a K sp value of 6 x making it less soluble than ZnS which has a K sp value of 2 x

Qualitative Analysis Qualitative Analysis determines the presence or absence of a particular metal ion, whereas quantitative analysis determines how much of a certain substance is present or produced. There are 5 main groups of metal ions: –Insoluble Chlorides, Acid-insoluble sulfides, base- insoluble sulfides and hydroxides, insoluble phosphates, and the alkali metal ions and NH 4 +1 These can be found on page 673 in your textbook