Additional Aspects of Aqueous Equilibria Chapter 17 Common

Additional Aspects of Aqueous Equilibria Chapter 17

Common Ion Effect the dissociation of a weak electrolyte is decreased by the addition of a common in the form of a strong electrolyte Calculate the p. H of 0. 30 mol HC 2 H 3 O 2 and 0. 30 mol Na. C 2 H 3 O 2 to make a 1. 0 L sol’n answer: p. H = 4. 74 p. H of sol’n containing just 0. 30 mol acetic acid is 2. 64 the addition of extra acetate reduces the ionization of acetic acid (notice ka & [H+] are identical when [acid] and [common ion] are the same)

Buffered Solutions resist changes in p. H as small amounts of acid or base are added to a solution made of a weak acid and its conjugate base HX ↔ H+ + X- When an acid & its conj. base are present in equal [ ]s, ka = [H+]

Buffered Solutions adding acid: causes decrease in [X-] and an increase in [HX] adding base: causes increase in [X-] and a decrease in [HX] Buffers are most resistant to change from either direction when [HX] = [X-] Buffer selection depends upon ka value pka should be near the desired p. H of the buffer

Buffer Capacity amount of acid or base that can be neutralized by a buffer before the p. H changes drastically depends on amount of HX and X- in sol’n increasing [HX] and [X-] causes an increase in buffer capacity p. H (. 1 M HF/. 1 M F-) = p. H(. 01 M HF/. 01 M F-) buffer capacity sol’n > buffer capacity sol’n

Henderson-Hasselbalch Eq’n rearranging ka eq’n yields: if [acid] = [base], then p. H = pka

problem: caluculate the p. H of a solution with 0. 12 M HC 3 H 5 O 3 and 0. 10 M Na. C 3 H 5 O 3. ka(HC 3 H 5 O 3) = 1. 4 x 10 -4 Answer: p. H = 3. 77

Adding Strong Acids or Bases to Buffers 1. 2. 3. 4. determine effect of neutralization rxn on [X-] and [HX] stoichiometry use ka and new [HX] and new [X-] to determine [H+] equilibrium calculation use new [H+] to calculate p. H ex: a buffer is made of 0. 300 mol acetic acid and 0. 300 mol sodium acetate to form 1. 00 L sol’n. Buffer p. H from earlier problem = 4. 74. Find p. H after adding 0. 020 mol Na. OH assuming there is no change in volume.

p. H changes in buffered sol’ns answer to example: p. H = 4. 8 a difference of only 0. 06 is observed in a buffered solution adding 0. 020 M Na. OH to pure water would produce a p. H of 12. 3

acid-base titrations a sol’n of known concentration is added to a sol’n of unknown concentration equivalence point – equal amounts of H+ and OHin sol’n titration curve – graph of p. H vs volume added; can be used to determine equivalence point, select an indicator, and determine ka or kb of a weak acid or base end point – the point at which an indicator changes color

Strong Acid-Strong Base Titrations strong base added to a strong acid; general shape is a pronounced s-curve 1. initial p. H is low due to presence of a strong acid 2. slow increase in p. H, then a rapid rise to the equivalence point 3. equivalence point for a strong acid-strong base titration is always at p. H =7 (no hydrolysis) 4. after eq point – p. H is determined by excess OH- in solution

problem: what is the p. H when 49. 00 m. L of 0. 100 M Na. OH are added to 50. 00 m. L of 0. 100 M HCl? Answer: p. H = 3. 00

indicator selection an indicator may be used to approximate the equivalence point as long as its color change coincides with the vertical portion of the titration curve during vertical portion, change in volume of base added is so small that the end point will closely approximate the equivalence point

weak acid-strong base titrations smaller vertical rise to the s-shaped curve 1. initial p. H is due to ionization of the weak acid 2. p. H gradually rises – neutralization of weak acid by a strong base; calculate p. H using [eq] of the buffer pair (weak acid + conj. base) 3. equivalence point p. H > 7 due to basic nature of the conjugate pair of the weak acid 4. after eq pt [OH-] >> [X-]; calculate p. H using [OH-]

3 major differences between SA/SB and WA/SB titrations 1. 2. 3. 4. 5. 6. 7. weak acid sol’n has a higher p. H than strong acid sol’n of the same concentration smaller vertical section in weak acid curve than in the strong acid curve weak acid eq pt > 7; strong acid eq pt = 7 As ka becomes smaller: 1. equivalence point has a higher p. H 2. vertical portion of the curve gets smaller 3. becomes more difficult to select an

titrations of polyprotic acids diprotic – two equivalence points triprotic – three equivalence points may appear to be as two or three titration curves in sequence

solubility equilbria solubility product constant (ksp) – used to express sol’n equilibrium between dissolved and undissolved solute Ca. CO 3(s) ↔ Ca 2+(aq) + CO 32 -(aq) ksp = [Ca 2+(aq)][CO 32 -(aq)] As ksp increases, the solubility of a substance at a given temperature increases

problem: silver chromate is added to pure water at 25 o. C. Some solid remains undissolved. After stirring the mixture to ensure that equilibrium has been reached, analysis shows that silver ion has a concentration of 1. 3 x 10 -4 M. Calculate the ksp for this compound assuming the silver chromate dissociates completely in water and there are no other important equilibria involving the silver ion and chromate ion in solution. Answer: 1. 1 x 10 -12

factors affecting solubility common ion effect – the addition of a common ion will cause a shift in the equilibrium to the left; producing more precipitate and decreasing the solubility of the substance 2. p. H – any species with a basic anion will be affected by the p. H of the solution; the more acidic a solution is, the better a salt with a basic anion will dissolve (anions with negligible basicity are unaffected by p. H 1.

factors affecting solubility 3. forming complex ions – ligands that interact more strongly with the metal ion than water can affect the solubility of the metal ion; ligands must be able to displace solvating water molecules the stability of a complex ion in aqueous solution is measured by the formation constant (kf) Ag+(aq) + 2 CN-(aq) ↔ [Ag(CN)2]-(aq)

factors affecting solubility 4. amphoterism – many metal oxides and hydroxides that are nearly insoluble in neutral sol’n will readily dissolve in strongly acidic or strongly basic sol’n ex: Al, Cr, Zn, Sn are common amphoteric metal oxides and hydroxides complex ions form using the metal cation and the hydroxide ion as the ligand; as the OH- ion replaces water, the charge of the complex ion becomes increasingly negative, making it harder to add more OH- ions ex: Al(OH)3 + OH- ↔ Al(OH)4 -

Precipitation and Separation of Ions the reaction quotient (Q) may be used to determine whether a sol’n is at equilibrium Q > ksp ; supersaturated sol’n; precipitation occurs until Q = ksp ; saturated sol’n ; currently @ equilibrium Q < ksp ; unsaturated sol’n ; solid dissolves until Q = ksp

selective precipitation using reagents to form precipitates with one or a few dissolved ions ex: Cu. S precipitates from an acidified solution but Zn. S will not precipitate from an acidified sol’n this allows the removal of copper (II) ion by the process of filtration from the remaining dissolved ions in the mixture

qualitative analysis determines the presence or absence of a particular metal ions are separated into broad groups based upon solubility properties ions within each group are selectively separated by more specific solubility properties ions may be positively identified via specific reactivity tests

Sample Scheme for qualitative analysis of metal ions insoluble chlorides include silver, mercury (I) and lead (II) only; usually found with the addition of dilute HCl; color of ppt can ID ion 2. acid-insoluble sulfides are found via the addition of hydrogen sulfide: copper (II), bismuth (III), cadmium (II), lead (II), mercury (II), arsenic (III), antimony (III), tin (IV) 3. Base-insoluble sulfides are found by increasing the p. H of the sol’n (appr 8) via addition of (NH 4)2 S: aluminum, chromium (III), iron (III), zinc, nickel (II), cobalt (II), and manganese (II) 1.

Sample Scheme for qualitative analysis of metal ions insoluble phosphates are found by adding (NH 4)2 HPO 4 to the basic sol’n: magnesium, calcium, strontium, and barium 5. the alkali metals and ammonium will not precipitate, but can be tested for individually – typically by flame test color 4.