 # Aqueous Equilibria Chapter 17 HW problems: 3, 5, 14, 15, 16, 23, 24, 27a, 28a, 31, 37, 43, 45, 51, 57.

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Aqueous Equilibria Chapter 17 HW problems: 3, 5, 14, 15, 16, 23, 24, 27a, 28a, 31, 37, 43, 45, 51, 57

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion

A buffer solution is a solution of: 1.A weak acid or a weak base and 2.The salt of the weak acid or weak base Both must be present! (and should NOT neutralize EACH OTHER!!!!) A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa Adding more acid creates a shift left IF enough acetate ions are present

Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HCl is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Mixture of weak acid and conjugate base! K a for HCOOH = 1.8 x 10 -4 [H + ] [HCOO - ] K a = [HCOOH] x = 1.038 X 10 -4 pH = 3.98

OR…… Use the Henderson-Hasselbach equation Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbach equation pH = pK a + log [conjugate base] [acid]

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = 3.74 + log [0.52] [0.30] = 3.98 Mixture of weak acid and conjugate base!

HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl -

pH= 9.18 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Initial End 0.30 0.360 0.30 - x0.36 + xx [NH 4 + ] [OH - ] [NH 3 ] K b = Change- x + x = 1.8 X 10 -5 (.36 + x)(x) (.30 – x) 1.8 X 10 -5 = 1.8 X 10 -5  0.36x 0.30 x = 1.5 X 10 -5 pOH = 4.82

pH = 9.20 What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (M) end (M) 0.29 0.01 0.24 0.280.00.25 final volume = 80.0 mL + 20.0 mL = 100 mL NH 4 + 0.36 M x 0.080 L = 0.029 mol /.1 L = 0.29 M OH - 0.050 x 0.020 L = 0.001 mol /.1 L = 0.01M NH 3 0.30 M x 0.080 = 0.024 mol /.1 L = 0.24M K a = = 5.6 X 10 -10 [H + ] [NH 3 ] [NH 4 + ] = 5.6 X 10 -10 [H + ] 0.25 0.28 [H + ] = 6.27 X 10 -10 Equilibrium Calculation NH 4 + (aq) H + (aq) + NH 3 (aq)

= 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (M) end (M) 0.29 0.01 0.24 0.280.00.25 pH = 9.25 + log [0.25] [0.28] final volume = 80.0 mL + 20.0 mL = 100 mL

Chemistry In Action: Maintaining the pH of Blood

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at the endpoint (hopefully close to the equivalence point) Slowly add base to unknown acid UNTIL The indicator changes color (pink)

Titration Curve A titration curve is a plot of pH vs. the amount of titrant added. Typically the titrant is a strong (completely) dissociated acid or base. Such curves are useful for determining endpoints and dissociation constants of weak acids or bases.

Features of the Strong Acid-Strong Base Titration Curve 1.The pH starts out low, reflecting the high [H 3 O + ] of the strong acid and increases gradually as acid is neutralized by the added base. 2.Suddenly the pH rises steeply. This occurs in the immediate vicinity of the equivalence point. For this type of titration the pH is 7.0 at the equivalence point. 3.Beyond this steep portion, the pH increases slowly as more base is added.

Sample Calculation: Strong Acid-Strong Base Titration Curve Consider the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH. Region 1. Before the equivalence point, after adding 20.0 mL of 0.100 M NaOH. (Half way to the equivalence point.) Initial moles of H 3 O + = 0.0400 L x 0.100 M = 0.00400 mol H 3 O + - Moles of OH - added = 0.0200 L x 0.100 M =0.00200 mol OH - pH=1.477

Sample Calculation: Strong Acid-Strong Base Titration Curve (Cont. I) Region 2. At the equivalence point, after adding 40.0 mL of 0.100 M NaOH. Initial moles of H 3 O + = 0.0400 L x 0.100 M = 0.00400 mol H 3 O + - Moles of OH - added = 0.0400 L x 0.100 M =0.00400 mol OH - An equal number of moles of NaOH and HCl have reacted, leaving only a solution of their salt (NaCl). The pH=7.00 because the cation of the strong base and the anion of the strong acid have no effect on pH.

Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. II) Region 3. After the equivalence point, after adding 50.0 mL of 0.100 M NaOH. (Now calculate excess OH - ) Total moles of OH - = 0.0500 L x 0.100 M = 0.00500 mol OH - -Moles of H 3 O + consumed = 0.0400 L x 0.100 M =0.00400 mol pOH=1.954 pH=12.046

Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7):

The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve 1.The initial pH is higher. 2.A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point. 3.The pH at the equivalence point is greater than 7.00. 4.The steep rise interval is less pronounced.

Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): H + (aq) + NH 3 (aq) NH 4 Cl (aq)

The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak Base-Strong Acid Titration Curve 1.The initial pH is above 7.00. 2.A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point. 3.The pH at the equivalence point is less than 7.00. 4.Thereafter, the pH decreases slowly as excess strong acid is added.

Features of the Titration of a Polyprotic Acid with a Strong Base 1.The loss of each mole of H + shows up as separate equivalence point (but only if the two pK a s are separated by more than 3 pK units). 2.The pH at the midpoint of the buffer region is equal to the pK a of that acid species. 3.The same volume of added base is required to remove each mole of H +.

Acid-Base Indicators

The titration curve of a strong acid with a strong base. 16.5

Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein

Finding the Equivalence Point (calculation method) Strong Acid vs. Strong Base –100 % ionized! pH = 7 No equilibrium! Weak Acid vs. Strong Base –Acid is neutralized; Need K b for conjugate base equilibrium Strong Acid vs. Weak Base –Base is neutralized; Need K a for conjugate acid equilibrium Weak Acid vs. Weak Base –Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7

Exactly 100 mL of 0.10 M HNO 2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 Equilibrium Calculation NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

Solubility Equilibria Looking at dissolution or precipitation reactions of ionic compounds K sp (solubility product) is the equilibrium constant for the equilibrium reaction between a solid and its saturated ions – Indicates how soluble the solid is when in water Solids, liquids and solvents do not appear in the equilibrium equation (just like before)

Solubility and K sp Solubility is how much actually dissolves while K sp is the equilibrium constant. – pH, temperature, and the presence of other ions in solution affect solubility – Only temperature affects K sp Can use K sp to calculate solubility but need to be careful…not always correct!

Factors that Affect Solubility (4) 1.Common Ion Effect presence of other ions reduces solubility presence of 2 nd solute that gives a common ion reduces solubility 2.pH dissolving increases when pH is made more acidic (more of the basic ion X - will be made to react with extra H + ions) the more basic the anion, the more its solubility is affected by pH 3.Formation of a Complex Ion 4.Amphoterism

Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- Co(H 2 O) 6 2+ CoCl 4 2- Where they “smush “ it all together Metal ions can act as Lewis acids Can react with water Can react with other Lewis bases (must be able interact more than water can) K eq becomes K f (formation constant) K f ’s size shows stability of complex ion in solution

Complex Ion Formation These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH 3 ) 4 2+ (ammonia is used as a test for Cu 2+ ions), and Ag(NH 3 ) 2 +. Memorize the common ligands.

Common Ligands LigandsNames used in the ion H2OH2Oaqua NH 3 ammine OH-hydroxy Cl-chloro Br-bromo CN-cyano SCN-thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

Names Names: ligand first, then cation Examples: –tetraamminecopper(II) ion: Cu(NH 3 ) 4 2+ –diamminesilver(I) ion: Ag(NH 3 ) 2 +. –tetrahydroxyzinc(II) ion: Zn(OH) 4 2- The charge is the sum of the parts (2+) + 4(-1)= -2.

When Complexes Form Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H 2 O) 6 3+ Transitional metals, such as Iron, Zinc and Chromium, can form complex ions. The odd complex ion, FeSCN 2+, shows up once in a while Acid-base reactions may change NH 3 into NH 4 + (or vice versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu 2+ + a little NH 4 OH will form the light blue precipitate, Cu(OH) 2. With excess ammonia, the complex, Cu(NH 3 ) 4 2+, forms.

Factors that Affect Solubility (cont.) 4.Amphoterism Some insoluble, metal hydroxides will dissolve in strong acids or strong bases -will dissolve in acid because of basic anion -Will dissolve in base because a complex ion forms Solubility varies with the metal bonded

Precipitation & Separation of Ions Q > K sp get precipitation until Q = K sp Q = K sp at equilibrium Q < K sp solid dissolves until Q = K sp Selective Precipitation of ions – Separate ions in an aqueous solution using a reagent that forms a precipitate 1 or few ions in solution – Ex. Ag + + Cu 2+ + HCl AgCl + Cu 2+ + H +

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