 # Aqueous Equilibria. The __________________________ is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved.

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Aqueous Equilibria

The __________________________ 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. and 2. Both must be present! (and should NOT neutralize EACH OTHER!!!!) A buffer solution has the ability to ________________ 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) (b) (c)

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) Mixture of weak acid and conjugate base!

OR…… Use the Henderson-Hasselbalch 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-Hasselbalch 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) Common ion effect pH = pK a + log [HCOO - ] [HCOOH] 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

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 + ] 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 __________________ 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. ________________ – 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 _________________ ____________________. 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 _________, 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 ________________. This occurs in the immediate vicinity of the ____________________. For this type of titration the pH is _____ at the equivalence point. 3.Beyond this steep portion, the pH ______________ 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 -

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

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 ___________. 2.A gradually rising portion of the curve, called the ____________, appears before the steep rise to the equivalence point. 3.The pH at the equivalence point is _____ ___________________ 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 ______________. 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 _______ ____________. 4.Thereafter, the pH _______________ 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 _______ of that acid species. 3.The same volume of added base is required to remove each mole of H +.

Acid-Base Indicators

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) 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 - ]

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 ________, _______ and ________ 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. – ____, __________, and the presence of ________ ________________in solution affect solubility – Only ________________ 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 ________________ presence of 2 nd solute that gives a common ion ___________________ 2.pH dissolving increases when pH is made __________ (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 ________________________ 3.Formation of a Complex Ion 4.Amphoterism

Complex Ion Equilibria and Solubility A complex ion is an ion containing a central ____________ bonded to ________________________________. 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 _________ Can react with water (a Lewis base) 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 __________________ -Will dissolve in base because a __________________ 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

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