Aqueous Equilibria The Common-Ion Effect The Common-Ion Effect • Consider a solution of acetic acid: • NaC 2 H 3 O 2 • Le Châtelier says the equilibrium will shift to the ______. • As a result, what happens to [H + ]? pH? HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 − (aq)
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Aqueous Equilibria The Common-Ion Effect Consider a solution of acetic acid: NaC 2 H 3 O 2 Le Châtelier says the equilibrium will shift to the ______.
Aqueous Equilibria The Common-Ion Effect Calculate the pH and % dissociation of a 0.2M HF solution. K a = 7.2 10 −4 Calculate the pH and % dissociation of a solution containing 0.20 M HF and 0.10 M of NaF Because NaF is ionic, it will dissociate completely, so the initial [F - ] is not 0, but rather 0.10 M
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AqueousEquilibria
The Common-Ion EffectThe Common-Ion Effect• Consider a solution of acetic acid:
• NaC2H3O2
• Le Châtelier says the equilibrium will shift to the ______.
• As a result, what happens to [H+]? pH?
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
AqueousEquilibria
The Common-Ion EffectThe Common-Ion Effect
• Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction
AqueousEquilibria
The Common-Ion EffectThe Common-Ion EffectCalculate the pH and % dissociation of a 0.2M HF solution. Ka = 7.2 10−4
Calculate the pH and % dissociation of a solution containing 0.20 M HF and 0.10 M of NaFBecause NaF is ionic, it will dissociate completely, so the initial [F-] is not 0, but rather 0.10 M
AqueousEquilibria
Buffered SolutionsBuffered Solutions• Solutions of a weak
conjugate acid-base pair.
• They are particularly resistant to pH changes, even when strong acid or base is added.
• After addition of strong acid or base, deal with stoichiometry first, then the equilibrium.
AqueousEquilibria
Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.
AqueousEquilibria
Buffers
If acid is added, the F− reacts to form HF and water.
…it is safe to assume that all of the strong acid or base is consumed in the reaction.
AqueousEquilibria
Buffer example• Calculate the pH of a solution containing
0.5M acetic acid (HC2H3O2, Ka=1.8x10-5) and 0.5M sodium acetate (NaC2H3O2).
AqueousEquilibria
• Calculate the change in pH that occurs when 0.010 mole of NaOH is added to 1.0 L of the previous solution. (0.5M acetic acid (HC2H3O2, Ka=1.8x10-5) and 0.5M sodium acetate (NaC2H3O2))
Step 1 – use stoichiometry to calculate how much acetic acid remains in solution
Step 2 – calculate [H+] and pH
AqueousEquilibria
Buffer Capacity• Amount of acid/base that can be neutralized before pH changes
significantly• A buffer with large capacity contains large concentrations of the buffering
• The pH range is the range of pH values over which a buffer system works effectively.
• It is best to choose an acid with a pKa close to the desired pH.
AqueousEquilibria
Calculating pH Changes in Buffers
A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after:
a)0.020 mol of NaOH is added.b)0.020 mol of HCl is added.
AqueousEquilibria
Calculating pH Changes in BuffersThe 0.020 mol NaOH will react with 0.020 mol of the acetic acid:
HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)
HC2H3O2 C2H3O2− OH−
Before reaction 0.300 mol 0.300 mol 0.020 mol
After reaction 0.280 mol 0.320 mol 0.000 mol
AqueousEquilibria
Calculating pH Changes in BuffersNow use the Henderson–Hasselbalch equation to calculate the new pH:
pH = 4.74 + log (0.320)(0. 280)
pH = 4.74 + 0.06
pH = 4.80
AqueousEquilibria
Calculating pH Changes in BuffersThe 0.020 mol HCl will react with 0.020 mol of the acetic ion:
C2H3O2 −(aq) + H3O+(aq) HC2H3O2(aq) + H2O(l)
C2H3O2− H+ HC2H3O2
Before reaction 0.300 mol 0.020 mol 0.300 mol
After reaction 0.280 mol 0.000 mol 0.320 mol
AqueousEquilibria
Calculating pH Changes in BuffersNow use the Henderson–Hasselbalch equation to calculate the new pH:
pH = 4.74 + log (0.280)(0. 320)
pH = 4.74 - 0.058
pH = 4.62
AqueousEquilibria
Acid-Base TitrationsAcid-Base Titrations
A known concentration of base (or acid) is slowly added to a solution of acid (or base).
(titrant)
(analyte)
AqueousEquilibria
TitrationA pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.
AqueousEquilibria
Titration Curves• Titrant volume vs. pH• 4 regions:Initial pHBefore equivalence pointEquivalence pointAfter equivalence point
AqueousEquilibria
Titration CurveStrong base added to strong acid
From the start of the titration to near the equivalence point, the pH goes up slowly.
AqueousEquilibria
Titration CurveStrong base added to strong acid
Just before (and after) the equivalence point, the pH increases rapidly.
AqueousEquilibria
Titration CurveStrong base added to strong acid
At eq. point, moles acid = moles base. Solution contains only water & salt from the cation of the base and the anion of the acid.
AqueousEquilibria
Titration CurveStrong base added to strong acid
As more base is added, the increase in pH again levels off.
AqueousEquilibria
Titration CurveStrong base added to weak acid
• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
• The pH at the equivalence point will be >7.
• Phenolphthalein is commonly used as an indicator in these titrations.
AqueousEquilibria
Titration of a Weak Acid with a Strong Base
At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.
AqueousEquilibria
Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
AqueousEquilibria
Weak base & strong
acid
AqueousEquilibria
Titrations of Polyprotic Acids
In these cases there is an equivalence point for each dissociation.
AqueousEquilibria
• Marks the end point of a titration by changing color.
• The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible).