The Common-Ion Effect
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Transcript of The Common-Ion Effect
Prepared byProf. Odyssa Natividad R.M. Molo
Consider a solution that contains not only a weak acid (HC2H3O2) but also a soluble salt (NaC2H3O2) of that acid.
NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)
HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
What happens when salt is added? (Remember Le Chatelier’s Principle)Shifts equilibriumpH of solution increases[H+] decreases
The Common-Ion Effect“The dissociation of a weak electrolyte is
decreased by adding to the solution a strong electrolyte that has an ion common with the weak electrolyte”Plays an important role in determining the pH
of the solution & the solubility of a slightly soluble salt
Despite distinctive name, this is simply a special case of Le Chatelier’s principle
Sample Problem1) What is the
pH of a soln made by adding a 0.30 mol acetic acid & 0.30 mol sodium acetate to enough water to make 1.0L of soln?
STEPS:1)Identify the major species
in the soln & consider their acidity & basicity
2)Identify the important eqlbm rxn
3)Calculate the initial & eqlbm conc of each species that participate in the eqlbm
4)Calculate the pH from the eqlbm conc
Practice Exercise1) Calculate the pH of a solution containing
0.085M nitrous acid (Ka = 4.5 x 10-4) & 0.10M potassium nitrite. What would be its pH if no salt were present.
2) Calculate the pH of a solution containing 0.20 M acetic acid & 0.30M sodium acetate. What would be the pH if no salt were present.
Buffered Solution/BuffersSolution that resist a drastic change in pH
upon addition of small amounts of acid or base
Contain weak conjugate acid-base pairsExample:
Human blood: bicarbonate-carbonic acid buffer Normal pH: 7.35 – 7.45 Acidosis: condition when pH falls below 7.35 Alkalosis: when pH rises above 7.45 Death may result if blood pH < 6.8 or > 7.8
Composition of BuffersA buffer resist in pH because it contains both
an acidic specie to neutralize OH- ions & a basic one to neutralize H+ ions. The acidic & basic species that make up the buffer, however, must not consume each other through a neutralization rxn. These requirements are fulfilled by a weak acid-base conjugate pair.
Composition cont….Buffers are often prepared by mixing a WA or
a WB with a salt of that acid or base.Example:
HC2H3O2 - C2H3O2- buffer (NaC2H3O2 & HC2H3O2)
NH4+ - NH3 buffer (NH4Cl & NH3)
By choosing appropriate components & adjusting their relative concentration, one can buffer a solution at virtually any pH
How Buffer WorksEx: Buffer composed of weak acid (HX) &
its salt (MX)HX(aq) H+(aq) + X-(aq)
If OH- ions are added, it reacts with the acid component of the buffer to produce water & its base component (X-)
OH-(aq) + HX(aq) H2O(l) + X-(aq)Result: [HX] dec & [X-] inc
How Buffer Works cont…If H+ ions are added, it reacts with the
base component of the buffer to produce water & its acid component (HX)
H+(aq) + X-(aq) HX(aq) ORH3O+(aq) + X-(aq) H2O(l) + HX(aq)
Result: [HX] inc & [X-] dec
Buffer pHHX(aq) H+(aq) + X-(aq)
so
pH is determined by 2 factors: (1) value of Ka(2) ratio of conjugate acid-base pair
If [HX] = [X-], [H+] = Ka, pH = pKaResult: select a buffer whose acid form has a
pKa close to desired pH
2 important characteristics of buffer:
(1) its capacity & (2) its pHBuffer capacity
Is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree
Depends on the amount of Acid & Base from which the buffer is made
pH of buffer depends onThe Ka for the acid & on the relative
concentration of A & B that comprise the buffer
To solve buffer pHUse the same procedure to treat the common-ion
effect OR the Henderson-Hasselbalch equation
Sample Problem: Calculate the pH of a buffer that is 0.12M lactic acid (Ka = 1.4 x 10-4) & 0.10M sodium lactate.
Practice: Calculate the pH of a buffer composed of 0.12M benzoic acid, HC7H5O2, (Ka = 6.3 x 10-5) & 0.20M sodium benzoate
Addition of SA/SB to BuffersWhen a SA is added, the H+ is consumed
by X- to produce HX; Result:[HX] inc & [X-] dec
When a SB is added, the OH- is consumed by HX to produce X-
Result: [HX] dec & [X-] inc
Steps to calculate pH1) Identify the neutralization rxn (Strong
Acid & Weak Base or Strong Base & Weak Acid)
2) Analyze and set-up condition before & after neutralization
3) Calculate pH based on what is left during equilibrium condition.
Sample ExerciseA buffer is made by adding 0.300 mol acetic
acid & 0.300 mol sodium acetate to enough water to make 1.00L soln. Calculate its pH (a) at the start; (b) after 0.020 mol KOH is added; (c) after 0.010 mol HCl is added. For comparison, what is the the pH of (d) 0.020 mol NaOH and (e) 0.010 mol HCl added in 1.00L pure water.
Practice Exercise1) What are the effects on the pH of adding (0.0060
mol HCl and (b) 0.0060 mol NaOH to 0.300L of a buffer solution that is 0.250M acetic acid (Ka = 1.8 x 10-5) & 0.560 M sodium acetate. For comparison, what is the pH of (c) 0.0060 mol HCl & (d) 0.0060 mol NaOH in 0.300L water.
2) A 1.00 L volume of buffer is made with concentrations of 0.350 M NaCHO2 (sodium formate) and 0.550M HCHO2 (Ka = 1.8 x 10-4). (a) What is the initial pH? (b) What is the pH after the addition of 0.0050 mol HCl? (assume that the volume remains 1.00L) (c) What would be the pH after the addition of 0.0050 mol NaOH to the original buffer?