# Aqueous Equilibria Bicarbonate buffer system. Aqueous Equilibria Buffers: What are they????? Solutions of a weak acid and its conjugate Or a weak base/conjugate.

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Aqueous Equilibria Bicarbonate buffer system

Aqueous Equilibria Buffers: What are they????? Solutions of a weak acid and its conjugate Or a weak base/conjugate

Aqueous Equilibria Buffers: Why are they important???? They are resistant to a change in pH when small quantities of acid and bases are added

Aqueous Equilibria How do buffers work??? Since a buffer consists of both an weak acid or base and its conjugate, an acid and a base are present in all buffer solutions. If a small amount of strong acid is added to the buffer, there is a base component ready and waiting to neutralize the “invader”.

Aqueous Equilibria Things to remember… HA has A- as its conjugate NaA is not only the salt, but that any soluble salt releases the conjugate Ka × Kb = Kw Any titration question involving a weak A/B is really a buffer problem.

Aqueous Equilibria 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.

Aqueous Equilibria Buffers If acid is added, the F − reacts to form HF and water.

Aqueous Equilibria Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A − ] [HA] K a = HA + H 2 OH 3 O + + A −

Aqueous Equilibria Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] [HA] [A-] [H 3 O + ] = K a base acid

Aqueous Equilibria If acid is added to the buffer, simply add acid to the numerator AND subtract the same quantity from the base since it was self-sacrificing and neutralized the acid. If base is added, simply add the base to the denominator and subtract from the numerator. Add or subtract in moles NOT molarity!!!!

Aqueous Equilibria When equal concentrations (or moles) of Acid and Base are present [which occurs at the ½ equivalence point of a titration] the ratio of acid to base equals ONE and therefore, the pH = pKa.

Aqueous Equilibria Buffer Calculations or: pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.

Aqueous Equilibria Calculating pH Changes in Buffers Ex 1 What is the pH of a buffer solution composed of 0.50 M formic acid and 0.70 M sodium formate before and after adding 10.0 mL of 1.00 M HCl. The Ka of formic acid is 1.8 × 10–4

Aqueous Equilibria Before: H + = K a [acid] [base] pH = (1.8  10 −4 ) (0.50) (0.70) pH H + = 1.286 x 10 -4 [pp pH =3.89

Aqueous Equilibria After: I added ACID!!!New Molarities!!! H + = K a [acid] [base] = (1.8  10 −4 ) ( 0.50L X.50 mole+.01Lx1M/.51 L) ( 0.50L x.70 M -.01Lx1M/.51L ) H + = (1.8  10 −4 ) [.26 moles] [.34 moles] pH H + = 1.38 x 10 -4 [pp pH =3.86

Aqueous Equilibria Calculating pH Changes in Buffers EX 2 A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 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 0.020 mol of NaOH is added. The K a of acetic acid is 1.8x 10 -5

Aqueous Equilibria Before: H + = K a [acid] [base] pH = (1.8  10 −5 ) (0.30) pH pppH =pKa [pp pH =4.74

Aqueous Equilibria After: I added BASE!!!New Molarities!!! H + = K a [acid] [base] = (1.8  10 −5 ) (.30 mole-.2mol) (.30 mole+.2 mol) H + = (1.8  10 −5 ) [.10 moles] [.50 moles] pH H + = 3.6 x 10 -5 [pp pH =4.80 (not much of a change!

Aqueous Equilibria Example 3 A buffered solution contains 0.25 M NH3 ( Kb = 1.8 × 10–5) and 0.40 M NH4Cl. Calculate the pH of this solution. b) Calculate the pH when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered solution.

Aqueous Equilibria pH Range 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 pK a close to the desired pH, the pH range of a buffer is generally within one pH unit of the pK a The greater the concentrations of the conjugate aced/base pair, the greater the buffering capacity

Aqueous Equilibria Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

Aqueous Equilibria Titration A 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.

Aqueous Equilibria

Aqueous Equilibria

Aqueous Equilibria

Aqueous Equilibria

Aqueous Equilibria

Aqueous Equilibria

Aqueous Equilibria Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

Aqueous Equilibria Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

Aqueous Equilibria Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

Aqueous Equilibria Titration of a Weak Acid with a Strong Base 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.

Aqueous Equilibria 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.

Aqueous Equilibria 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.

Aqueous Equilibria Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

Aqueous Equilibria Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.

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