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Applications of aqueous equilibria Neutralization Common-Ion effect Buffers Titration curves Solubility and K sp

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Learning objectives Determine extent of neutralization Construct buffer solutions Derive Henderson-Hasselbalch equation Apply HH to calculate pH of buffer solutions Calculate pH titration curves Write solubility product expressions Identify factors that affect solubility

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Neutralization Acid + Base → Salt + Water Extent → depends on type of acid and base What are the major species left when base neutralizes an acid? Four combinations: Strong-strong Strong-strong Strong-weak Strong-weak Weak-strong Weak-strong Weak-weak Weak-weak

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Strong-strong Net ionic equation K n = 1/K w = 10 14 Neutralization goes to completion Na +, Cl - and H 2 O

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Manipulating equilibria expressions If reaction A can be written as the sum of reactions B + C + D +...+ N (ΣR i ) Then K is product over ΠK i This approach is used routinely in solution equilibria problems

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Weak-strong Acetic acid is not completely ionized Net ionic equation: What is K? Write K in terms equilbria we know: K = 1.8x10 -5 x1.0x10 14 = 1.8x10 9 Goes to completion – attraction of OH - for protons Na +, Ac -, H 2 O (v. small amount OH - )

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Strong-weak Net ionic equation for neutralization of ammonia with HCl What is K? Write in terms of known equilibria: Add 1 and 2: K 1+2 = K 1 K 2 = 1.8x10 -5 x1.0x10 14 = 1.8x10 9 K 1+2 >>1 - Neutralization complete NH 4 +, Cl -, H 2 O and v. small amount of H 3 O +

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Weak-weak Net ionic equation for neutralization of acetic acid by ammonia – neither ionized Obtain K n from equilibria we know K 1+2+3 = K 1 K 2 K 3 = 1.8x10 -5 x1.8x10 -5 x1.0x10 14 = 3.2x10 4 NH 4 +, Ac -, H 2 O (v. small amounts of HAc and NH 3 )

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Common ion effect - buffering Solutions of weak acid and conjugate bases have important applications for “buffering” pH – resisting change to pH from added acid or base. (weak base and conjugate acid perform the same function) The pH of operation will depend on the dissociation constants for the particular acid (base).

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Use weak acid strategy for calculating pH in buffer Acetic acid and sodium acetate 0.1 M. Initial species are HAc, Na +, Ac - and H 2 O Two proton exchange reactions, but one does not alter concentrations

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The Big Table of concentrations Determination of final concentrations in terms of initial concentrations Note difference between acid case and buffer case: [Ac - ] i is 0 with HAc only [Ac - ] i is 0 with HAc only [Ac - ] i is > 0 in the buffer [Ac - ] i is > 0 in the buffer Principal process HAc(aq) + H 2 O = H 3 O + (aq) + Ac - (aq) Initial conc 0.1000.10 Change-xxx Equilibrium conc 0.10 - x x 0.10 + x

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Solving for x Put concentrations into expression for K a 0.10 – x ≈ 0.10 ≈ 0.10 + x x = K a = 1.8 x10 -5 M (life is good) pH = 4.74 Note: when [HAc] = [Ac - ], pH = pKa In all buffers pH = pK a when [HB] = [B - ] X << 0.1

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Common-ion effect – Le Chatelier in action Without added acetate ion the pH of 0.10 M acetic acid is 2.89. Addition of Ac - causes [H 3 O + ] to decrease Consequence of Le Chatelier: increasing [product] equilibrium goes to reactants pH changes in different buffers

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Modulating pH in acetic acid by adding acetate ion

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Do buffers really work? Compare effect of adding OH - to solution of given pH 1. Buffer solution 2. Strong acid at same initial pH

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Adding base to buffer Analytical proof in HAc/Ac - system Initial pH = 4.74 What happens when 0.01 mol of NaOH is added to I L Principal process HAc(aq) + H 2 O = H 3 O + (aq) + Ac - (aq) Equilibrium conc/M 0.10 - x x 0.10 + x

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What happens when 0.01 mol of NaOH is added? Addition of OH - causes conversion of HAc into Ac - Neutralization process HAc(aq) + OH - = H 2 O (aq) + Ac - (aq) Initial conc 0.100.010.10 Change-0.01-0.010.01 Equilibrium conc 0.0900.11

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After addition of OH -, recompute [H 3 O + ] [H 3 O + ] = 1.8x10 -5 x0.09/0.11 = 1.5x10 -5 M pH = 4.82 – a change of + 0.08 pH units

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Same deal when adding acid Almost all the added H + is converted into HAc In same way as with base, new [HAc] = 0.11 M and new [Ac - ] = 0.09 M [H 3 O + ] = 1.8x10 -5 x0.11/0.09 = 2.2 x 10 -5 M pH = 4.66 – a change of -0.08 pH units

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Adding OH - to strong acid at pH = 4.74 What is change in pH if added to a solution of HCl at pH 4.74? [H + ] = 1.8 x 10 -5 M HCl removes 1.8x10 -5 mol of OH -. Excess OH - = 0.01 - 1.8x10 -5 mol of OH - ≈ 0.01 mol. New pH = 12 – a change of 7 pH units In buffer at same initial pH, change was only 0.08 units In buffer at same initial pH, change was only 0.08 units

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Buffer capacity A measure of the amount of acid or base that a buffer solution can absorb before a significant change in pH occurs. Or – the amount of acid or base required to yield a given change in pH. Messin’ with buffers

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Henderson-Hasselbalch: Send the pain below Buffer calculation short-cut Derivation Assumes that [H + ] << [HA]

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Applications Able to predict percent dissociation of an acid from the difference between pH and pK a Calculate relative quantities of acid and conjugate base required to achieve a given pH (pK a must be reasonably close to pH)

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The glass half-neutered Instead of using a source of weak acid and a conjugate base, take a weak acid and neutralize with strong base to make the conjugate base. (HAc and NaOH). When half the HAc is neutralized [HAc] = [Ac - ] pKa = pH Useful buffers can be made in pH range of pKa ± 1 pH units

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To calculate the new pH, use the Henderson- Hasselbalch equation: 1141.

To calculate the new pH, use the Henderson- Hasselbalch equation: 1141.

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