Last week, we discussed the Brønsted Lowry concept of acids and bases. According to this model:

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3 Last week, we discussed the Brønsted Lowry concept of acids and bases This model is not limited to aqueous solutions; it can be extended to reactions in the gas phase! According to this model: Acids are proton (H + ) donors Bases are proton (H + ) acceptors

4 Result of this type of acid/base reaction is a conjugate acid-base pair There is an important connection between the strength of an acid and that of its conjugate base A strong acid has an equilibrium that lies far to the right; thus, almost all the original HA is dissociated at equilibrium A strong acid yields a weak conjugate base A weak conjugate base has a low affinity for a proton Conversely, a weak acid has an equilibrium that lies far to the left; thus, most of the HA originally placed in the solution is still present as HA A weak acid yields a strong conjugate base A strong conjugate base has a high affinity for a proton

5 There is also a connection between the strength of a base and that of its conjugate acid A strong base has an equilibrium that lies far to the right; thus, almost all the original BOH is dissociated at equilibrium A strong base yields a weak conjugate acid Conversely, a weak base has an equilibrium that lies far to the left; thus, most of the B originally placed in the solution is still present as B A weak base yields a strong conjugate acid

6 During a Brønsted Lowry acid/base reaction, a salt is formed Recall, a salt is an ionic compound that dissolves and completely dissociates in water into its ions, which move about independently Salts do not always produce neutral solutions when dissolved in water Under certain conditions, these ions behave as acids or bases

7 Look at the salt and ask yourself: Which acid and which base reacted to form it? Is the acid strong or weak? Is the base strong or weak? Embrace the fact that STRONG WINS and is a spectator The remaining ion of the salt reacts with water If you predict a basic solution, YOU MUST WRITE: H 2 O OH If you predict an acidic solution, YOU MUST WRITE: H 2 O H +

8 We just discussed how the conjugate base of a strong acid has virtually no affinity for protons in water Thus, when anions such as Cl - and NO 3 - are placed in water, they do not combine with H+ and have no effect on ph Cations such as K + and Na + from strong bases have no affinity for H +, nor can they produce H +, so they too have no effect on the ph So, salts that consist of the cations of strong bases and the anions of strong acids have no effect on [H + ] when dissolved in water

9 Salts that are formed from the cation of a strong base and the conjugate base (anion) of a weak acid will produce a basic aqueous solution

10 Salts that are formed from the conjugate acid (cation) of a weak base and the anion of a strong acid will produce an acidic aqueous solution

11 #1(a), #1(c)

12 Can predict whether the solution will be basic, acidic, or neutral by comparing K a value for the acidic ion with the K b value for the basic ion K a = K b Neutral solution K a > K b Acidic solution K a < K b Basic solution

13 Type of Salt Examples Comment ph of Solution Cation is from strong base; anion is from strong acid KCl, KNO 3, NaCl, NaNO 3 Neither acts as an acid or a base Neutral Cation is from strong base; anion is conjugate base from weak acid NaC 2 H 3 O 2, KCN, NaF Anion acts as a base; cation has no effect on ph Basic Cation is conjugate acid of weak base; anion is from strong acid NH 4 Cl, NH 4 NO 3 Cation acts as an acid; anion has no effect on ph Acidic Cation is conjugate acid of weak base; anion is conjugate base of weak acid NH 4 C 2 H 3 O 2, NH 4 CN Cation acts as an acid; anion acts as a base Neutral if K a = K b Acidic if K a > K b Basic if K a < K b

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15 The most important application of acid-base solutions containing a common ion is for buffering A buffered solution is one that resists a change in its ph when either hydroxide ions or protons are added HUGE EXAMPLE OF A BUFFERED SOLUTION Our blood! Blood can absorb the acids and bases produced in biologic reactions without changing its ph Vital because cells can survive only in a very narrow ph range! Buffering system in blood involves HCO 3 - and H 2 CO 3

16 Defined as 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 In other words, a mixture of NaC 2 H 3 O 2 and HC 2 H 3 O 2 is less acidic than a solution of HC 2 H 3 O 2 alone CH 3 COONa (s) CH 3 COOH (aq) Na + (aq) + CH 3 COO - (aq) H + (aq) + CH 3 COO - (aq) Common Ion

17 A buffer solution can consist of: A weak acid and its salt HF and NaF OR a weak base and its salt NH 3 and NH 4 Cl How do we calculate the ph of a buffer solution? Solving these types of problems is very similar to the common ion problems from last chapter!

18 All groups work on #18 (a) and #19 (a) Be prepared to share out your solutions with the rest of the class!

19 Remember, buffered solutions resist ph change with addition of a STRONG acid or base In a buffered solution, the ph is governed by the ratio [HA]/[A-] As long as this ratio remains virtually constant, the ph will remain virtually constant

20 For a weak acid buffer (ph < 7) Added strong acid (H + ) reacts with conjugate base Added strong base (OH - ) reacts with weak acid For a weak base buffer (ph > 7): Added strong acid (H + ) reacts with weak base (B) Added strong base (OH - ) reacts with conjugate acid (BH + )

21 The acid dissociation equilibrium expression can be rearranged to provide a useful equation when calculating ph: K a [ H ][ A ] [ H K HA ] [ ] a [ HA] [ A ] HA log[ H ] log Ka log [ ] [ A ] ph = pk a + log A HA = pk a + log base acid This log form of the expression for K a is called the Henderson-Hasselbach equation and is useful for calculating the ph of solutions when the ratio [HA]/[A-] is known

22 So, how do we go about calculating the ph of a buffered solution once a strong acid or base is added? Solution #1 Do a Before Reaction Change After Reaction table using MOLES ONLY for stoichiometry calculations to determine new concentrations Assume reaction with H+/OH- goes to completion Set-up an ICE table with new concentrations to determine equilibrium concentrations Calculate ph from [H+] using K a expression Solution #2 Do a Before Reaction Change After Reaction table for stoichiometry calculations to determine new concentrations Assume reaction with H+/OH- goes to completion Use Henderson-Hasselbach equation to solve for ph

23 All groups work on #18 (c) and #18 (d) Be prepared to share out your solutions with the rest of the class!

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25 Buffer capacity represents the amount of H + or OH - the buffer can absorb with a significant change in ph A good buffer can withstand the addition of relatively large amounts of H + or OH - without changing ph! The most effective buffer occurs when the ratio of [A-] to [HA] or [B] to [BH+] is 1 If this ratio is 1, then according to the Henderson-Hasselbalch equation: ph = pk a

26 Work backwards Choose a weak acid whose pk a is close to the desired ph You can: Calculate [H + ] from ph Substitute this value with K a value of weak acid into the equation: H + = K a [HA] [A ] Choose the ratio [HA]/[A-] that is closest to 1 OR: Substitute pk a (-log K a ) and ph values into the Henderson-Hasselbach Equation This will give a ratio of [A-]/[HA] Then, convert ratio to molar quantities

27 All groups work on #17 (a) and (b) Be prepared to share out your solutions with the rest of the class!

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