Jeffrey Mack California State University, Sacramento Chapter 18 Principles of Chemical Reactivity:...

Jeffrey Mack California State University, Sacramento Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

More About Chemical Equilibria: Acid Base & Precipitation Reactions

Stomach Acidity & AcidBase Reactions

In the previous chapter, you examined the behavior of weak acids and bases in terms of equilibrium involving conjugate pairs. The pH of a solution was found via K a or K b. What would happen if you started with a solution of acid that was mixed with a solution of its conjugate base? The change of pH when a significant ammout of conjugate base is present is an example of the Common Ion Effect. The Common Ion Effect

What is the effect on the pH of a 0.25M NH 3 (aq) solution when NH 4 Cl is added? NH 4 + is an ion that is COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the left to reduce the disturbance. This results in a reduciton of the hydroxide ion concentration, which will lower the pH. Hint: NH 4 + is an acid! The Common Ion Effect

First lets find the pH of a 0.25M NH 3 (aq) Solution: [NH 3 ] [NH 4 + ] [OH - ] Initial0.2500 Change x +x Equilibrium 0.025 x xx The Common Ion Effect

First lets find the pH of a 0.25M NH 3 (aq) Solution: Assuming x is

Consider the acetic acid / acetate buffer system. The ability for the acid to consume OH is seen from the reverse of the base hydrolysis: K rev is >> 1, indicating that the reaction is product favored. An hydroxide added will immediately react with the acid so long as it is present. Controlling pH: Buffer Solutions

Consider the acetic acid / acetate buffer system. Similarly, the conjugate base (acetate) is readily capable of consuming H 3 O + K rev is >> 1, indicating that the reaction is product favored. An hydronium ion added will immediately react wit the acid so long as it is present. Controlling pH: Buffer Solutions

Problem: What is the pH of a buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2 ] = 0.600 M? Buffer Solutions

Problem: What is the pH of a buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2 ] = 0.600 M? Since the concentration of acid is greater than the base, equilibrium will move the reaction to the right. Buffer Solutions

Problem: What is the pH of a buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2 ] = 0.600 M? Assuming that x

Problem: 100. mL of a 0.025 M solution of benzoic acid is titrated with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?Solution: At the equivalence point, all of the HBz is converted to Bz - by the strong base. The conjugate base of a weak acid (Bz - ) will hydrolyze to reform the weak acid (K b ). The pH will be > 7 This will yield the [H 3 O + ] and pH.

Problem: 100. mL of a 0.025 M solution of benzoic acid is titrated with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Volume of OH - added to the eq. point: HBz (aq) + OH (aq) Bz (aq) + H 2 O(l) HBz (aq) + OH (aq) Bz (aq) + H 2 O(l) The new total volume of the solution is 125 mL Titration of a Weak Acid with a Strong Base

Problem: 100. mL of a 0.025 M solution of benzoic acid is titrated with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Moles of OH - & Bz - at the eq. point: HBz (aq) + OH (aq) Bz (aq) + H 2 O(l) The concentration of Bz - at the eq. point is: Titration of a Weak Acid with a Strong Base

Problem: 100. mL of a 0.025 M solution of benzoic acid is titrated with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? [OH - ] at the eq. point: [Bz - ][HBz][OH - ] Initial0.02000 Change- x+ x Equilibrium0.020 - xxx Titration of a Weak Acid with a Strong Base

Problem: 100. mL of a 0.025 M solution of benzoic acid is titrated with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? [OH - ] at the eq. point: Assuming that x

Conclusion: Conclusion: At the equivalence point of the titration, unlike the titration of a strong acid and strong base, the pH is > 7. This is due to the production of the conjugate base of a week acid. Equivalence point pH = 8.25 Equivalence point pH = 8.25 Titration of a Weak Acid with a Strong Base

Conclusion: Conclusion: What would the pH equal at the half-way point of the titration? Hint: Only of the moles of weak acid have been converted to its conjugate base! Half-way point pH = ?? Half-way point pH = ?? Titration of a Weak Acid with a Strong Base

Problem: 100. mL of a 0.025 M solution of benzoic acid is titrated with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? At the half-way point, moles of Hbz and Bz - are equal. This is a BUFFER SOLUTION!

Problem: 100. mL of a 0.025 M solution of benzoic acid is titrated with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? At the half-way point, moles of Hbz and Bz - are equal. This is a BUFFER SOLUTION! Titration of a Weak Acid with a Strong Base

Acetic Acid Titrated with NaOH

n equivalence points In the case of a titration of a weak polyprotic acid (H n A) there are n equivalence points. two In the case of the diprotic oxalic acid, (H 2 C 2 O 4 ) there are two equivalence points. Titration of a Weak Polyprotic Acid with a Strong Base

The titration of a polyprotic weak acid follows the same process a monoprotic weak acid. As the acid is titrated, buffering occurs until the last eq. point is reached. The pH is relative to the amounts of conjugate acids and bases. At the second eq. point all of the acid has been converted to A 2-, pH is determined by K b. Titration of a Weak Polyprotic Acid with a Strong Base

In the case of a titration of a weak base, the process follows that of a weak acid in reverse. There exists a region of buffering followed by a rapid drop in pH at the eq. point. Titration of a Weak Base with a Strong Acid

pH Indicators for AcidBase Titrations

An acid/base indicator is a substance that changes color at a specific pH. HInd (acid) has another color than Ind (base) These are usually organic compounds that have conjugated pi-bonds, often they are dyes or compounds that occur in nature such as red cabbage pigment or tannins in tea. end point of the titrationCare must be taken when choosing an appropriate indicator so that the color change (end point of the titration) is close to the steep portion of the titration curve where the equivalence point is found. pH Indicators for AcidBase Titrations

Neutral pH pH7 Natural Indicators: Red Rose Extract in Methanol

Prior to this chapter, exchange reactions which formed ionic salts were governed by the solubility rules. A compound was either soluble, insoluble or slightly soluble. So how do we differentiate between these? The answer lies in equilibrium. It turns out that equilibrium governs the solubility of inorganic salts. Solubility of Salts Lead(II) iodide

K sp The extent of solubility can be measured by the equilibrium process of the salts ion concentrations in solution, K sp. K sp is called the solubility constant for an ionic compound. It is the product of the ions solubilities. For the salt: A x B y (s) xA y+ (aq) + yB x- (aq) K sp = [A y+ ] x [B x- ] y Solubility of Salts

Consider the solubility of a salt MX: If MX is added to water then: Generally speaking, If K sp >> 1 then MX is considered to be soluble If K sp

Jeffrey Mack California State University, Sacramento Chapter 18 Principles of Chemical Reactivity:...

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