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

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 & Acid–Base 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 let’s 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 let’s find the pH of a 0.25M NH 3 (aq) Solution: Assuming x is << 0.25, x = [OH  ] = pOH = 2.67 pH = 14.00  2.67 = 11.33 for 0.25 M NH 3 The Common Ion Effect

What is the pH of a solution made by adding equal volumes of 0.25M NH 3 (aq) and 0.10M NH 4 Cl(aq)? Since the solutions are mixed with one another, the effect of dilution is cancelled out. One can use the initial concentrations of each species in the reaction without calculating new molarities. This also works with mole ratios. The Common Ion Effect

What is the pH of a solution made by adding equal volumes of 0.25 M NH 3 (aq) and 0.10 M NH 4 Cl(aq)? Since there is more ammonia than ammonium present, the RXN will proceed to the right. [NH 3 ] [NH 4 + ] [OH - ] Initial0.250.100 Change  x + x Equilibrium 0.025  x 0.10 + xx The Common Ion Effect

What is the pH of a solution made by adding equal volumes of 0.25 M NH 3 (aq) and 0.10 M NH 4 Cl(aq)? [NH 3 ] [NH 4 + ] [OH - ] Initial0.250.100 Change  x + x Equilibrium 0.025  x 0.10 + xx The Common Ion Effect

What is the pH of a solution made by adding equal volumes of 0.25 M NH 3 (aq) and 0.10 M NH 4 Cl(aq)? Assuming x is << 0.25 or 0.10 pOH = 4.35pH = 9.65 The pH drops from11.33 due to the common ion! The Common Ion Effect

HCl is added to pure water. HCl is added to a solution of a weak acid H 2 PO 4 - and its conjugate base HPO 4 2-. Controlling pH: Buffer Solutions

A “Buffer Solution” is an example of the common ion effect. From an acid/base standpoint, buffers are solutions that resist changes to pH. A buffer solution requires two components that do not react with one another: 1.An acid capable of consuming OH  2.The acid’s conjugate base capable of consuming H 3 O + Controlling pH: Buffer Solutions

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 << 0.700 and 0.600, we find: [CH 3 CO 2 H][CH 3 CO 2  ][H 3 O + ] Initial0.7000.6000 Change  x + x Equilibrium 0.700  x 0.600 + xx 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 [H 3 O + ] = 2.1  10  5 pH = 4.68

The expression for calculating the [H + ] of the buffer reduces to: The H 3 O + concentration depends only K a and the ratio of acid to base. Buffer Solutions

Similarly for a basic solution the [OH  ] of the buffer reduces to: The OH  concentration depends only K b and the ratio of base to acid. Similarly for a basic solution the [OH  ] of the buffer reduces to: The OH  concentration depends only K b and the ratio of base to acid. Buffer Solutions

The result is known as the “Henderson-Hasselbalch” equation. The pH of a buffer can be adjusted by manipulating the ratio of acid to base. Buffer Solutions: The Henderson- Hasselbalch Equation

24 What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = 3.77 + log [0.52] [0.30] = 4.01 Mixture of weak acid and conjugate base!

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? 5.0  10  5 MA pH of 4.30 corresponds to an [H 3 O + ] = 5.0  10  5 M Preparing a Buffer

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? 5.0  10  5 MA pH of 4.30 corresponds to an [H 3 O + ] = 5.0  10  5 M First choose an acid with a pK a close to the desired pH

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? 5.0  10  5 MA pH of 4.30 corresponds to an [H 3 O + ] = 5.0  10  5 M First choose an acid with a pK a close to the desired pH Next adjust the ratio of acid to conjugate base to achieve the desired pH. Preparing a Buffer

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? 5.0  10  5 MA pH of 4.30 corresponds to an [H 3 O + ] = 5.0  10  5 M First choose an acid with a pK a close to the desired pH Acetic acid is the best choice. Possible AcidsKaKa 1.2  10  2 1.8  10  5 4.0  10  10 Preparing a Buffer

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? 5.0  10  5 MA pH of 4.30 corresponds to an [H 3 O + ] = 5.0  10  5 M Preparing a Buffer

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? 5.0  10  5 MA pH of 4.30 corresponds to an [H 3 O + ] = 5.0  10  5 M Therefore, if you start with 0.100 mol of acetate ion then add 0.278 mol of acetic acid, will result in a solution with a pH of 4.30. Preparing a Buffer

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? Therefore, if you start with 0.100 mol of acetate ion then add 0.278 mol of acetic acid, will result in a solution with a pH of 4.30. Since both species are in the same solution (the same volume), the mole ratios are equal to the concentration ratios!

Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? So, by adding 8.20 g of sodium acetate to 2780 ml of a 0.100 M acetic acid solution, one would make a buffer of pH 4.30. Preparing a Buffer

Buffer prepared from 8.4 g NaHCO 3 weak acid 16.0 g Na 2 CO 3 conjugate base What is the pH? Preparing a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to: a)1.00 L of pure water b)1.00 L of buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2  ] = 0.600 M (pH = 4.68) Adding an Acid to a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to: a)1.00 L of pure water mL  L  mols H 3 O +  M(H 3 O + )  pH Adding an Acid to a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to: b)1.00 L of buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2  ] = 0.600 M (pH = 4.68) 1.The acid added will immediately be consumed by the acetate ion. 2.This in turn increases the acetic acid concentration. 3.The acetic acid then will react with water to reestablish equilibrium.

What is the pH when 1.00 mL of 1.00 M HCl is added to: b)1.00 L of buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2  ] = 0.600 M (pH = 4.68) 1.The acid added will immediately be consumed by the acetate ion. Adding an Acid to a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to b)1.00 L of buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2  ] = 0.600 M (pH = 4.68) [H 3 O + ][CH 3 CO 2  ][CH 3 CO 2 H] Before0.001000.6000.700 Change  0.00100 + 0.00100 After00.5990.701 Adding an Acid to a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to: b)1.00 L of buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2  ] = 0.600 M (pH = 4.68) 2.This in turn increases the acetic acid concentration. 3.The acetic acid then will react with water to reestablish equilibrium. Adding an Acid to a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to b)1.00 L of buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2  ] = 0.600 M (pH = 4.68) [CH 3 CO 2 H][H 3 O + ][CH 3 CO 2  ] Initial0.70100.599 Change  x + x Equilibrium 0.700  x x0.599 + x Adding an Acid to a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to b)1.00 L of buffer that has [CH 3 CO 2 H] = 0.700 M and [CH 3 CO 2  ] = 0.600 M (pH = 4.68) The pH does not change! The solution absorbs the added acid. It is a buffer! Adding an Acid to a Buffer

The solid acid and conjugate base in the packet are mixed with water to give the specified pH. Note that the quantity of water does not affect the pH of the buffer. Commercial Buffers

43 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is its conjugate acid buffer solution

46 = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) 0.0290.001 0.024 0.0280.00.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH 3 ] = 0.025 0.10

Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly. Acid–Base Titrations

Additional NaOH is added. pH rises as equivalence point is approached. Acid–Base Titrations

Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point. Acid–Base Titrations

Titration of a Strong Acid with a Strong Base

The reaction of a strong acid and strong base produces a salt and water. The Net Ionic Equation is: At the equivalence point, the moles of base added equal the moles of acid titrated. Titration of a Strong 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? Titration of a Weak Acid with a Strong Base

Equivalence Point (moles base = moles acid) At the equivalence point, all of the acid is converted to its conjugate base. The conjugate base will then react with water to reestablish equilibrium. The pH can be determined from K b. Titration of a Weak Acid with a Strong Base

HBz (aq) + OH  (aq)  Bz  (aq) + H 2 O(l) C 6 H 5 CO 2 H = HBzBenzoate ion = Bz - 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?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 << 0.020, Titration of a Weak Acid with a Strong Base

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 ½ 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 Acid–Base 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 Acid–Base Titrations

Neutral pH pH<<7pH >7 Buffer <7 pH >>7 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 salt’s ion concentrations in solution, K sp. K sp is called the solubility constant for an ionic compound. It is the product of the ion’s 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 << 1 then MX is considered to be insoluble If K sp  1then MX is slightly soluble Solubility of Salts

INSOLUBLE.All salts formed in this experiment are said to be INSOLUBLE..They form when mixing moderately concentrated solutions of the metal ion with chloride ions. Analysis of Silver Group

Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) e Ag + (aq)  Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED. Analysis of Silver Group

AgCl(s)  Ag + (aq) + Cl - (aq) When solution is SATURATED, expt. shows that [Ag + ] = 1.67 x 10 -5 M. This is equivalent to the SOLUBILITY of AgCl. What is [Cl - ]? [Cl - ] = [Ag + ] = 1.67 x 10 -5 M Analysis of Silver Group

AgCl(s)  Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = 1.67  10 -5 M Use this to calculate K c K c = [Ag + ] [Cl - ] = (1.67  10 -5 )(1.67  10 -5 ) = 2.79  10 -10 Analysis of Silver Group

AgCl(s)  Ag + (aq) + Cl - (aq)  K c = [Ag + ] [Cl - ] = 2.79  10 -10 Because this is the product of “solubilities”, we call it: K sp = solubility product constant Analysis of Silver Group

PbCl 2 (s)  Pb 2+ (aq) + 2 Cl - (aq)  K sp = 1.9  10 -5 = [Pb 2+ ][Cl – ] 2 Lead(II) Chloride

Problem: The solubility of lead (II) iodide is found to be 0.00130M. What is the K sp for PbI 2 ? Relating Solubility & K sp

Problem: The solubility of lead (II) iodide is found to be 0.00130M. What is the K sp for PbI 2 ? Recall that lead (II) iodide dissociates via: Relating Solubility & K sp

Problem: The solubility of lead (II) iodide is found to be 0.00130M. What is the K sp for PbI 2 ? From the reaction stoichiometry, [Pb 2+ ] = 0.00130M & [I  ] = 2  0.00130M = 0.00260M Relating Solubility & K sp

Problem: The solubility of lead (II) iodide is found to be 0.00130M. What is the K sp for PbI 2 ? Entering these values into the K sp expression, For PbI 2, K sp = 4  (solubility) 3 Relating Solubility & K sp

K sp for MgF 2 = 5.2  10  11. Calculate the solubility in: (a) moles/L & (b) in g/L Relating Solubility & K sp

K sp for MgF 2 = 5.2  10  11. Calculate the solubility in: (a) moles/L & (b) in g/L (a) The solubility of the salt is governed by equilibrium so let’s set up an ICE table: Relating Solubility & K sp [MgF 2 (s)][Mg 2+ ][F  ] Initial-00 Change-+ x+ 2x Equilibrium-x2x

K sp for MgF 2 = 5.2  10  11. Calculate the solubility in: (a) moles/L & (b) in g/L (a) Entering the values into the K sp expression: The solubility of MgF 2 = 2.4  10  4 mols/L Relating Solubility & K sp

K sp for MgF 2 = 5.2  10  11. Calculate the solubility in: (a) moles/L & (b) in g/L (b) the solubility of the salt in g/L is found using the formula weight: Relating Solubility & K sp

What is the maximum [Cl  ] in solution with 0.010 M Hg 2 2+ without forming Hg 2 Cl 2 (s)? Relating Solubility & K sp

What is the maximum [Cl  ] in solution with 0.010 M Hg 2 2+ without forming Hg 2 Cl 2 (s)? Precipitation will initiate when the product of the concentrations exceeds the K sp. Relating Solubility & K sp

What is the maximum [Cl  ] in solution with 0.010 M Hg 2 2+ without forming Hg 2 Cl 2 (s)? The maximum chloride concentration can be found from the K sp expression. Relating Solubility & K sp

Adding an ion “common” to an equilibrium causes the equilibrium to shift towards reactants according to Le Chatelier’s principle. Solubility & the Common Ion Effect

Common Ion Effect

What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2 ? K sp for BaSO 4 = 1.1  10  10 Solubility & the Common Ion Effect

What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2 ? K sp for BaSO 4 = 1.1  10  10 (a) Solubility & the Common Ion Effect

What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2 ? K sp for BaSO 4 = 1.1  10  10 (b) Solubility & the Common Ion Effect [BaSO 4 (s)][Ba 2+ ][SO 4 2  ] Initial-0.0100 Change-+ x Equilibrium-0.010 + xx

What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2 ? K sp for BaSO 4 = 1.1  10  10 (b) Since x << 0.010, Solubility & the Common Ion Effect

What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2 ? K sp for BaSO 4 = 1.1  10  10 (b) Since x << 0.010, Solubility & the Common Ion Effect Solubility is significantly decreased by the presence of a common ion.

Some anions of precipitates are the conjugate bases of weak acids. In solution these anions can hydrolyze to form the weak acid. Addition of additional base can shift equilibrium to the right thus reducing the concentration of X . This in turn would increase the solubility of the ppt. Effect of Basic Salts on Solubility

Decrease the concentration of HX Add a base: Effect of Basic Salts on Solubility

Eq. Shifts to the right Effect of Basic Salts on Solubility

Eq. Shifts to the right The concentration of X  drops Effect of Basic Salts on Solubility

Eq. Shifts to the right The solubility of MX increases! Effect of Basic Salts on Solubility

Relating Q to K sp Q = K sp The solution is saturated Q < K sp The solution is not saturated Q = K sp The solution is over saturated, precipitation will occur K sp & the Reaction Quotient

Suppose you have a solution that is 1.5  10  6 M Mg 2+. Enough NaOH(s) is added to give a [OH  ] = 1.0  10  6 M. Will a precipitate form? K sp & the Reaction Quotient

Suppose you have a solution that is 1.5  10  6 M Mg 2+. Enough NaOH(s) is added to give a [OH  ] = 1.0  10  6 M. Will a precipitate form? Solution:Calculate “Q” and compare to K sp.

1.5  10  6 M Mg 2+ &[OH  ] = 1.0  10  4 M. Solution:Calculate “Q” and compare to K sp. K sp & the Reaction Quotient

1.5  10  6 M Mg 2+ &[OH  ] = 1.0  10  4 M. Solution:Calculate “Q” and compare to K sp. Since Q < K sp, no precipitation will occur. K sp & the Reaction Quotient

What concentration of hydroxide ion will precipitate a 1.5  10  6 M Mg 2+ solution. K sp & the Reaction Quotient

complex ions Metal ions exist in solution as complex ions. ligands A complex ion involves a metal ion bound to molecules or ions called “ligands”. Ligands are Lewis bases that form “coordinate covalent bonds” with the metal. Examples are: Equilibria & Complex Ions

The equilibrium constants for complex ion are very product favored: K f is known as the formation equilibrium constant. Equilibria & Complex Ions

The extent of dissociation of a complex ions is given by the “dissociation constant”, K D Equilibria & Complex Ions

The presence of a ligand dramatically affect the solubility of a precipitate: Dissolving Precipitates by forming Complex Ions

The presence of a ligand dramatically affect the solubility of a precipitate: Solubility of Complex Ions

What is the solubility of AgCl(s) in grams per liter in a 0.010M NH 3 (aq) solution: Solubility of Complex Ions

What is the solubility of AgCl(s) in grams per liter in a 0.010M NH 3 (aq) solution: [NH 3 ][[Ag(NH 3 ) 2 ] + ][Cl  ] Initial0.01000 Change- 2x+ x Equilibrium0.010 – 2xxx Solubility of Complex Ions

What is the solubility of AgCl(s) in grams per liter in a 0.010M NH 3 (aq) solution: Solving using the quadratic equation: X = 0.0032M = [Cl  ] Solubility of Complex Ions

What is the solubility of AgCl(s) in grams per liter in a 0.010M NH 3 (aq) solution: In pure water, the solubility of AgCl is only 0.0019 g/L Solubility of Complex Ions

K sp Values AgCl1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 Separating Metal Ions Cu 2+, Ag +, Pb 2+

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

Similar presentations

Presentation on theme: "Jeffrey Mack California State University, Sacramento Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

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

Similar presentations

Presentation on theme: "Jeffrey Mack California State University, Sacramento Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria."— Presentation transcript:

Similar presentations

About project

Feedback