Ionic Equilibria in Aqueous Systems

Ionic Equilibria in Aqueous Systems Chapter Nineteen AP Chemistry

There are buffers in our blood that keep the ph of our blood at a constant level. The foods that we eat are often acidic or basic. This ingestion of foods would cause changes (0.1) in ph in our blood and death could result.

Another Example Buffered Aspirin Aspirin is acetylsalicylic acid. Extra acid in the stomach by the ingestion of aspirin affects some people with sensitive stomachs. Buffered aspirin helps by using chemistry to lessen the affect.

The buffers in the bloodstream resist large changes in ph which would cause death! Starting ph

Unbuffered Solution Acid Added Base Added

Buffered Solution Acid Added Base Added

Acid-Base Buffers What is a BUFFER? SOMETHING THAT LESSENS THE IMPACT OF AN EXTERNAL FORCE

An acid-base buffer is a solution that lessens the changes of ph (amount of H 3 O + ) of a solution when small amounts of [H + ] or [OH - ] are added. The components of a buffer are typically a conjugate acid-base pair of a weak acid.

Common-ion effect and LeChatelier s Principle. Common-ion effect occurs when a reactant containing a given ion is added to an equilibrium mixture that already contains that ion and the position of equilibrium shifts away from forming more of it.

Essential Features of a Buffer Consists of high concentrations of the acidic HA and the basic A - components or basic B and acidic BH + components.

When small amounts of H 3 O + or OH - are added to the buffer, they cause a small amount of one buffer component to convert into the other. This changes the relative concentrations of the two parts.

As long as the amount of H 3 O + or OH - added is much smaller than the amounts of HA or BH + and A - or B originally present, there is little ph change. A - or B consumes any added H 3 O +. HA or BH + consumes any added OH -.

Dissolve some acetic acid (HC 2 H 3 O 2 ) in water. (HC 2 H 3 O 2 = HAc) HAc + H 2 O H 3 O + + Ac - Q. What would happen if we added some Ac - ion to the system?

HAc + H 2 O H 3 O + + Ac - Adding Ac - will cause the following changes to the system. System shifts to the left. () Amount of H 3 O + decreases. Amount of HAc and water increases.

The Ac - can be provided by adding the salt NaAc. By adding the extra salt, NaAc, to the acid solution, the amount of hydronium ion is reduced.

If this new solution (with the extra Ac - ion) is now subjected to the addition of some added acid, the extra Ac - ion will absorb the acid, keeping the equilibrium from shifting to the right.

How a buffer works Buffer after addition of H 3 O + Buffer with equal [ ] of conjugate base and acid H 3 O + H 2 O + CH 3 COOH H 3 O + + CH 3 COO -

Buffer with equal [ ] of conjugate base and acid Buffer after addition of OH - OH - CH 3 COOH + OH - H 2 O + CH 3 COO -

Still confused? Let s let the math show us the way! Acid Buffer Definition: An acid and its dissociated (ionized) negative ion (called the conjugate base) at equilibrium.

Calculating the Effect of Added H 3 O + and OH - on Buffer ph Add 0.500 mole of HAc (K a = 1.80 X 10-5 ) and 0.500 mole of solid NaAc to make 1.00 L solution. Calculate the ph.

Acid Conjugate Base HAc + H 2 O H 3 O + + Ac - M = 0.500 mole 1.00 L = 0.500 M for both the acid and the salt.

HAc + H 2 O H 3 O + + Ac - [ ] [HAc] [H 3 O + ] [Ac - ] initial 0.500 0 0.500 - X + X + X.500 - X X.500 + X

[H 3 O + ][CH 3 COO - ] K a = [CH 3 COOH] [CH 3 COOH] [H 3 O + ] = K a [CH 3 COO - ]

[H 3 O + ] = 1.80 X 10-5 x [0.500 x] [0.500 + x] H 3 O + = 1.80 X 10-5 = X Check the assumption: 1.8 x 10-5 /0.500 X 100 = 3.60 x 10-3 % ph= -log 1.80 X 10-5 = 4.744

Calculate the ph after adding 0.020 mole of solid NaOH to 1.0 L of the buffer solution. Before any reaction occurs, the solution contains the following major species. Na +, OH -, H 2 O, CH 3 COOH, CH 3 COO -

What reaction can occur? Na + will not react with CH 3 COO -. OH - has a great affinity for protons (H + ) Therefore OH - will react with CH 3 COOH to form CH 3 COO -. OH - will react with the acid.

Reaction Table for the Stoichiometry of adding OH - to HAc. CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l)

Stoichiometry Table HAc OH - Ac - Before 0.500 0 0.500 Addition Addition 0.020 After 0.48 0 0.52 Addition

Reaction Table For The Buffer Using New [ ] Setting up a new reaction table for the buffer reaction, using these newly determined concentrations. HAc + H 2 O H 3 O + + Ac -

HAc + H 2 O H 3 O + + Ac - [ ] [HAc] [H 3 O + ] [Ac - ] initial 0.480 0 0.520 - X + X + X.480 - X X.520 + X

[H 3 O + ][CH 3 COO - ] K a = [CH 3 COOH] [CH 3 COOH] [H 3 O + ] = K a [CH 3 COO - ]

[H 3 O + ] = 1.80 X 10-5 x [0.480 x] [0.520 + x] H 3 O + = 1.66 X 10-5 = X ph= -log 1.66 X 10-5 = 4.78

The addition of the strong base increased the [ ] of the basic buffer component at the expense of the acidic buffer component. ph changed from 4.74 to 4.78.

Calculate the ph after adding 0.020 mole of HCl to 1.0 L of the buffer solution. HAc + H 2 O H 3 O + + Ac - Before any reaction occurs, the solution contains the following major species. H 3 O +, Cl -, CH 3 COOH, H 2 O, CH 3 COO -

What reaction can occur? H 3 O + will not react with CH 3 COOH. CH 3 COO - (salt of a weak acid) has a great affinity for protons (H + ) Therefore H 3 O + will react with CH 3 COO - to form CH 3 COOH. H 3 O + will react with the conjugate base.

Reaction Table for the Stoichiometry of adding H 3 O + to Buffer. CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O (l)

Stoichiometry Table Ac - H 3 O + HAc Before 0.500 0 0.500 Addition Addition 0.020 After 0.48 0 0.52 Addition

Reaction Table For The Buffer Using New [ ] Setting up a new reaction table for the buffer reaction, using these newly determined concentrations. CH 3 COOH(aq) + H 2 O (l) H 3 O + (aq) + CH 3 COO - (aq)

CH 3 COOH(aq) + H 2 O (l) H 3 O + (aq) + CH 3 COO - (aq) [ ] [HAc] [H 3 O + ] [Ac - ] initial 0.520 0 0.480 - X + X + X.520 - X X.480 + X

[CH 3 COOH] [H 3 O + ] = K a [CH 3 COO - ] [H 3 O + ] = 1.80 X 10-5 x [0.520- X] [0.480 + X] [H 3 O + ] = 2.00 x 10-5 = X ph = -log 2.00 X 10-5 = 4.70

The addition of the strong acid increased the [ ] of the acidic buffer component at the expense of the basic buffer component. ph changed from 4.74 to 4.70.

The Henderson-Hasselbalch Equation This equation was designed to help you arrive at the answers to these kinds of buffer problems more speedily. ph = pk a + log [conjugate base] [acid]

From the earlier problem Add 0.500 mole of HAc (K a = 1.80 X 10-5 ) and 0.500 mole of solid NaAc to make 1.00 L solution. Calculate the ph.

Acid Conjugate Base HAc + H 2 O H 3 O + + Ac - [ ] [HAc] [H 3 O + ] [Ac - ] initial 0.500 0 0.500 - X + X + X.500 - X X.500 + X

ph = pk a + log [conjugate base] [acid] ph = -log 1.80 X 10-5 + log [0.50] [0.50] ph = 4.744 + 0 = 4.744

Mix 1.72 mole of NH 3 and 0.830 mole of NH 4 Cl to make 2.72 L of solution. Find the ph. M = 1.72 moles NH 3 2.72 L = 0.632 M M =.830 moles NH 4 Cl = 0.305 M 2.72 L

Base NH 3 Conjugate Acid + H 2 O NH + 4 + OH - [ ] [NH 3 ] [NH 4+ ] [OH - ] initial 0.632 0.305 0 - X + X + X.632 - X.305 + X X

1.76 X 10-5 = [0.305 + X ] [X] [0.632 X] K b X = 3.78 X 10-5 = [OH - ] poh = -log 3.65 X 10-5 = 4.438 Therefore ph = 9.562

HH equation: ph = pk a + log [conjugate base] [acid] This is the HH for a weak acid what do we have? A base. Use the conjugate system as was done in the last chapter.

K a = K w / K b = 1.00 X 10-14 1.76 X 10-5 K a = 5.68 X 10-10 ph = pk a + log [conjugate base] [acid] ph = -log 5.68 X 10-10 + log [ NH 3 ] [NH 4+ ]

ph = -log 5.68 X 10-10 + log [0.632 ] [0.305] ph = 9.246 + 0.316 = 9.562

For Acids ph = pk a + log [conjugate base] [acid] For Bases poh = pk b + log [conjugate acid] [base]

poh = -log 1.76 X 10-5 + log 0.305 0.632 poh = 4.754 + log 0.48259 poh = 4.754 + (-0.31642) = 4.437 ph = 14.000 4.437 = 9.562

HH in Review An acetic acid/acetate buffer system is made with 0.456 mole of the acid and 0.391 mole of the NaC 2 H 3 O 2. Both are mixed to make 1.25 L of solution. Calculate the ph of the system if K a = 1.80 X 10-5.

Make the equation for the weak acid-common ion buffer. Acid Conjugate Base HAc + H 2 O H 3 O + + Ac - M = 0.456 mole HAc =.365 M HAc 1.25 L M = 0.391 mole Ac 1.25 L =.313 M Ac -

This is a weak acid system and hence will choose the acid H-H equation. ph = pk a + log [conjugate base] [acid] ph = -log 1.80 X 10-5 + log [.313] [.365] ph = 4.745 + (-0.0667) = 4.678

What About Basic Buffers Calculate the ph of a 0.169 M NH 3 and 0.183 M NH 4 Cl solution. Base Conjugate Acid NH 3 + HOH NH + 4 + OH -

poh = pk b + log [conjugate acid] [base] = - log 1.76 X 10-5 + log [.183] [.169] poh = 4.754 + 0.0345 = 4.788 ph = 9.211

Adding A Base To Basic Buffers Calculate the ph when you add 20.0 ml of a 0.100 M NaOH to 80.0 ml of the buffer consisting of 0.169 M NH 3 and 0.183 M NH 4 Cl. Assume the volumes are additive. K b = 1.76 X 10-5.

NH 3 + HOH NH + 4 + OH - Before any reaction occurs, the solution contains the following major species. NH 3, Na +, OH -, NH + 4, Cl -, H 2 O What reaction can occur? Na + will not react with Cl -.

OH - has a great affinity for protons (H + ) Therefore OH - will react with NH 4+ to form NH 3. OH - will react with the conjugate acid. NH 4 + + OH - NH 3 + HOH

Do the stoichiometry. 0.0200 L x 0.100 M NaOH = 0.00200 mole OH - 0.0800 L x 0.183 M NH 4+ = 0.0146 mole NH + 4 0.0800 L x 0.169 M NH 3 = 0.0135 mole NH 3

Stoichiometry Table NH 4 + + OH - NH 3 + HOH NH 4 + OH - NH 3 Before 0.0146 0 0.0135 Addition Addition 0.002 After 0.0126 0 0.0155 Addition

Add the 20.0 ml of the base to the 80.0 ml of the buffer! Add 20.0 ml to 80.0 ml of buffer = 100.0 ml solution M = 0.0155 mole NH 3.100 L = 0.155 M

M = 0.0126 mole NH 4 +.100 L = 0.126 M

Reaction Table For The Buffer Using The New [ ] NH 3 + HOH NH + 4 + OH - [ ] [NH 3 ] [NH 4+ ] [OH - ] initial 0.155 0.126 0 - X + X + X 0.155 - X 0.126 + X X

Recalculate the ph of the buffer. poh = pk b + log [conjugate acid] [base] -log 1.76 X 10-5 + log [.126] [.155] poh = 4.754 + (- 0.0899) = 4.644 ph = 9.336

Adding An Acid to a Base Buffer Calculate the ph when you add 20.0 ml of a 0.100 M HCl to 80.0 ml of the buffer consisting of 0.169 M NH 3 and 0.183 M NH 4 Cl. Assume the volumes are additive. K b = 1.76 X 10-5.

NH 3 + HOH NH + 4 + OH - Before any reaction occurs, the solution contains the following major species. NH 3, H 3 O +, Cl -, NH + 4, H 2 O What reaction can occur? H 3 O + will not react with Cl -.

NH 3 has an affinity for protons (H + ). Therefore NH 3 will react with H 3 O + to form more NH 4+. H 3 O + will react with the base. NH 3 + H 3 O + NH 4 + + HOH

Do the stoichiometry. 0.0200 L x 0.100 M HCl = 0.00200 mole H 3 O + 0.0800 L x 0.183 M NH 4+ = 0.0146 mole NH + 4 0.0800 L x 0.169 M NH 3 = 0.0135 mole NH 3

Stoichiometry Table NH 3 + H 3 O + NH 4 + + HOH NH 3 H 3 O + NH 4 + Before 0.0135 0 0.0146 Addition Addition 0.002 After 0.0115 0 0.0166 Addition

Add 20.0 ml HCl to 80.0 ml of buffer = 100.0 ml solution M = 0.0115 mole NH 3.100 L = 0.115 M M = 0.0166 mole NH 4 +.100 L = 0.166 M

Reaction Table For Buffer Using The New [ ] NH 3 + H 3 O + NH + 4 + HOH [ ] [NH 3 ] [NH 4+ ] [OH - ] initial 0.115 0.166 0 - X + X + X 0.115 - X 0.166 + X X

poh = pk b + log [conjugate acid] [base] -log 1.76 X 10-5 + log [.166] [.115] poh = 4.754 + (.159) = 4.913 ph = 9.087

Summary of Characteristics of Buffered Solution. 1. Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. It can involve a weak acid (HA) and the conjugate base (A - ) or a weak base (B) and the conjugate acid (BH + ).

2. When H + is added to a buffered solution, it reacts essentially to completion with the weak base that is present. H + + A - HA H + + B BH +

3. When OH - is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH - + HA A - + H 2 O OH - + BH + B + H 2 O

4. As long as the ratio between the [weak acid] and [weak base] is small, the ph will remain virtually constant. This remains true as long as the [buffering materials] HA and A - or B and BH + are large compared with the amounts of H + or OH - added.

Buffer Capacity and Buffer Range Buffers resist a ph change as long as the [ ] of the buffer components are large compared to the amount of strong acid or base added. The more concentrated the components of a buffer, the greater buffer capacity.

The ph of a buffer is distinct from its buffer capacity. A buffer of equal volumes of 1.0 M HA and 1.0 M A - has the same ph (4.74) as a 0.1 M HA and 0.1 M A -. But, the 1.0 M combination has a greater ability for resisting a ph change.

Buffer capacity is also affected by the relative [ ] of the buffer components. Since the ratio of the [ ] determines the ph, the less the ratio changes, the less the ph changes. The [ ] ratio changes less for similar buffer component [ ] than it does for different [ ].

Add 0.010 mole of OH - to the 1.000 M HA and 1.000 M A - in 1.00 L of solution. [HA] decreases to.990 M and [A - ] increases to 1.010 M. [A - ] initial = [HA] initial 1.000 M 1.000 M = 1.00

[A - ] final = [HA] final 1.010 M.990 M = 1.02 % Change = 1.02 1.000 x 100 = 1.000 2%

Add 0.010 mole of OH - to the 0.250 M HA and 1.750 M A - in 1.00 L of solution. [HA] decreases to.240 M and [A - ] increases to 1.760 M. [A - ] initial = [HA] initial 1.750 M 0.250 M = 7.00

[A - ] final = [HA] final 1.760 M.240 M = 7.33 % Change = 7.33 7.00 x 100 = 7.00 4.7% The buffer component ratio gets larger when the initial component [ ] are very different.

A buffer has the highest capacity when the component concentrations are equal. ph = pk a + log [X] [X] ph = pk a + log 1 ph = pk a A buffer whose ph is equal to or near the pk a of its acid component has the highest buffer capacity.

Buffer range is the ph range over which the buffer acts effectively. If the [A - ]/[HA] ratio is greater than 10 or less than 0.1, the buffering action is poor. Buffers have a useable range within +/- 1 ph unit of the pk a of the acid component.

ph = pk a + log [10] [ 1 ] ph = pk a + 1 ph = pk a + log [ 1 ] [10] ph = pk a - 1

The Relation Between Buffer Capacity and ph Change

Preparing a Buffer 1. Choose the conjugate acid-base pair. Decision is determined by the desired ph. Ratio of its component [ ] should be close to 1. ph pk a

Common buffer solutions: acetic acid and sodium acetate phosphoric acid and potassium phosphate oxalic acid and lithium oxalate carbonic acid and sodium carbonate ammonium hydroxide and ammonium nitrate

A buffer of ph = 3.9 is needed. pk a should be close to 3.9 or K a = 10-3.90 = 1.3 x 10-4 Table 18.2 Formic acid K a = 1.8 x 10-4 pk a = 3.74

Buffer components: Formic acid HCOOH Formate ion HCOO - (Conjugate base) Supplied by a soluble salt sodium formate, HCOONa

2. Calculate the ratio of buffer component concentrations. Find the [A - ]/[HA] ratio that gives the needed ph. ph = pk a + log [conjugate base] [acid] 3.90 = 3.74 + log [HCOO- ] [HCOOH]

0.16 = log [HCOO- ] [HCOOH] [HCOO - ] [HCOOH] = 100.16 = 1.4 1 mole HCOOH = 1.4 mole HCOONa

3. Determine the buffer [ ]. How concentrated should the buffer be? The greater the [ ], the greater the buffer capacity. Most times 0.5 M are suitable.

If 0.40 M HCOOH is available and a 1.0 L buffer solution is needed, how much HCOONa is needed? 1.0 L X 0.40 mole HCOOH 1.0 L X 1.4 mole HCOONa 1.0 mole HCOOH = 0.56 mole HCOONa

0.56 mole HCOONa X 68.01 g HCOONa 1 mole HCOONa = 38 g HCOONa

4. Mix the solution and adjust ph. Dissolve 38 g HCOONa in 0.40 M HCOOH to a total volume of 1.0 L. If ph needs adjusting, add some strong acid or base while monitoring with a ph meter.

Preparing a Buffer An environmental chemist needs a carbonate buffer of ph 10.00 to study the effects of the acid rain on limestone-rich soils. How many grams of Na 2 CO 3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO 3 to make the buffer? K a of HCO 3- is 4.7x10-11.

We know the K a and the conjugate acid-base pair. Convert ph to [H 3 O + ], find the number of moles of carbonate and convert to mass. HCO 3- (aq) + H 2 O(l) Acid CO 3 2- (aq) + H 3 O + (aq) Conjugate Base

K a = [CO 3 2- ][H 3 O + ] [HCO 3- ] ph = 10.00; [H 3 O + ] = 1.0 x 10-10 4.7 x 10-11 = [CO 3 2- ][1.0 x 10-10 ] [0.20] [CO 3 2- ] = 0.094 M

moles of Na 2 CO 3 = (1.5 L)(0.094 moles) = 0.14 moles L 0.14 moles Na 2 CO 3 X 105.99g mole = 15 g Na 2 CO 3

Acid Base Titration Curves Plot of ph vs titrant added. ph meters and /or acid-base indicators are used to monitor ph. An acid-base indicator is a weak organic acid (HIn) that has a different color than its conjugate base (In - ).

The color change should be obvious and occur over a narrow ph range Selecting an indicator requires that the approximate ph of the titration end point be known. Indicators typically change color over about 2 ph units.

Colors and approximate ph range of some common acid-base indicators

Color change of indicator bromthymol blue acidic basic Change occurs over ~ 2pH units.

Acid-Base Neutralizations Titrations NaOH + HCl NaCl + HOH Notice a neutral salt Note: Notice the mole ratio in the balanced equation.

When exact amounts of OH - and H + react, the stoichiometric point or equivalence point or end point has been reached. The visible change in color of the indicator at the end point signals the invisible point when moles of base added = the original moles of acid.

25.0 ml of unknown [HCl] is titrated with 18.43 ml of 0.614 M NaOH to its equivalence point. Determine the unknown [acid]. 0.614 M NaOH x 0.01843 L = 0.0113 mole NaOH

0.0113 mole NaOH x 1 mole HCl 1 mole NaOH = 0.0113 mole HCl M = 0.0113 mole HCl =.453 M HCl.0250 L

Or, note because of the 1:1 acid-base ratio M a X V a = M b X V b 0.614 M x 0.01843 L = M b x 0.0250 L M b =.453 M HCl

Strong Acid Strong Base (H 2 SO 4 / NaOH) ph starts low high [H 3 O + ] from the strong acid. ph then increases gradually until moles of OH- added mole of H 3 O + where it rises steeply. ph then increases gradually as OH - becomes in excess.

37.0 ml of an unknown [NaOH] is used to reach the equivalence point of a 25.0 ml of a 0.300 M H 2 SO 4 solution. Calculate the (a) concentration of the unknown base and (b) calculate the final ph of the system.

(a) H 2 SO 4 + 2 NaOHNa 2 SO 4 + 2 H 2 O At the equivalence point, equal numbers of H + and OH - must have reacted. Mole H 2 SO 4 = 0.02500 L X 0.300 M = 0.0075 mole H 2 SO 4

The balanced reaction has the acid:base ratio of 1:2 0.00750 mole H 2 SO 4 x 2 mole NaOH 1 mole H 2 SO 4 = 0.0150 mole NaOH M base = 0.0150 moles NaOH 0.0370 L =.405 M

(b) The resulting salt, Na 2 SO 4 has both a cation of a strong base and an anion of a strong acid. Therefore, there is no hydrolysis ph = 7

Find the ph before the End Point. Find the ph after adding 20.00 ml of the 0.405 M NaOH. Moles of H 3 O + remaining = Initial moles of H 3 O + - Moles of H 3 O + reacted

Initial moles of H 3 O + = 0.0075 mole H 2 SO 4 X 2 mole H 3 O + 1 mole H 2 SO 4 = 0.015 mole H 3 O + H 3 O + reacted =.02 L x.405 mole OH - x 1 mole H 3 O + L 1 mole OH - = 0.0081 H 3 O +

0.015 H 3 O + - 0.0081 H 3 O + =.0069 mole H 3 O + remaining Calculate [H 3 O + ] taking the total volume into account..0069 mole H 3 O + =.153 M H 3 O + (.025 L +.02 L) ph =.814

Find the ph after the End Point. Find the ph after adding 50.00 ml of the 0.405 M NaOH. Moles of NaOH is in excess. Moles of OH - remaining = Moles of OH - added - Moles of OH - reacted

Moles of OH - added =.05 L x.405 mole OH - = L.02025 mole OH - OH - reacted = 0.0150 mole OH - From earlier problem

0.02025 mole OH - - 0.0150 mole OH - 0.00525 mole OH - remaining [OH - ] = 0.00525 mole =.07 M (.025 L +.05 L) poh = 1.15 ph = 12.85 Similar calculations yield a titration curve.

Strong acid-strong base Titration

Does 15.75 ml of 0.413 M HCl react exactly with 21.35 ml of 0.395 M NaOH? If not, what will be the final ph? Mole HCl = 0.01575 L X 0.413 M = 0.00650 mole HCl Mole NaOH = 0.02135 L X 0.395 M = 0.00843 mole NaOH

Note: HCl is the limiting reagent! 0.00650 mole HCl x 1 mole NaOH 1 mole HCl = 0.00650 mole NaOH reacted 0.00843 mole NaOH - 0.00650 mole NaOH reacted 0.00193 mole NaOH unreacted

The ph? M = 0.00193 mole NaOH (.01575 +.02135)L =.0520 M NaOH poh = -log 0.0520 = 1.284 ph = 12.716

Weak Acid Strong Base (HAc / NaOH) The initial ph is higher because the weak acid only dissociates slightly. A gradually rising portion of the curve, called the buffer region, appears before a steep rise to the equivalence point.

As HAc reacts with the strong base, a significant amount of the conjugate base (Ac - ) forms. This creates an HAc/Ac - buffer. Note the midpoint of the buffer region. At this point, half the original HAc has reacted so [HAc] = [Ac - ].

Weak acid-strong base titration pk a of HPr = 4.89 ph = 8.80 at equivalence point [HPr] = [Pr - ] methyl red Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH

Or [ Ac - ] = 1 [HAc] Therefore, at the midpoint of the buffer region, the ph = pk a of the acid. ph = pk a + log [Ac-] = [HAc] pk a + log 1 = pk a + 0 = pk a

Observing ph at the midpoint of the buffer region is a common experimental method for estimating pk a of an unknown acid.

ph at the equivalence point > 7.0. Why? The solution contains the strong base cation Na +, which does not react with water, and the weak Ac - anion which acts as a base to accept a proton from H 2 O and yield OH -.

(a) What volume of 0.750 M NaOH is necessary to neutralize 20.00 ml of 0.500 M acetic acid? (Note: both are mono H and OH) HAc + NaOHNaAc + HOH M a X V a = M b X V b 0.500 M X 20.00 ml = 0.750 M X V b V b = 13.33 ml

(b) Will the resulting solution have a ph of 7.00? No Way! The salt, sodium acetate has a strong ion (Na + ) but it also has an anion of a weak acid which will therefore hydrolyze and produce a basic solution hence the final ph should be > 7.

HAc + NaOHNaAc + HOH Mole HAc = 0.02000 L X 0.500 M = 0.0100 mole HAc 0.0100 mole HAc x 1 mole NaAc 1 mole HAc = 0.0100 mole NaAc

NaAc will dissociate completely. Therefore: 0.0100 mole Ac - upon dissociation M = 0.0100 mole Ac - =.0300 M Ac- (.0200 +.01333)L Ac - + HOH HAc + OH -

Ac - + HOH HAc + OH - [ ] [Ac - ] [HAc] [OH - ] initial 0.0300 0 0 - X + X + X 0.0300-X + X + X

K b = [HAc][OH - ] Must Find [Ac - ] K b = [X][X] [0.300 - X] K b = K w Ka = 1.00 X 10-14 1.80 X 10-5 From HAc K b = 5.6 X 10-10

5.6 X 10-10 = [X][X] [0.300 - X] X = 1.3 X 10-5 = [OH - ] A basic solution! poh = 4.89 ph = 9.11

Weak Acid-Strong Base Titration Calculate the ph during the titration of 40.00 ml of 0.1000M propanoic acid (HPr; K a = 1.3x10-5 ) after adding the following volumes of 0.1000M NaOH: PLAN: (a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mL The amounts of HPr and Pr - will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION: (a) Find the starting ph using the methods of Chapter 18. K a = [Pr - ][H 3 O + ]/[HPr] [Pr - ] = x = [H 3 O + ] [Pr - ] = x = [H 3 O + ] x = (1.3x10 5 )(0.10) x = 1.1x10-3 ; ph = 2.96 (b) Amount (mol) Before addition Addition After addition HPr(aq) + OH - (aq) Pr - (aq) + H 2 O (l) 0.04000-0 - - 0.03000 - - 0.01000 0 0.03000 -

continued [H 3 O + ] = 1.3x10-5 0.001000 mol 0.003000 mol = 4.3x10-6 M ph = 5.37 (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be (0.004000 mol) = 0.05000M (0.004000L) + (0.004000L) K a x K b = K w K b = K w /K a = 1.0x10-14 /1.3x10-5 = 7.7x10-10 [H 3 O + ] = K w / = 1.6x10-9 M K b x[pr ] ph = 8.80 (d) 50.00mL of NaOH will produce an excess of OH -. mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol [H 3 O + ] = 1.0x10-14 /0.01111 = 9.0x10-11 M M = (0.00100) (0.0900L) M = 0.01111 ph = 12.05

Weak Base Strong Acid (NH 3 / HCl ) The titration curve has the same shape as the weak acid/strong base but it is inverted. The ph decreases during the process.

Weak base-strong acid titration

ph starts above 7.00 ph decreases gradually in the buffered region where significant amounts of base (NH 3 ) and conjugate acid (NH 4+ ) exists. At the midpoint of the buffer region, ph = pk a of the ammonium ion.

After the buffer region, the ph drops to the equivalence point. The solution contains only NH 4+ and Cl -. The ph is below 7.00 because Cl - does not react with water and NH 4 + in water produces an acidic solution. Beyond the equivalence point, the ph decreases slowly as excess H 3 O + is added.

Observing ph at the midpoint of the buffer region is a common experimental method for estimating pk b of an unknown base. 14 ph = poh = pk b

(a) What volume of 0.200 M HCl is necessary to titrate to the equivalence point a 0.200 M NH 3 solution of 25.00 ml? (b) What will be the final ph of the solution?

(a) From the HCl NH 3 + H 3 O + NH + 4 + HOH M a X V a = M b X V b 0.200 M X 0.02500 L = 0.200 M X V b V b = 0.0250 L = 25.00 ml

(b) From an acid salt NH 3 + H 3 O + NH + 4 + HOH Mole HCl = 0.02500 L X 0.200 M = 0.0050 mole H 3 O + 0.0050 mole H 3 O + x 1 mole NH 3 1 mole H 3 O + = 0.0050 mole NH 3

Therefore: 0.0050 mole NH 4+ upon reaction NH + 4 + H 2 O NH 3 + H 3 O + M = 0.0050 mole NH 4 + (.0250 +.0250)L =.100 M NH 4 +

NH + 4 + H 2 O NH 3 + H 3 O + [ ] [NH 4+ ] [NH 3 ] [H 3 O + ] initial 0.100 0 0 - X + X + X 0.100 - X + X + X

K a = Must Find [NH 3 ][H 3 O + ] [NH 4+ ] K a = [X][X] [0.100 - X] K a = K w Kb = 1.00 X 10-14 1.76 X 10-5 From NH 3 K a = 5.7 X 10-10

5.7 X 10-10 = [X][X] [0.100 - X] X = 7.5 X 10-6 = [H 3 O + ] ph = 5.12 An acid solution!

Titration Curve for Polyprotic Acid Polyprotic acids have more than one ionizable proton. Except for sulfuric acid, the common polyprotic acids are weak acids. Successive K a values differ by several orders of magnitude.

This means that the first H + is lost more easily than the next ones. The titration curve shows the loss of each mole of H + as a separate equivalence point and buffer region. Note: Same volume of base is required to remove each H +.

Titration of a weak polyprotic acid. Titration of 40.00 ml of 0.1000 M H 2 SO 3 with 0.1000 M NaOH pk a = 7.19 pk a = 1.85

Equilibria of Slightly Soluble Ionic Compounds In a saturated solution, an equilibrium exists between dissolved and undissolved solute. Slightly soluble (insoluble) ionic compounds have relatively low solubilities.

Therefore, equilibrium is reached when relatively little solute is dissolved. Assume that when any amount of ionic solid dissolves in water, it dissociates completely into ions whether it is highly soluble or not..

Ion Product Expression (Q sp ) and the Solubility-Product Constant (K sp ) Equilibrium exists between solid solute and aqueous ions. PbSO 4(s) Pb 2+ (aq) + SO 4 2- (aq) Q c = [Pb 2+ ][SO 4 2- ] [PbSO 4 ]

Combine constant concentration of the solid [PbSO 4 ] with the value of Q c and eliminate it. Q sp = Q [Pb 2+ ][SO 2-4 ] c [PbSO 4 ] = Solubility product When the solution reaches equilibrium, Q sp reaches a constant value.

The new equilibrium constant is called solubility-product constant K sp. K sp depends on temperature, not the individual ion concentrations. Q sp is identical to other reaction quotients.

Find Q sp of M p X q at equilibrium. M p X q(s) pmn+ (aq) + qx z- (aq) Q sp = [M n+ ] p [X z- ] q = K sp

Cu(OH) 2(s) Cu 2+ (aq) + 2 OH - (aq) K sp = [Cu 2+ ][OH - ] 2 Insoluble metal sulfides present a slightly different case. The sulfide ion is so basic that it is not stable in water and reacts completely to form the hydrogen sulfide ion (HS - ) and the OH - ion.

MnS (s) Mn 2+ (aq) + S 2- (aq) S 2- (aq) + H 2 O (l) HS - (aq) + OH - (aq) MnS (s) + H 2 O (l) Mn 2+ (aq) + HS - (aq) + OH - (aq) K sp = [Mn 2+ ][HS - ][OH - ]

Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds Write the ion-product expression for each of the following: (a) Magnesium carbonate MgCO 3 (s) Mg 2+ (aq) + CO 3 2- (aq) K sp = [Mg 2+ ][CO 3 2- ]

(b) Iron (II) hydroxide Fe(OH) 2 (s) Fe 2+ (aq) + 2OH - (aq) K sp = [Fe 2+ ][OH - ] 2 (c) Calcium phosphate Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 3-4 (aq) K sp = [Ca 2+ ] 3 [PO 3-4 ] 2

(d) Silver sulfide Ag 2 S (s) 2Ag + (aq) + S2- (aq) S 2- (aq) + H 2 O (l) HS - (aq) + OH- (aq) Ag 2 S (s) + H 2 O (l) 2Ag + (aq) + HS- (aq) + OH- (aq) K sp = [Ag + ] 2 [HS - ][OH - ]

Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds Problem: Write the ion-product expression for (a) silver bromide; (b) strontium phosphate; (c) aluminum carbonate; (d) nickel(iii) sulfide. Plan: Write the equation for a saturated solution, then write the expression for the solubility product. Solution: (a) Silver bromide: AgBr (s) Ag + (aq) + Br - (aq) K sp = [Ag + ] [Br - ] (b) Strontium phosphate: Sr 3 (PO 4 ) (s) K sp = [Sr 2+ ] 3 [PO 3-4 ] 2 3 Sr 2+ (aq) + 2 PO 4 3- (aq) (c) Aluminum carbonate: Al 2 (CO 3 ) 3 (s) K sp = [Al 3+ ] 2 [CO 2-3 ] 3 (d) Nickel(III) sulfide: Ni 2 S 3 (s) + 3 H 2 O (l) 2 Ni 3+ (aq) + 3 HS - (aq) + 3 OH- (aq) K sp =[Ni 3+ ] 2 [HS - ] 3 [OH - ] 3 2 Al 3+ (aq) + 3 CO 3 2- (aq)

Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, Formula Aluminum hydroxide, Al(OH) 3 Cobalt (II) carbonate, CoCO 3 Iron (II) hydroxide, Fe(OH) 2 Lead (II) fluoride, PbF 2 Lead (II) sulfate, PbSO 4 Mercury (I) iodide, Hg 2 I 2 Silver sulfide, Ag 2 S Zinc iodate, Zn(IO 3 ) 2 K sp 3 x 10-34 1.0 x 10-10 4.1 x 10-15 3.6 x 10-8 1.6 x 10-8 4.7 x 10-29 8 x 10-48 3.9 x 10-6

Calculations Involving the Solubility-Product Constant Most solubility values are given in units of grams of solute dissolved per 100 grams of water. (Assume 100 ml of solution)

Convert the solubility from grams of solute per 100 ml of solution to molar solubility. Molar solubility is the amount of moles of solute per liter of solution or molarity.

Determining K sp from Solubility (a) Lead (II) sulfate is a component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25 x 10-3 g/100 ml solution. What is the K sp of PbSO 4?

Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. PbSO 4 (s) Pb 2+ (aq) + SO 4 2- (aq) Note: 1:1 ion ratio

4.25 x10-3 g X 100 ml 1000mL L X mole PbSO 4 303.3g PbSO 4 = 1.40 x 10-4 M PbSO 4 K sp = [Pb 2+ ][SO 4 2- ] = [1.40 x 10-4 ][1.40 x 10-4 ] = 1.96 x 10-8

(b) When lead (II) fluoride (PbF 2 ) is shaken with pure water at 25 0 C, the solubility is found to be 0.64 g/l. Calculate the K sp of PbF 2. PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2

0.64 g X mole PbF 2 L 245.2 g PbF 2 Note: 1:2 ion ratio = 2.6 x 10-3 M K sp = [2.6 x 10-3 ][5.2 x 10-3 ] 2 = 7.0 x 10-8

Determining Solubility from K sp Use an approach similar to one used for weak acids. Define the unknown molar solubility as S. Define the ion concentrations in terms of this unknown in a reaction table and the solve for S.

Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water if the K sp is 6.5 x 10-6.

Write out a dissociation equation and K sp expression; Find the molar solubility (S) using a table. Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) K sp = [Ca 2+ ][OH - ] 2

[ ] [Ca(OH) 2 ] [Ca 2+ ] [2OH - ] initial 0 0 + S + 2 S + S + 2 S

K sp = [S][2S] 2 = 6.5 x 10-6 4 S 3 = 6.5 x 10-6 S = 3 6.5x10 6 4 = 1.2 x 10-2 M

Determining K sp from Solubility Problem: Lead chromate is an insoluble compound that at one time was used as the pigment in the yellow stripes on highways. It s solubility is 5.8 x 10-6 g/100ml water. What is the K sp? Plan: We write an equation for the dissolution of the compound to see the number of ions formed, then write the ion-product expression. Solution: PbCrO 4 (s) Pb 2+ (aq) + CrO 2-4 (aq) Molar solubility of PbCrO 4 = 5.8 x 10-6 g x 1000 ml x 100 ml 1 L 1mol PbCrO 4 323.2 g = 1.79 x 10-7 M PbCrO 4 1 Mole PbCrO 4 = 1 mole Pb 2+ and 1 mole CrO 2-4 Therefore [Pb 2+ ] = [CrO 2-4 ] = 1.79 x 10-7 M K sp = [Pb 2+ ] [CrO 2-4 ] = (1.79 x 10-7 M) 2 = 3.20 x 10-14

Determining Solubility from K sp Problem: Lead chromate used to be used as the pigment for the yellow lines on roads, and is a very insoluble compound. Calculate the solubility of PbCrO 4 in water if the K sp is equal to 2.3 x 10-13. Plan: We write the dissolution equation, and the ion-product expression. Solution: Writing the dissolution equation, and the ion-product expression: PbCrO 4 (s) Pb 2+ (aq) + CrO 2-4 (aq) K sp = 2.3 x 10-13 = [Pb 2+ ] [CrO 2-4 ] Concentration (M) PbCrO 4 Pb 2+ CrO 4 2- Initial ---------- 0 0 Change ---------- +x +x Equilibrium ---------- x x K sp = [Pb 2+ ] [CrO 4 2- ] = (x)(x ) = 2.3 x 10-13 x = 4.80 x 10-7 Therefore the solubility of PbCrO 4 in water is 4.8 x 10-7 M

Using K sp Values to Compare Solubilities K sp values are a guide to relative solubilities as long as compounds whose formulas contain the same total number of ions are compared. The higher the K sp, the greater the solubility.

Relationship Between K sp and Solubility at 25 0 C No. of Ions Formula Cation:Anion K sp Solubility (M) 2 MgCO 3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO 4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO 4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH) 2 1:2 5.5 x 10-6 1.2 x 10-2 3 BaF 2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF 2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag 2 CrO 4 2:1 2.6 x 10-12 8.7 x 10-5

Predicting the Formation of a Precipitate: Q sp vs. K sp The solubility produce constant, K sp, can be compared to the ion-product constant, Q sp to understand the characteristics of a solution with respect to forming a precipitate.

Q sp = K sp : When a solution becomes saturated, no more solute will dissolve, and the solution is called saturated. There will be no changes that will occur. Q sp > K sp : Precipitates will form until the solution becomes saturated. Q sp < K sp : Solution is unsaturated, and no precipitate will form.

Predicting Whether a Precipitate Will Form Will a precipitate form when 0.100 L of a solution containing 0.055 M barium nitrate is added to 200.00 ml of a 0.100 M solution of sodium chromate?

First see if the solutions will yield soluble ions, then we calculate the concentrations, adding the two volumes together to get the total volume of the solution, then calculate the product constant (Q sp ), and compare it to the solubility product constant to see if a precipitate will form.

Both Na 2 CrO 4 and Ba(NO 3 ) 2 are soluble, so we will have Na +, CrO 2-4, Ba 2+ and NO 3- ions present in 0.300 L of solution. We change partners, look up solubilities, and we find that BaCrO 4 would be insoluble, so we calculate it s ionproduct constant and compare it to the solubility product constant of 2.1 x 10-10.

For Ba 2+ : (0.100 L Ba(NO 3 ) 2 ) (0.55 M) = 0.055 mole Ba 2+ [Ba 2+ ] = 0.055 mole Ba2+ 0.300 L = 0.183 M in Ba 2+

For CrO 2-4 : (0.100 M Na 2 CrO 4 ) (0.200 L) = 0.0200 mole CrO 2-4 [CrO 2-4 ] = 0.0200 mole CrO 4 2-0.300 liters = 0.0667 M in CrO 2-4

Q sp = [Ba 2+ ] [CrO 4 2- ] = [0.183 M Ba 2+ ][0.0667 M CrO 4 2- ] = 0.0121 Since K sp = 2.1 x 10-10 and Q sp = 0.0121 Q sp >> K sp a precipitate will form.

Predicting Whether a Precipitate Will Form PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO 3 ) 2 is mixed with 0.200L of 0.060M NaF? PLAN: Write out a reaction equation to see which salt would be formed. Look up the K sp values in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = 3.2x10-11 mol Ca 2+ = 0.100L(0.30mol/L) = 0.030mol [Ca 2+ ] = 0.030mol/0.300L = 0.10M mol F - = 0.200L(0.060mol/L) = 0.012mol [F - ] = 0.012mol/0.300L = 0.040M Q = [Ca 2+ ][F - ] 2 = (0.10)(0.040) 2 = 1.6x10-4 Q is >> K sp and the CaF 2 WILL precipitate.

Effect of Common Ions on Solubility The presence of a common ion decreases solubility of a slightly soluble ionic compound. Le Chateliers Principle helps explain this effect.

Effect of a Common Ion on Solubility PbCrO 4(s) Pb 2+ (aq) + CrO 4 2- (aq) CrO 4 2- added

Calculating the Effect of a Common Ion on Solubility What is the solubility of Ca(OH) 2 in 0.10 M Ca(NO 3 ) 2? K sp of Ca(OH) 2 is 6.5 x 10-6. Ca(NO 3 ) 2(s) Ca 2+ (aq) + 2 NO 3 - (aq) Therefore 0.10 M Ca 2+ (aq) In original solution

Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) [ ] [Ca(OH) 2 ] [Ca 2+ ] [2OH - ] initial 0.10 0 + S + 2 S 0.10 + S + 2 S

K sp = [Ca 2+ ][OH - ] 2 K sp = 6.5 x 10-6 = [0.10 + S][2 S] 2 Assume S << 0.10 6.5 x 10-6 = [0.10][2 S] 2 S = 4.0 x 10-3 M

Check the assumption: 4.0 x 10-3 M x 100 = 4.0% 0.10 M

Effect of ph on Solubility [H 3 O + ] can have an effect on the solubility of an ionic compound. If the compound contains the anion of a weak acid, adding H 3 O + from a strong acid increases its solubility. Le Chatelier s Principle explains this.

CaCO 3(s) Ca 2+ (aq) + CO 3 2- (aq) Adding HCl increases [H 3 O + ] which reacts with the CO 2-3 to form a weak acid HCO 3-. CO 3 2- (aq) + H 3 O + (aq) HCO 3 - (aq) + H 2 O (l)

If more HCl is added, [H 3 O + ] increases and a further reaction occurs. HCO - 3 (aq) + H 3 O + (aq) H 2 CO 3(aq) + H 2 O (l) Leaves the system CO 2(g) + 2 H 2 O (l)

Adding H 3 O + shifts the equilibrium to the right causing more CaCO 3 to dissolve. However, adding H 3 O + to a solution with a strong acid anion has no effect on equilibrium.

AgCl (s) Ag + (aq) + Cl - (aq) The Cl - ion is a conjugate base of a strong acid HCl. The Cl - ion can exist with the H 3 O + without reacting and it does not leave the system.

Calculating the Effect of a Common Ion on Solubility Problem: What is the solubility of silver chromate in 0.0600 M silver nitrate solution? K sp = 2.6 x 10-12. Plan: From the equation and the ion-product expression for Ag 2 CrO 4, we predict that the addition of silver ion will decrease the solubility. Solution: Writing the equation and ion-product expression: Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq) K sp = [Ag + ] 2 [CrO 4 2- ] Concentration (M) Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq) Initial --------- 0.0600 0 Change --------- +2x +x Equilibrium --------- 0.0600 + 2x x Assuming that K sp is small, 0.0600 M + 2x = 0.600 M K sp = 2.6 x 10-12 = (0.0600) 2 (x) x = 7.22 x 10-10 M Therefore, the solubility of silver chromate is 7.22 x 10-10 M

Application of Ionic Equilibria to Chemical Analysis Selective precipitation is a process where a solution of precipitating ion is added until the Q sp of the more soluble compound is almost at its K sp value.

Ensures that the K sp value of the less soluble compound is exceeded by the maximum amount. The maximum amount of the less soluble compound precipitates but none of the more soluble compound does.

Separating Ions by Selective Precipitation A solution consists of 0.20 M MgCl 2 and 0.10 M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3 x 10-10 and K sp of Cu(OH) 2 is 2.2 x 10-20.

Both precipitates are of the same ion ratio, 1:2, so we can compare their K sp values to determine which has the greater solubility. Cu(OH) 2 precipitates first. Calculate the [OH - ] needed for a saturated solution of Mg(OH) 2. This should ensure Mg(OH) 2 does not precipitate. Then, check how much Cu 2+ remains in solution.

Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = 6.3 x 10-10 Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) K sp = 2.2 x 10-20

[OH - ] needed for a saturated Mg(OH) 2 solution = K sp [Mg 2+ ] [OH - ] = 6.3x10 10 0.20 = 5.6 x 10-5 M

Use the K sp for Cu(OH) 2 to find the amount of Cu remaining. [Cu 2+ ] = K sp = 2.2 x 10-20 = [OH - ] 2 [5.6 x 10-5 ] 2 [Cu 2+ ] = 7.0 x 10-12 M Since the solution was 0.10 M CuCl 2, virtually none of the Cu 2+ remains in solution.

Separating Ions by Selective Precipitation Problem: A solution consists of 0.10 M AgNO 3 and 0.15 M Cu(NO 3 ) 2. Calculate the [I - ] that would separate the metal ions as their iodides. K sp of AgI = 8.3 x 10-17 ; K sp of CuI = 1.0 x 10-12. Plan: Since the two iodides have the same formula type (1:1), we compare their K sp values and we see that CuI is about 100,000 times more soluble than AgI. Therefore, AgI precipitates first, and we solve for [I - ] that will give a saturated solution of AgI. Solution: Writing chemical equations and ion-product expressions: H 2 O AgI (s) Ag + (aq) + I - (aq) K sp = [Ag + ][I - ] H 2 O CuI (s) Cu + (aq) + I - (aq) K sp = [Cu + ][I - ] Calculating the quantity of iodide needed to give a saturated solution of CuI: K sp [I - ] = = 1.0 x 10-12 = 1.0 x 10-11 M [Cu + ] 0.10 M

Two or more types of ionic equilibria can be controlled simultaneously to precipitate ions selectively. Separating ions as their sulfides so HS - is the precipitating ion. Control [HS - ] to exceed the K sp value of one metal sulfide but not the other.

Controlling H 2 S dissociation controls [HS - ]. Control H 2 S dissociation by adjusting [H 3 O + ]. H 2 S (aq) + H 2 O (l) H 3 O+ (aq) + HS- (aq) Adding a strong acid, H 3 O +, shifts the equilibrium to the left.

This lowers the [HS - ] which allows the less soluble sulfide to precipitate first. Adding a strong base, OH - lowers the [H 3 O + ] and shifts the equilibrium to the right. [HS - ] increases and the more soluble sulfide precipitates.

Qualitative Analysis: ID Ions in Complex Mixtures. Inorganic qualitative analysis is the separation and ID of ions in a mixture. In general, the following method is used.

Separation into Ion Groups Separate the unknown solution into ion groups. (Ions groups has nothing to do with periodic table groups.) Mixture of metal ions is treated with a solution that precipitates a certain group of ions. Others are left in solution.

Filtration or a centrifuge is used to separate the precipitate. Remaining ion solution is decanted off and treated with a different solution to precipitate a different group of ions. Usually a known solution of ions and a blank of distilled water are treated the same way so that results are easier to judge.

General procedure for separating ions in qualitative analysis Add precipitating ion Add precipitating ion Centrifuge Centrifuge

Ion Group I Insoluble chlorides Treat solution with 6 M HCl. Since most metal chlorides are soluble. If a precipitate appears, it contains: Ag +, Hg 2 2+, Pb 2+ Centrifuge and decant remaining solution.

Ion Group II Acid- Insoluble Sulfides Decanted solution is already acidic. Adjust ph to 0.5 and treat with aqueous H 2 S. H 3 O + keeps the [HS - ] very low. H 2 S (aq) + H 2 O (l) H 3 O+ (aq) + HS- (aq)

Precipitate contains the least soluble sulfides. Cu 2+, Cd 2+, Hg 2+, As 3+, Bi 3+, Sn 2+, Sn 4+, and perhaps Pb 2+ Centrifuge and decant remaining solution.

Ion Group III Base-Insoluble Sulfides and Hydroxides Make decanted solution slightly basic with a buffer of NH 3 /NH 4+. OH - increases [HS - ] which precipitates the more soluble sulfides and some hydroxides.

Zn 2+, Mn 2+, Ni 2+, Fe 2+, Co 2+ as sulfides and Al 3+, Cr 3+ as hydroxides Centrifuge and decant remaining solution.

Ion Group IV Insoluble Phosphates Solution is slightly basic. Add (NH 4 ) 2 HPO 4. This precipitates alkaline earth ions as phosphates. Or, add Na 2 CO 3. This precipitates alkaline earth ions as carbonates.

Mg 2+, Ca 2+, Ba 2+ Centrifuge and decant remaining solution.

Ion Group V Alkali Metals and Ammonium Ions The final solution contains: Na +, K +, NH + 4 ID of this group of ions is done through flame and color tests. Na + produces yellow-orange flame. K + produces violet flame.

Tests to Determine the Presence of Cations in Ion Group 5 Na + ions K + ions Plus litmus paper OH - + NH 4+ NH 3 + H 2 O

Acid-base behavior ID s NH 4+. Make solution basic by adding NaOH and hold moist red litmus paper over it. If paper turns blue NH + 4 is present since OH - reacts with NH 4+ which reacts with the water on the paper.

NH 4 + (aq) + OH - (aq) NH 3 (g) + H 2 O (l) NH 3 (g) + H 2 O (l) NH 4 + (aq) + OH- (aq) Turns moist red litmus paper blue.

The End