Ch 17 notes

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Transcript Ch 17 notes

More Acid/Base Equilibria!!!

17.1 – 17.3

17.1 The Common Ion Effect

Consider the ionization of a weak acid, acetic acid: HC 2 H 3 O 2 (

)  H + (

) + C 2 H 3 O 2

–

(

) If we increase the [C 2 H 3 O 2

–

] ions by adding NaC 2 H 3 O 2 , the equilibrium will shift to the left. (Le Chatelier) This reduces the [H + ] and raises the pH (less acidic) This phenomenon is called the

common-ion effect

. Common ion equilibrium problems are solved following the same pattern as other equilibrium problems (ICE charts)

EXCEPT

the initial concentration of the common ion must be considered (it is

NOT

zero).

Example 1: Does the pH increase, decrease, or stay the same on addition of each of the following?

(a) NaNO 2 to a solution of HNO 2 (b) (CH 3 NH 3 )Cl to a solution of CH 3 NH 2 (c) sodium formate to a solution of formic acid (d) potassium bromide to a solution of hydrobromic acid (e) HCl to a solution of NaC 2 H 3 O 2

(a) (b) HNO CH 3 2



NH (c) HCHO 2 2



H + + NO 2 increases + H 2 O



CH 3 NH 3 + decreases H + + CHO 2 increases (d) HBr



H + + Br (e) C 2 H 3 O 2 -1 + OH no change + H 2 O



HC 2 H 3 O 2 decreases + OH -1

Example 2 : Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC 3 H 5 O 2 and 0.085 M HC 3 H 5 O 2

HC H O (aq) 3 5 2



+1 3 5 -1 H (aq) + C H O (aq) K = 1.3 x 2 a 10 -5 in itial 0.085 0 0 .060

change: -x +x +x Equilibrium: 0.085-x x .060+ x (.060

 (.085



)  1.3*10  5 x 2 +.060013x-(1.105*10 -6 )=0 x=1.84*10 -5 pH = -log(1.84*10 -5 ) = 4.74

17. 2 Buffered Solutions

A buffered solution or buffer is a solution that resists a change in pH after addition of small amounts of strong acid or strong base. A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X – ) or weak base (B) and its conjugate acid (HB + ) Thus a buffer contains both: an acidic species to neutralize added OH – OH – When a small amount of OH – is added to the buffer solution, the reacts with the acid in the buffer solution.

a basic species to neutralize added H + When a small amount of H + is added to the buffer solution, the H + reacts with the base in the buffer solution.

Composition of a Buffer 4 ways to make a buffer solution:

1.) Weak acid + salt of the acid

HCN

and Na

weak acid: HCN weak base: CN -1 2.) Weak base + salt of the base

NH 3

and

NH 4

Cl weak acid: NH 4 +1 weak base: NH 3 3.)

EXCESS

Weak acid + strong base 2 mol HCN + 1 mol NaOH  1 mol

HCN

+ 1 mol Na

weak acid: HCN 2 mol NH 4 Cl + 1 mol NaOH weak base: CN -1  1 mol

NH 4

Cl + 1 mol weak acid: NH 4 +1 weak base: NH 3

NH 3

+ H 2 O + NaCl 4.)

EXCESS

Weak base + strong acid 2 mol NH 3 + 1 mol HCl  1 mol weak acid: NH 4 +1 2 mol NaF + 1 mol HCl 

weak base: 1 mol Na

F 3

and 1 mol NH 3 + 1 mole

NH HF 4

Cl + NaCl weak acid: HF weak base: F -1

Example 3: Explain why a mixture of HCl and KCl does not function as a buffer, whereas a mixture of HC NaC 2 H 3 O 2 does.

2 H 3 O 2 and HCl is a strong acid -

Cl -1 is a negligible base and will NOT react with added H + - added H + will significantly change the pH of the solution HC 2 H 3 O 2 and C 2 H 3 O 2 -1 are a weak conjugate acid/base pair which act as a buffer HC 2 H 3 O 2 reacts with added base C 2 H 3 O 2 -1 reacts with added acid leaving the [H +1 ] and pH relatively unchanged

Buffer Capacity and pH

Buffer capacity

is the amount of acid or base that can be neutralized by the buffer before there is a significant change in pH.

Buffer capacity depends on the concentrations of the components of the buffer -

the greater the concentrations of the conjugate acid-base pair, the greater the buffer capacity

The pH of the buffer is related to

a and to the relative concentrations of the acid and base.

Henderson-Hasselbalch equation

– used for buffer solutions (on AP equation sheet!!)

A -



  

base acid

 

acid form HB +

 

acid base

 

base form These equations technically use the equilibrium concentrations of the acid (base) and the conjugate base (acid).

However, since the acid/base in the buffer is WEAK – the amount of the conjugate produced by dissociation is generally small compared to the amount of the conjugate added as a salt. IF this is true (it is for all AP buffer problems!) we do not need to do an equilibrium problem – just use the INITIAL concentrations .

Example 4 (Example 2 again!): Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC 3 H 5 O 2 and 0.085 M HC 3 H 5 O 2

HC H O (aq) 3 5 2



+1 3 5 -1 H (aq) + C H O (aq) K = 1.3 x 2 a 10 -5 in itial 0.085 0 0 .060

this is a BUFFER solution - use the Henerdson-Hasslebalch equation use the initial concentrations !

pH = pK + log a [base] [acid] = -5 - log(1.3 x 10 ) + log 0.060

0.085

= 4.73

Example 5: Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate

2 +1 -1 HCHO (aq) H (aq) + CHO (aq) K 2 a



1.4 x 1 0 -4 in iti al 0.12 Using the Henderson-Hasselbalch equation (and initial conc) pH = pK + log a [base] [acid] 4 = - log(1.4 x 10 ) + log 0.11

0.12

= 3.85 - .038 = 3.81

you would get the same answer doing a complete ICE chart

Example 6:

A buffer is prepared by adding 20.0 g of acetic acid, HC 2 H 3 O 2 20.0 g of sodium acetate to enough water to form 2.00 L of solution. and (a) Determine the pH of the buffer (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide are added to the buffer

(a) [HC H O ] = 2 3 2 20.0 g HC H O 2 3 2 2.00 L x 1 mol HC H O 2 3 60.0 g HC H O 2 3 2 2 = 0.167 M -1 [C H O ] = 2 3 2 20.0 g NaC H O 2 3 2 2.00 L x 1 mol NaC H O 2 3 2 82.0

g NaC 2 H O 3 2 = 0.1

22 M 2 3 2 +1 2 3 -1 HC H O (aq) H (aq) + C H O (aq) K = 1.8 2 a x 10 -5 in i tial 0.167 0.122

pH = pK + log a [base] [acid] = -5 - log(1.8 x 10 ) + log 0.122

0.16

7 = 4.74 - .14 = 4.60

(b) C 2 (c) HC H 2 3 H O 3 2 O -1 2 (aq) + H +1 (aq) + Cl (aq) + Na +1 -1 (aq) (aq) + OH



-1 HC 2 H 3 O 2 (aq)



C 2 (aq) + Cl H 3 O 2 -1 -1 (aq) (aq) + H 2 O (l) + Na +1 (aq)

17.3 Acid-Base Titrations – Titration Curves

In an acid-base titration: A solution of base (or acid) of known concentration (called standard) is added to an acid (or base).

Acid-base indicators or a pH meter are used to signal the equivalence point (when moles acid = moles base). The plot of pH versus volume during a titration is called a

pH titration curve

Starts high Ends low Equivalence point = 7 equal moles of acid and base present end

Strong acid

added

start

to strong base

Starts low Ends high Equivalence point = 7 end

Strong base

added

start

to strong acid

Starts med-high Ends low Equivalence point < 7 Buffer area – in this area there is weak base and some salt of the weak base Actual pH depends on the salt formed but it will be < 7

Strong acid

added

to weak base

Starts low Ends med-high Equivalence point < 7 end

Weak base

added

start

to strong acid

Starts high Ends med-low Equivalence point > 7 Actual pH depends on the salt formed but it will be > 7 end

Weak acid

added

start

to strong base

Starts med-high Ends med-low Equivalence point = 7 end

Weak acid

added

start

weak base

Strong base

added

to strong

diprotic

acid (H

SO

)

Example 7: Predict whether the equivalence point of each of the following titrations is below, above or at pH 7: a) NaHCO 3 titrated with NaOH b) NH 3 titrated with HCl c) KOH titrated with HBr At the equivalence point, only products are present in solution, so determine the products of the reaction and then determine if the solution is acidic, basic or neutral

a) NaHCO 3 + NaOH



Na 2 CO 3 + H 2 O weak acid strong base pH > 7 CO 3 -2 is basic, Na + is neutral, H 2 O is neutral b) NH 3 + HCl weak base strong acid



NH 4 Cl pH < 7 NH 4 +1 is acidic, Cl is neutral c) KOH + HBr



strong base strong acid K + and Br are both neutral KBr + H 2 O pH = 7

Example 8: How many mL of 0.0850 M NaOH solution is required to titrate 40.0 mL of 0.0900 M HNO 3 ?

? mL 40.0 mL 0.0850 M 0.0900 M NaOH + HNO 3 1 mole 1 mole



H 2 O + NaNO 3 0.090 mol HNO 3 x 1 L HNO 3 1 mol NaOH x 1 mol HNO 3 1 L NaOH 0.0850 mol NaOH = 0.0423 L or 42.3 mL

(a)

Example 9: A 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the pH of the solution after the following volumes of base solution have been added: (a) 15.0 mL (b) 19.9 mL (c) 20.0 mL (d) 20.1 mL (e) 35.0 mL

mL HBr mL NaOH 20.0

15.0

(b) (c) 20.0

20.0

(d) 20.0

(e) 20.0

19.9

20.0

20.1

35.0

mL Total 35.0

mol H +1 (M) (V) 0.00400

mol OH (M) (V) 0.00300

-1 M of excess ion (mol / tot vol) 0.0286 M H +1 (.200)(.0200) (.200)(.0150)

(.004 .003)

mol

0.0350 L pH 1.544

- log(.0286) 39.9

40.0

40.1

55.0

0.00400

0.00398

0.00400

0.00402

0.00700

0.0005 M H +1 1 x 10 -7 M H +1* 0.0005 M OH -1 0.0545 M OH -1 3.3

7.0

10.7

12.736

When molarity of H + (or OH ) is less than 10 -6 we must consider the autoionization of water! (H + = 1.0*10 -7 )

Example 10: (b) NH 2 OH Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr: (a) NaOH

(a) strong acid/strong base titration so pH = 7 (b) HBr strong acid + NH 2 OH



weak base Br + NH 2 OH 2 + .200M

.200M

all product at equivalence point – no excess & Br is neutral and will have no affect on pH Volume doubles (equal molarity and 1:1 stoich ratio) so molarity halves [NH 2 OH 2 + ] = 0.200mol / 2 = 0.100 M I NH 2 OH 2 + .100



H +1 + NH 2 OH K b = 1.1 x 10 -8 (appendix) 0 0 C -x +x +x E K a 0.100 – x x = K w / K b = 1



10 -14 x / 1.1



10 -8 = 9.1 (x 2 ) / (0.100-x) = 9.1 x 10 -7



10 -7 x = 3.0 x 10 -4 M = [H +1 ] pH = - log(3.0 x 10 -4 ) = 3.52

Acid – Base Indicators

The equivalence point of an acid-base titration can be determined by measuring pH, but it can also be determined by using an acid-base indicator which marks the end point of a titration by changing color. Although the equivalence point (defined by the stoichiometry) is not necessarily the same as the end point (where the indicator changes color), careful selection of the indicator can ensure that the difference between them is negligible.

Acid-base indicators are complex molecules that are themselves,

weak acids

(represented by HIn). They exhibit one color when the proton is attached and a different color when the proton is absent.

Acid – Base Indicators

Bromthymol Blue Indicator In Acid In Base

Acid – Base Indicators

Methyl Orange Indicator In Acid In Base

Acid – Base Indicators

Phenolphthalein color at different pH values pH values 5 6 7 8 9

Acid – Base Indicators

Consider a hypothetical indicator, HIn, a weak acid with K a =1.0x10

-8 . It has a red color in acid and a blue color in base.

HIn(aq)  H +1 (aq) + In -1 (aq) red blue Re arranging, we get K = a   [H ][In ] K a [H ] [HIn] [In ] = [HIn] or [HIn] = K a [H ] Suppose we add a few drops of indicator to an acid solution   1 whose pH = 1.0 ([H ] = 1.0 x 10 )  [In ] [HIn] = K a  [H ] = 1.0 x 10  8 1.0 x 10  1 =  7 10 = 1 10,000,000

Acid – Base Indicators

HIn(aq)  H +1 (aq) + In -1 (aq) red blue 1 10,000,000  [In ] = [HIn] This ratio shows that the predominant form of the indicator is HIn, resulting in a red solution. As OH -1 is added (like in a titration) [H +1 ] decreases and the equilibrium shifts to the right, changing HIn to In . At some point in the titration, enough of the In form will be present so we start to notice a color change.

Acid – Base Indicators

It can be shown (using the Henderson-Hasselbalch equation) that for a typical acid-base indicator with dissociation constant, K a , the color transition occurs over a range of pH values given by pK a ± 1. For example, bromthymol blue with K a = 1.0 x 10 -7 (pK a = 7), would have a useful pH range of 7 ± 1 or from 6 to 8. You want to select an indicator whose pKa value is close to the pH you want to detect (usually the pH at the equivalence point)

Acid – Base Indicators

The pH curve for the titration of 100.0 mL of 0.10

HCl with 0.10

NaOH. Neither of the indicators shown would be useful for a titration. Bromthymol blue (pK a =7) would be useful.

The pH curve for the titration of 50 mL of 0.1

HC 2 H 3 O 2 with 0.1

NaOH. Here, phenolphthalein is the indicator of choice. It has a pK a value of about 9.

Example 11 : Use the following table to determine which of the following would be the best indicator to use to indicate the equivalence point of the titrations described in Example 10.

Indicator

Methyl Yellow Methyl Red Bromthymol Blue Phenolpthalein

1*10 -4 1*10 -5 1*10 -7 1*10 -9 a. pH at equivalence point was 7.0

Bromthymol Blue b. pH at equivalence point was 3.52

Methyl Yellow

Solubility Equilibria & Complex Ions

17.4 – 17.6

Know your solubility rules: SOLUBILITY GUIDELINES Soluble Compounds Exceptions NOT precipitates PRECIPITATES

Nitrates Acetates Chlorates Chlorides Bromides Iodides Sulfates None We classify these based on the None Solubility - maximum None Ag +1 , Hg 2 +2 , Pb +2 Ag +1 , Hg 2 +2 , Pb +2 amount of solute that dissolves in water. Ag +1 , Hg 2 +2 , Pb +2 Ca +2 , Sr +2 , Ba +2 , Hg 2 +2 , Pb +2

Insoluble Compounds PRECIPITATES

Sulfides Carbonates Phosphates Hydroxides Chromates

Exceptions NOT Precipitates

NH 4 +1 , Li +1 , Na +1 , K +1 , Ca +2 , Sr +2 , Ba +2 NH 4 +1 , Li +1 , Na +1 , K +1 NH 4 +1 , Li +1 , Na +1 , K +1 Li +1 , Na +1 , K +1 , Ca +2 , Sr +2 , Ba +2 NH 4 +1 , Li +1 , Na +1 , K +1 , Ca +2 , Mg +2

17.4 Solubility Equilibria The Solubility-Product Constant, K sp

Consider a saturated solution of BaSO 4 in contact with solid BaSO solid. 4 . We can write an equilibrium expression for the dissolving of the BaSO 4 (

)  Ba 2+ (

) + SO 4 2 – (

) Since BaSO 4 (

) is a pure solid, the equilibrium expression depends only on the concentration of the ions.

sp = [Ba 2+ ][ SO 4 2 – ]

sp is the equilibrium constant for the equilibrium between an ionic solid solute and its saturated aqueous solution.

sp is called the

solubility-product constant

• In general: the solubility product is equal to the product of the molar concentration of ions raised to powers corresponding to their stoichiometric coefficients.

Al

(CO

)



2 Al

+ 3 CO

3 -2 K sp = [Al +3 ] 2 [CO 3 -2 ] 3

Solubility and K

Solubility

is the amount of substance that dissolves to form a saturated solution.

This can be expressed as grams of solid that will dissolve per liter of solution.

Molar solubility

- the number of moles of solute that dissolve to form a liter of saturated solution.

Solubility can be used to find K sp and K sp used to find solubility (see problems) can be

Example 1: a. If the molar solubility of CaF 2 at this temperature?

at 35 o C is 1.24



10 -3 mol/L, what is K sp CaF 2



E .00124M actually dissolves Ca +2 + 2 F -1 .00124 M 2(.00124) = .00248 M K sp = [Ca +2 ] [F -1 ] 2 = (.00124 M)(.00248 M) 2 = 7.63 x 10 -9 b. It is found that 1.1



10 -2 g of SrF 2 dissolves per 100 mL of aqueous solution at 25 o C. Calculate the solubility product of SrF 2 .

[SrF 2 ] = (.011 g / 125.6 g/mole) / .100 L = .00088 M SrF 2



Sr +2 + 2 F E .00088 M .00088 M -1 2(.00088) = .00176 M K sp = [Sr +2 ] [F -1 ] 2 = (.00088 M)(.00176) 2 = 2.7 x 10 -9

c. The K

of Ba(IO

)

at 25

C is 6.0



10 What is the molar solubility of Ba(IO

)

?

-10

. Ba(IO

)

E x



Ba

+2 x

+ 2 IO

3 -1 2 x

(x) (2x)

= 4 x

= 6.0 x 10

-10

x = 5.3 x 10

-4

M

17.5 Factors That Affect Solubility

Factors that have a significant impact on solubility are: - The presence of a common ion - The pH of the solution

Common-Ion Effect

Solubility is decreased when a common ion is added.

This is an application of Le Châtelier’s principle: Consider the solubility of CaF 2 : CaF 2 (

)  Ca 2+ (

) + 2F – (

) If more F – is added (say by the addition of NaF), the equilibrium shifts left to offset the increase.

Therefore, more CaF 2 (

) is formed (precipitation occurs).

Example 2: Using Appendix D, calculate the molar solubility of AgBr in (a) pure water (b) 3.0



10 -2 M AgNO 3 solution (c) 0.50 M NaBr solution (a) AgBr



Ag +1 + Br -1 E x x x 5.0



10 -13 x = 7.1



= x 10 -7 2 M K sp = 5.0



10 -13 (b) AgBr



Ag +1 + Br E x .030 + x x 5.0



10 -13 = (.030+x) x x = 1.7



10 -11 M -1 K sp = 5.0



10 -13 (c) AgBr



Ag +1 + Br -1 x x .50 + x 5.0



10 -13 x = 1.0



= x (.50+x) 10 -12 M K sp = 5.0



10 -13 notice the DECREASED solubility with the common ion in (b) and (c)

pH effects

Consider: Mg(OH) 2 (

)  Mg 2+ (

) + 2 OH – (

) If OH – is removed, then the equilibrium shifts right and Mg(OH) 2 dissolves.

OH – can be removed by adding a strong acid (lowering the pH):

OH – (

) + H + (

)  H 2 O(

) Another example: CaF 2 (

)  Ca 2+ (

) + 2 F – (

) If the F – is removed, then the equilibrium shifts right and CaF 2 dissolves.

F – can be removed by adding a strong acid (or lowering pH):

F – (

) + H + (

)  HF(

)

Example 3: Calculate the molar solubility of Mn(OH) 2 at (a) pH 7.0 (b) pH 9.5 (c) pH 11.8

the [OH -1 ] is set by the pH (or pOH) (a) pH = 7.0 so pOH = 7.0 so [OH -1 ] = 1.00



Mn(OH) 2



Mn +2 + 2 OH -1 x x 1.00 x 10 -7 10 -7 1.6



K sp 10 = [Mn +2 ][OH -1 ] 2 -13 = (x) (1.00



10 -7 ) 2 x = 16 M K sp = 1.6



10 -13 (b) pH = 9.5 so pOH = 4.5 so [OH -1 ] = 3.16



Mn(OH) 2



Mn +2 + 2 OH -1 x x 3.16 x 10 -5 10 -5 1.6



K sp = [Mn 10 -13 +2 = (x) (3.16 x = 1.7



][OH 10 -4 -1 M



] 2 10 -5 ) 2 K sp = 1.6



10 -13 (c) pH = 11.8 so pOH = 2.2 so [OH -1 ] = 6.31



Mn(OH) 2



Mn +2 + 2 OH -1 x x 6.31 x 10 -3 10 -3 1.6



K sp = [Mn 10 -13 +2 = (x) (6.31 x = 4.0



][OH 10 -9 -1 M



] 2 10 -3 ) 2 K sp = 1.6



10 -13 Common ion effect – increasing [OH ] decreases solubility

Example 4: Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO 3 (b) ZnS (c) BiI 3 (d) AgCN (e) Ba 3 (PO 4 ) 2 If the anion of the salt is the conjugate base of a weak acid, it will combine with H +1 , reducing the concentration of the anion and making the salt more soluble ZnCO 3 the CO 3 -2



Zn +2 + CO 3 -2 ion will react with the added H + CO 3 -2 + H +



HCO 3 -1 Le Chatelier effect of removing CO 3 -2 more soluble in acid: ZnCO 3 , ZnS, AgCN, Ba 3 (PO 4 ) 2

17.6 Precipitation and Separation of Ions

Consider the following: BaSO 4 (

)  At any instant in time,

Ba 2+ (

) + SO 4 2 – (

) = [Ba 2+ ][ SO 4 2 – ] If

sp , =

(too many ions)

precipitation occurs until equilibrium exists (saturated solution) <

sp ,

(not enough ions)

solid dissolves until

Q Q

sp . sp

Selective Precipitation of Ions

Removal of one metal ion from a solution of two or more metal ions is called

selective precipitation.

Ions can be separated from each other based on the solubilities of their salt compounds.

Example: If HCl is added to a solution containing Ag + the silver precipitates (as AgCl) while the Cu 2+ and Cu remains in 2+ , solution Generally, the less soluble ion is removed first!

Example 5 : Will Ca(OH) 2 a 0.050 M solution of CaCl 2 precipitate if the pH of is adjusted to 8.0?

if Q > than K sp then precipitation will occur pH = 8.0 so pOH = 6.0 so [OH -1 ] = 1.0



10 -6 Ca(OH) 2



Ca +2 + 2 OH -1 K sp = 6.5



.050 1.00 x 10 -6 M 10 -6 Q = [Ca +2 ] [OH -1 ] 2 Q = (.050)(1.0



10 -6 ) 2 = 5.0



10 -14 Q < K so no precipitation occurs

Example 6: A solution contains 0.00020 M Ag +1 0.0015 M Pb +2 . If NaI is added, will AgI or PbI 2 precipitate first? Specify the [I -1 ] needed to begin and precipitation for each cation.

the cation needing the lower [I -1 ] will precipitate first AgI



Ag +1 + I -1 8.3



4.2



.000200 x K sp 10 -17 = [Ag +1 ][x] = (.00020)[x] 10 -13 = x = [I -1 ] K sp = 8.3 x 10 -17 PbI 2



Pb +2 + 2 I -1 1.4 0.0015 x K sp = [Pb +2 ][x] 2



3.1



10 -8 10 -3 = (.0015)[x] = x = [I -1 ] 2 K sp = 1.4 x 10 AgI will precipitate first at an [I -1 ] = 4.2



10 -13 -8

Complex Ions

• Complex ion – a metal ion bonded to one or more Lewis bases. (We saw this with water in chapter 16)

• It can happen with other Lewis bases (things that have lone pairs of electrons)

• Rule of thumb:

The number of Lewis bases (ligands) that a metal ion attracts is equal to double its charge. (Works about 75% of the time!)

• Extra Stuff Below…

Acid – Base Indicators

How much In must be present for the human eye to detect that the color is different? For most indicators, about 1/10 of the initial form must be converted to the other form before a color change is apparent. We can assume that in the titration of an acid with a base, the color change will occur at a pH where [HIn] 1 = 10

Acid – Base Indicators

Bromthymol blue, an indicator with a K a HIn form and blue in its In = 1.0 x 10 -7 , is yellow in its form. Suppose we put some strong acid in a flask, add a few drops of bromthymol blue and titrate with NaOH. At what pH will the indicator color change first be visible?

HIn(aq)  H +1 (aq) + In -1 (aq) yellow blue K = 1.0 x 10 = a  7   [H ][In ] [HIn]  [In ] we assume that the color change is visible when [HIn] 1 = 10 thus K = 1 x 10 = a  7  [H ](1) (10)   6 and [H ] = 1.0 x 10 or pH = 6.00

Selective Precipitation of Ions (continued)

Sulfide ion is often used to separate metal ions. Example: Consider a mixture of Zn 2+ (

) and Cu 2+ (

). CuS (

sp = 6 x 10 –37 ) is less soluble than ZnS (

sp = 2 x 10 –25 ). Because

CuS is LESS SOLUBLE than ZnS, CuS will be removed from solution before ZnS.

As H 2 S is bubbled through the acidified green solution, black CuS forms. When the precipitate is removed, a colorless solution containing Zn 2+ (

) remains.

When more H 2 S is added to the solution, a second precipitate of white ZnS forms.

Formula Type of Formula of Hydrolysis equation Hydrolysis equation acid or conjugate of the acid of the base base acid or base HCl Strong acid Cl -1 HCl + H 2 O



H 3 O +1 + Cl -1 Cl -1 + H 2 O



X HOCl Weak acid OCl -1 HOCl + H 2 O



H 3 O +1 + OCl -1 OCl -1 + H 2 O



HOCl + OH -1 NH 3 Weak base NH 4 +1 NH 4 +1 + H 2 O



H 3 O +1 + NH 3 NH 3 + H 2 O



OH -1 + NH 4 +1 Ba(OH) 2 Strong base H 2 O KI neutral K +1 or I -1 H 2 O + H 2 O



H 3 O +1 + OH -1 K +1 + H 2 O



X OH -1 + H 2 O



X I -1 + H 2 O



X NaC 2 H 3 O 2 weak base HC 2 H 3 O 2 HC 2 H 3 O 2 + H 2 O



H 3 O +1 + C 2 H 3 O 2 -1 C 2 H 3 O 2 -1 + H 2 O



HC 2 H 3 O 2 + OH -1