More Acid/Base Equilibria!!! 17.1 – 17.3. 17.1 The Common Ion Effect Consider the ionization of a...

More Acid/Base

Equilibria!!!17.1 – 17.3

17.1 The Common Ion EffectConsider the ionization of a weak acid, acetic acid:

HC2H3O2(aq) H+(aq) + C2H3O2–(aq)

If we increase the [C2H3O2–] ions by adding NaC2H3O2, 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) NaNO2 to a solution of HNO2 (b) (CH3NH3)Cl to a solution of CH3NH2 (c) sodium formate to a solution of formic acid (d) potassium bromide to a solution of hydrobromic acid (e) HCl to a solution of NaC2H3O2

(a) HNO2 H+ + NO2-

increases(b) CH3NH2 + H2O CH3NH3

+ + OH-

decreases(c) HCHO2 H+ + CHO2

-

increases(d) HBr H+ + Br-

no change(e) C2H3O2

-1 + H2O HC2H3O2 + OH-1 decreases

Example 2: Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC3H5O2 and 0.085 M HC3H5O2

change: -x +x +x

Equilibrium: 0.085-x x .060+ x

x2+.060013x-(1.105*10-6)=0

x=1.84*10-5

pH = -log(1.84*10-5) = 4.74

+1 -1 -53 5 2 3 5 2 a HC H O (aq) H (aq) + C H O (aq) K = 1.3 x

0.085 0

10

in 0itial .060

5(.060 )( )1.3*10

(.085 )

x x

x

17. 2 Buffered SolutionsA 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–

When a small amount of OH– is added to the buffer solution, the OH– 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 acidHCN and NaCN

weak acid: HCN weak base: CN-1

2.) Weak base + salt of the baseNH3 and NH4 Cl

weak acid: NH4+1 weak base: NH3

3.) EXCESS Weak acid + strong base2 mol HCN + 1 mol NaOH 1 mol HCN + 1 mol NaCN

+ H2Oweak acid: HCN weak base: CN-1

2 mol NH4Cl + 1 mol NaOH 1 mol NH4Cl + 1 mol NH3 + NaCl


4.) EXCESS Weak base + strong acid 2 mol NH3 + 1 mol HCl 1 mol NH3 and 1 mol NH4Cl


2 mol NaF + 1 mol HCl 1 mol NaF + 1 mole HF + 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 HC2H3O2 and NaC2H3O2 does.

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

HC2H3O2 and C2H3O2-1 are a weak conjugate acid/base

pair which act as a buffer

HC2H3O2 reacts with added base

C2H3O2-1 reacts with added acid

leaving the [H+1] and pH relatively unchanged

Buffer Capacity and pHBuffer 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 Ka and to the relative concentrations of the acid and base.

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

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.

acid form

-

a a A base

pH = pK + log = pK + log HA acid

base form

+

b b HB acid

pOH = pK + log = pK + log B base

Example 4 (Example 2 again!): Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC3H5O2 and 0.085 M HC3H5O2

a

[base]pH = pK + log =

[acid]

this is a BUFFER solution - use the Henerdson-Hasslebalch equation

use the initial concentrations !

-5 0.060- log(1.3 x 10 ) + log =

0.0854.73

+1 -1 -53 5 2 3 5 2 a HC H O (aq) H (aq) + C H O (aq) K = 1.3 x

0.085 0

10

in 0itial .060

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

+1 -1 -42 2 a HCHO (aq) H (aq) + CHO (aq) K 1.4 x 1

0.12

0

in iti 1al 0. 1

D

-a

4 0.11 - log(1.4 x 10 ) + log = 3.85 - .038 = 3.81

0.12


[acid]

Using the Henderson-Hasselbalch equation (and initial conc)

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, HC2H3O2 and 20.0 g of sodium acetate to enough water to form 2.00 L of solution.

(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

(b) C2H3O2-1(aq) + H+1(aq) + Cl-1(aq) HC2H3O2 (aq) + Cl-1 (aq)

(c) HC2H3O2 (aq) + Na+1 (aq) + OH-1 (aq) C2H3O2-1 (aq) + H2O (l) + Na+1 (aq)

2 3 2 2 3 22 3 2

2 3 2

20.0 g HC H O 1 mol HC H O(a) [HC H O ] = x

2.00 L 60.0 g HC H O = 0.167 M

-1 2 3 2 2 3 22 3 2

2 3 2

20.0 g NaC H O 1 mol NaC H O [C H O ] = x

2.00 L 82.0 =

g NaC0.1

H O22 M

-5a

0.122 - log(1.8 x 10 ) + log = 4.74 - .14 = 4.60

0.16


[acid] 7

+1 -1 -52 3 2 2 3 2 a HC H O (aq) H (aq) + C H O (aq) K = 1.8

0.167

x 10

in i 0.122tial

D

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.

Strong acid added to strong base

equal moles of acid and base present

Starts highEnds lowEquivalence point = 7

startend

Strong base added to strong acid

Starts lowEnds highEquivalence point = 7

startend

Strong acid added to weak base

Buffer area – in this area there is weak base and some salt of the weak base

Actual pH depends on the salt formedbut it will be < 7

Starts med-highEnds lowEquivalence point < 7

Weak base added to strong acid

Starts lowEnds med-highEquivalence point < 7

startend

Weak acid added to strong base

Actual pH depends on the salt formedbut it will be > 7

Starts highEnds med-lowEquivalence point > 7

startend

Weak acid added weak base

Starts med-highEnds med-lowEquivalence point = 7

startend

Strong base added to strong diprotic acid (H2SO4)

Example 7: Predict whether the equivalence point of each of the following titrations is below, above or at pH 7:

a) NaHCO3 titrated with NaOH b) NH3 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) NaHCO3 + NaOH Na2CO3 + H2O

weak acid strong base pH > 7

CO3-2 is basic, Na+ is neutral, H2O is neutral

b) NH3 + HCl NH4Cl

weak base strong acid pH < 7

NH4+1 is acidic, Cl- is neutral

c) KOH + HBr KBr + H2O

strong base strong acid pH = 7

K+ and Br- are both neutral

Example 8: How many mL of 0.0850 M NaOH solution is required to titrate 40.0 mL of 0.0900 M HNO3?

? mL 40.0 mL

0.0850 M 0.0900 M

NaOH + HNO3 H2O + NaNO3

1 mole 1 mole

30.0400 L HNO x 3

3

0.090 mol HNO x

1 L HNO 3

1 mol NaOH x

1 mol HNO

1 L NaOH

0.0850 mol NaOH= 0.0423 L or 42.3 mL

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 mL mL mol H+1 mol OH-1 M of pH

HBr NaOH Total (M) (V) (M) (V) excess ion

(mol / tot vol)

(a) 20.0 15.0 35.0 0.00400 0.00300 0.0286 M H+1 1.544

(b) 20.0 19.9 39.9 0.00400 0.00398 0.0005 M H+1 3.3

(c) 20.0 20.0 40.0 0.00400 0.00400 1 x 10-7 M H+1* 7.0

(d) 20.0 20.1 40.1 0.00400 0.00402 0.0005 M OH-1 10.7

(e) 20.0 35.0 55.0 0.00400 0.00700 0.0545 M OH-1

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)

(.200)(.0200) (.200)(.0150) (.004 .003)mol0.0350 L

- log(.0286)

Example 10: 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 (b) NH2OH

(a) strong acid/strong base titration so pH = 7

(b) HBr + NH2OH Br- + NH2OH2+

strong acid weak base.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

[NH2OH2+] = 0.200mol / 2 = 0.100 M

NH2OH2+ H+1 + NH2OH Kb = 1.1 x 10-8

(appendix)

I .100 0 0

C -x +x +x

E 0.100 – x x x

Ka = Kw / Kb = 1 10-14 / 1.1 10-8 = 9.1 10-7

(x2) / (0.100-x) = 9.1 x 10-7

x = 3.0 x 10-4 M = [H+1]

pH = - log(3.0 x 10-4) = 3.52

Acid – Base IndicatorsThe 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


Methyl Orange Indicator

In Acid In Base


Phenolphthalein color at different pH values

pH values 5 6 7 8 9

Acid – Base IndicatorsConsider a hypothetical indicator, HIn, a weak acid with Ka=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

a

[H ][In ]K =

[HIn]

a aK K[In ] [In ]Rearranging,we get = or =

[H ] [HIn] [HIn] [H ]

1

Suppose we add a few drops of indicator to an acid solution

whose pH = 1.0 ([H ] = 1.0 x 10 )

8a

1

K[In ] 1.0 x 10 = = =

[HIn] [H ] 1.0 x 10

710 = 1

10,000,000



red blue

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.

1 [In ] =

10,000,000 [HIn]


It can be shown (using the Henderson-Hasselbalch equation) that for a typical acid-base indicator with dissociation constant, Ka, the color transition occurs over a range of pH

values given by pKa ± 1.

For example, bromthymol blue with Ka = 1.0 x 10-7 (pKa = 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)

The pH curve for the

titration of 100.0 mL of

0.10 M HCl with 0.10 M

NaOH. Neither of the

indicators shown would

be useful for a

titration.

Bromthymol

blue (pKa=7) would be

useful.

The pH curve for the

titration of 50 mL of

0.1 M HC2H3O2 with

0.1 M NaOH.

Here, phenolphthalein

is the indicator of

choice. It has a pKa

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.

a. pH at equivalence point was 7.0

Bromthymol Blue

b. pH at equivalence point was 3.52

Methyl Yellow

Indicator Ka

Methyl Yellow 1*10-4

Methyl Red 1*10-5

Bromthymol Blue 1*10-7

Phenolpthalein 1*10-9

Solubility Equilibria & Complex Ions

17.4 – 17.6

Know your solubility rules:SOLUBILITY GUIDELINES

Soluble Compounds Exceptions

NOT precipitates PRECIPITATES

Nitrates None

Acetates None

Chlorates None

Chlorides Ag+1, Hg2

+2, Pb+2

Bromides Ag+1, Hg2

+2, Pb+2

Iodides Ag+1, Hg2

+2, Pb+2

Sulfates Ca+2, Sr+2

, Ba+2, Hg2

+2, Pb+2

Insoluble Compounds Exceptions

PRECIPITATES NOT Precipitates

Sulfides NH4+1

, Li+1, Na+1

, K+1, Ca+2

, Sr+2, Ba+2

Carbonates NH4+1

, Li+1, Na+1

, K+1

Phosphates NH4+1

, Li+1, Na+1

, K+1

Hydroxides Li+1, Na+1

, K+1, Ca+2

, Sr+2, Ba+2

Chromates NH4+1

, Li+1, Na+1

, K+1, Ca+2

, Mg+2

We classify these based on the Solubility - maximum amount of solute that dissolves in water.

17.4 Solubility Equilibria

The Solubility-Product Constant, Ksp

Consider a saturated solution of BaSO4 in contact with solid BaSO4.

We can write an equilibrium expression for the dissolving of the solid.

BaSO4(s) Ba2+(aq) + SO42–(aq)

Since BaSO4(s) is a pure solid, the equilibrium expression depends only on the concentration of the ions.

Ksp = [Ba2+][ SO42–]

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

Ksp 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.

Al2(CO3)3 2 Al+3 + 3 CO3-2

Ksp = [Al+3]2 [CO3-2]3

Solubility and Ksp

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 Ksp and Ksp can be used to find solubility (see problems)

Example 1: a. If the molar solubility of CaF2 at 35oC is 1.24 10-3 mol/L, what is Ksp at

this temperature?

CaF2 Ca+2 + 2 F-1 E .00124M actually dissolves

Ksp = [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 SrF2 dissolves per 100 mL of aqueous solution at 25oC. Calculate the solubility product of

SrF2. [SrF2] = (.011 g / 125.6 g/mole) / .100 L = .00088 M

SrF2 Sr+2 + 2 F-1 E .00088 M .00088 M 2(.00088)

= .00176 M

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

.00124 M 2(.00124) = .00248 M

c. The Ksp of Ba(IO3)2 at 25oC is 6.0 10-10. What is the molar solubility of Ba(IO3)2?

Ba(IO3)2 Ba+2 + 2 IO3

-1

E x

(x) (2x)2 = 4 x3 = 6.0 x 10-10

x = 5.3 x 10-4 M

x 2 x

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 CaF2:

CaF2(s) Ca2+(aq) + 2F–(aq)

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

Therefore, more CaF2(s) 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 AgNO3 solution (c) 0.50 M NaBr solution

(a) AgBr Ag+1 + Br-1 Ksp = 5.0 10-13 E x x x 5.0 10-13 = x2

x = 7.1 10-7 M

(b) AgBr Ag+1 + Br-1 Ksp = 5.0 10-13 E x .030 + x x

5.0 10-13 = (.030+x) x x = 1.7 10-11 M

(c) AgBr Ag+1 + Br-1 Ksp = 5.0 10-13 x x .50 + x 5.0 10-13 = x (.50+x)

x = 1.0 10-12 M

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

pH effects

Consider: Mg(OH)2(s) Mg2+(aq) + 2 OH–(aq)

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–(aq) + H+(aq) H2O(aq)

Another example:

CaF2(s) Ca2+(aq) + 2 F–(aq)

If the F– is removed, then the equilibrium shifts right and CaF2 dissolves.

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

F–(aq) + H+(aq) HF(aq)

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 10-7 Mn(OH)2 Mn+2 + 2 OH-1 Ksp = 1.6 10-13

x x 1.00 x 10-7 Ksp = [Mn+2][OH-1]2

1.6 10-13 = (x) (1.00 10-7)2

x = 16 M

(b) pH = 9.5 so pOH = 4.5 so [OH-1] = 3.16 10-5 Mn(OH)2 Mn+2 + 2 OH-1 Ksp = 1.6 10-13

x x 3.16 x 10-5 Ksp = [Mn+2][OH-1]2

1.6 10-13 = (x) (3.16 10-5)2

x = 1.7 10-4 M

(c) pH = 11.8 so pOH = 2.2 so [OH-1] = 6.31 10-3 Mn(OH)2 Mn+2 + 2 OH-1 Ksp = 1.6 10-13

x x 6.31 x 10-3 Ksp = [Mn+2][OH-1]2

1.6 10-13 = (x) (6.31 10-3)2

x = 4.0 10-9 M

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) ZnCO3 (b) ZnS (c) BiI3 (d) AgCN (e) Ba3(PO4)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

ZnCO3 Zn+2 + CO3-2

the CO3-2 ion will react with the added H+

CO3-2 + H+ HCO3

-1 Le Chatelier effect of removing CO3

-2

more soluble in acid: ZnCO3, ZnS, AgCN, Ba3(PO4)2

17.6 Precipitation and Separation of Ions

Consider the following:

BaSO4(s) Ba2+(aq) + SO42–(aq)

At any instant in time, Q = [Ba2+][ SO42– ]

If Q > Ksp, (too many ions) precipitation occurs until Q = Ksp. If Q = Ksp equilibrium exists (saturated solution)

If Q < Ksp, (not enough ions) solid dissolves until Q = Ksp.

Selective Precipitation of IonsRemoval 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+ and Cu2+, the

silver precipitates (as AgCl) while the Cu2+ remains in solutionGenerally, the less soluble ion is removed first!

Example 5: Will Ca(OH)2 precipitate if the pH of a 0.050 M solution of CaCl2 is adjusted to 8.0?

if Q > than Ksp then precipitation will occurpH = 8.0 so pOH = 6.0 so [OH-1] = 1.0 10-6 M

Ca(OH)2 Ca+2 + 2 OH-1 Ksp = 6.5 10-6 .050 1.00 x 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 and 0.0015 M Pb+2 . If NaI is added, will AgI or PbI2 precipitate first? Specify the [I-1] needed to begin precipitation for each cation.

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

AgI Ag+1 + I-1 Ksp = 8.3 x 10-17

.000200 x Ksp = [Ag+1][x]

8.3 10-17 = (.00020)[x] 4.2 10-13 = x = [I-1]

PbI2 Pb+2 + 2 I-1 Ksp = 1.4 x 10-8

0.0015 xKsp = [Pb+2][x]2

1.4 10-8 = (.0015)[x]2

3.1 10-3 = x = [I-1]

AgI will precipitate first at an [I-1] = 4.2 10-13

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…


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

[In ] 1 =

[HIn] 10

Acid – Base IndicatorsBromthymol blue, an indicator with a Ka = 1.0 x 10-7, is yellow in its

HIn form and blue in its In- 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?


yellow blue

7a

[H ][In ] K = 1.0 x 10 =

[HIn]

[In ] 1we assume that the color change is visible when =

[HIn] 10

7a

[H ](1)thus K = 1 x 10 =

(10)

6and [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 Zn2+(aq) and Cu2+(aq).

CuS (Ksp= 6 x 10–37) is less soluble than ZnS (Ksp= 2 x 10–25).

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

As H2S is bubbled through the acidified green solution, black

CuS forms.

When the precipitate is removed, a colorless solution containing Zn2+(aq) remains.

When more H2S 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

HOCl

NH3

Ba(OH)2

KI

NaC2H3O2

Strong acid Cl-1 HCl + H2O H3O+1 + Cl-1 Cl-1 + H2O X

Weak acid OCl-1 HOCl + H2O H3O+1 + OCl-1 OCl-1 + H2O HOCl + OH-1

Weak base NH4+1 NH4

+1 + H2O H3O+1 + NH3 NH3 + H2O OH-1 + NH4+1

Strong base H2O H2O + H2O H3O+1 + OH-1 OH-1 + H2O X

neutral K+1 or I-1 K+1 + H2O X I-1 + H2O X

weak base HC2H3O2 HC2H3O2 + H2O H3O+1 + C2H3O2-1

C2H3O2-1 + H2O HC2H3O2 + OH-1

More Acid/Base Equilibria!!! 17.1 – 17.3. 17.1 The Common Ion Effect Consider the ionization of a...

Documents

Transcript of More Acid/Base Equilibria!!! 17.1 – 17.3. 17.1 The Common Ion Effect Consider the ionization of a...