chapter 14 powerpoint Cahng
Transcript of chapter 14 powerpoint Cahng
٠٩/١٢/١٤٣٥
١
Chemical EquilibriumAcid and bases
equilibriaequilibria
Chapter 5
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 5
Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when:
the rates of the forward and reverse reactions are equal and• the rates of the forward and reverse reactions are equal and
• the concentrations of the reactants and products remain constant
Physical equilibrium
H O (l) H O (g)H2O (l)
Chemical equilibrium
N2O4 (g)
14.1
H2O (g)
2NO2 (g)
٠٨/١٢/١٤٣٥
٢
N2O4 (g) 2NO2 (g)
equilibrium
equilibrium
equilibriumequilibrium
Start with NO2 Start with N2O4 Start with NO2 & N2O4
14.1
constant
14.1
٠٨/١٢/١٤٣٥
٣
N2O4 (g) 2NO2 (g)
= 4.63 x 10-3K = [NO2]2
[N2O4]
aA + bB cC + dD
K = [C]c[D]d
[A]a[B]bLaw of Mass Action
K >> 1
K << 1
Lie to the right Favor products
Lie to the left Favor reactants
Equilibrium Will
14.1
Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) 2NO2 (g)
[NO ]2 NOP2
Kc = [NO2]2
[N2O4]Kp =
NO2P
N2O4P
In most cases
Kc ≠ Kp
aA (g) + bB (g) cC (g) + dD (g)(g) (g) (g) (g)
14.2
Kp = Kc(RT)∆n
∆n = moles of gaseous products – moles of gaseous reactants= (c + d) – (a + b)
٠٨/١٢/١٤٣٥
٤
Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
[CH COO ][H O+]Kc =‘ [CH3COO-][H3O+]
[CH3COOH][H2O][H2O] = constant
Kc = [CH3COO-][H3O+]
[CH3COOH] = Kc [H2O]‘
General practice not to include units for the equilibrium constant.
14.2
The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] =0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)CO (g) + Cl2 (g) COCl2 (g)
Kc = [COCl2]
[CO][Cl2]=
0.140.012 x 0.054
= 220
Kp = Kc(RT)∆n
∆n = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
14.2
٠٨/١٢/١٤٣٥
٥
The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2
2NO2 (g) 2NO (g) + O2 (g)
PNO PO2
Kp = 2PNO PO
PNO2
2
PO2 = KpPNO
22
PNO2
14.2
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
Kc =‘ [CaO][CO2][CaCO3]
[CaCO3] = constant[CaO] = constant
Kc = [CO2] = Kc x‘ [CaCO3][CaO] Kp = PCO2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
14.2
٠٨/١٢/١٤٣٥
٦
CaCO3 (s) CaO (s) + CO2 (g)
PCO2 = Kp
PCO2 does not depend on the amount of CaCO3 or CaO
14.2
Consider the following equilibrium at 295 K:
The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = PNH3 H2SP = 0.265 x 0.265 = 0.0702
Kp = Kc(RT)∆n
Kc = Kp(RT)-∆n
∆n = 2 – 0 = 2 T = 295 K
Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4
14.2
٠٨/١٢/١٤٣٥
٧
A + B C + D
C + D E + F
A + B E + F
Kc =‘ [C][D][A][B] Kc =‘‘ [E][F]
[C][D]
[E][F][A][B]
Kc =
Kc‘Kc‘‘Kc
Kc = Kc‘‘Kc‘ x
If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the o erall reaction is gi en bconstant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
14.2
N2O4 (g) 2NO2 (g)
= 4.63 x 10-3K = [NO2]2
[N2O4]
2NO2 (g) N2O4 (g)
K = [N2O4][NO2]2
‘ =1K = 216
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
14.2
٠٨/١٢/١٤٣٥
٨
Writing Equilibrium Constant Expressions
• The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.
Th t ti f lid li id d l t• The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
• The equilibrium constant is a dimensionless quantity.
• In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
• If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
14.2
Chemical Kinetics and Chemical Equilibrium
A + 2B AB2
kf
kr
ratef = kf [A][B]2
rater = kr [AB2]
Equilibriumratef = rater
kf [A][B]2 = kr [AB2]
kf [AB2]= K =
14.3
kr [A][B]2= Kc =
٠٨/١٢/١٤٣٥
٩
The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.
IF• Qc > Kc system proceeds from right to left to reach equilibriumc c y g
• Qc = Kc the system is at equilibrium
• Qc < Kc system proceeds from left to right to reach equilibrium
14.4
Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, e s o e a co ce a o s a d a s g e u o ,which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium concentrations of all species.
14.4
٠٨/١٢/١٤٣٥
١٠
At 12800C the equilibrium constant (Kc) for the reaction
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.
Br2 (g) 2Br (g)
Br2 (g) 2Br (g)
Let x be the change in concentration of Br2
Initial (M)
Change (M)
0.063 0.012
-x +2xEquilibrium (M) 0.063 - x 0.012 + 2x
[Br]2[Br2]
Kc = Kc = (0.012 + 2x)2
0.063 - x = 1.1 x 10-3 Solve for x
14.4
Kc = (0.012 + 2x)2
0.063 - x = 1.1 x 10-3
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x4x2 + 0.0491x + 0.0000747 = 0
ax2 + bx + c =0 -b ± b2 – 4ac√2x =ax + bx + c 0 2ax
Br2 (g) 2Br (g)
Initial (M)
Change (M)
0.063 0.012
-x +2x
x = -0.00178x = -0.0105
Change (M)
Equilibrium (M)
x +2x0.063 - x 0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 MAt equilibrium, [Br2] = 0.062 – x = 0.0648 M
14.4
٠٨/١٢/١٤٣٥
١١
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Le Châtelier’s Principle
Ch i C t ti• Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
AddEquilibrium shifts left to NH3shifts left to offset stress
14.5
Le Châtelier’s Principle
• Changes in Concentration continuedAddAddRemove Remove
Change Shifts the Equilibrium
aA + bB cC + dD
Increase concentration of product(s) leftDecrease concentration of product(s) right
Decrease concentration of reactant(s)Increase concentration of reactant(s) right
left14.5
٠٨/١٢/١٤٣٥
١٢
Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g) C (g)
Change Shifts the Equilibrium
Increase pressure Side with fewest moles of gasDecrease pressure Side with most moles of gasI l Sid i h l fDecrease volumeIncrease volume Side with most moles of gas
Side with fewest moles of gas
14.5
Le Châtelier’s Principle
• Changes in Temperature
Change Exothermic Rx Endothermic Rx
Increase temperature K decreasesDecrease temperature K increases
K increasesK decreases
14.5colder hotter
٠٨/١٢/١٤٣٥
١٣
• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner
Le Châtelier’s Principle
y q
uncatalyzed catalyzed
14.5
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
Chemistry In Action
Life at High Altitudes and Hemoglobin Production
Kc = [HbO2]
[Hb][O2]
Hb (aq) + O2 (aq) HbO2 (aq)
[ ][ 2]
٠٨/١٢/١٤٣٥
١٤
Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g) 2NH3 (g) ∆H0 = -92.6 kJ/mol
Le Châtelier’s Principle
Change Shift EquilibriumChange Equilibrium
Constant
Concentration yes noConcentration yes no
Pressure yes no
Volume yes no
Temperature yes yes
Catalyst no noCatalyst no no
14.5
٠٨/١٢/١٤٣٥
١٥
Acids and BasesAcids and Bases
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acidfruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon dioxide gas
Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Bases
4.3
٠٨/١٢/١٤٣٥
١٦
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
4.3
A Brønsted acid is a proton donorA Brønsted base is a proton acceptor
acidbase acid base
15.1
acid conjugatebasebase conjugate
acid
٠٨/١٢/١٤٣٥
١٧
Acid-Base Properties of Water
H2O (l) H+ (aq) + OH- (aq)
autoionization of water
O
H
H + O
H
H O
H
H H OH-+[ ] +
base conjugate
H2O + H2O H3O+ + OH-
acid conjugatebase
base j gacid
15.2
H2O (l) H+ (aq) + OH- (aq)
The Ion Product of Water
Kc =[H+][OH-]
[H2O] [H2O] = constant
K [H2O] = K = [H+][OH-]Kc[H2O] Kw [H ][OH ]
The ion-product constant (Kw) is the product of the molar concentrations of H+ and OH- ions at a particular temperature.
[H+] [OH ]Solution Is
t lAt 250C
Kw = [H+][OH-] = 1.0 x 10-14
[H+] = [OH-][H+] > [OH-][H+] < [OH-]
neutralacidicbasic
15.2
٠٨/١٢/١٤٣٥
١٨
What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M?
Kw = [H+][OH-] = 1.0 x 10-14
[H+] = 1.3 M
[OH-] =Kw
[H+]1 x 10-14
1.3= = 7.7 x 10-15 M
15.2
pH – A Measure of Acidity
pH = -log [H+]
[H+] = [OH-][H+] > [OH-][H+] < [OH-]
Solution Isneutralacidicbasic
[H+] = 1 x 10-7
[H+] > 1 x 10-7
[H+] < 1 x 10-7
pH = 7pH < 7pH > 7
At 250C
pH [H+]
15.3
٠٨/١٢/١٤٣٥
١٩
OH l [OH ]pOH = -log [OH-]
[H+][OH-] = Kw = 1.0 x 10-14
-log [H+] – log [OH-] = 14.00
pH + pOH = 14 00
15.3
pH + pOH 14.00
The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?
pH = -log [H+]
[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M
The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?
pH + pOH = 14.00
pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60
pH = 14.00 – pOH = 14.00 – 6.60 = 7.40
15.3
٠٨/١٢/١٤٣٥
٢٠
Strong Electrolyte – 100% dissociation
NaCl (s) Na+ (aq) + Cl- (aq)H2O
Weak Electrolyte – not completely dissociatedWeak Electrolyte not completely dissociated
CH3COOH CH3COO- (aq) + H+ (aq)
Strong Acids are strong electrolytes
HCl (aq) + H O (l) H O+ (aq) + Cl- (aq)HCl (aq) + H2O (l) H3O (aq) + Cl (aq)
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)
H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)
15.4
HF (aq) + H2O (l) H3O+ (aq) + F- (aq)
Weak Acids are weak electrolytes
HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)
HSO H O H O SO 2HSO4- (aq) + H2O (l) H3O+ (aq) + SO4
2- (aq)
H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)
Strong Bases are strong electrolytes
NaOH (s) Na+ (aq) + OH- (aq)H2O( ) ( q) ( q)
KOH (s) K+ (aq) + OH- (aq)H2O
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O
15.4
٠٨/١٢/١٤٣٥
٢١
F- (aq) + H2O (l) OH- (aq) + HF (aq)
Weak Bases are weak electrolytes
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Conjugate acid-base pairs:
• The conjugate base of a strong acid has no measurable strength.
• H3O+ is the strongest acid that can exist in aqueous solution.
• The OH- ion is the strongest base that can exist in aqeous solution.
15.4
15.4
٠٨/١٢/١٤٣٥
٢٢
Strong Acid Weak Acid
15.4
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
Start 0.002 M 0.0 M 0.0 M3 ( q) 2 ( ) 3 ( q) 3 ( q)
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
End 0.002 M 0.002 M0.0 M
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.( )2 g
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)Start
End
0.018 M
0.018 M 0.036 M0.0 M
0.0 M 0.0 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.5615.4
٠٨/١٢/١٤٣٥
٢٣
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Weak Acids (HA) and Acid Ionization Constants
HA (aq) H+ (aq) + A- (aq)
Ka =[H+][A-]
[HA]
Ka is the acid ionization constant
Kaweak acidstrength
15.5
15.5
٠٨/١٢/١٤٣٥
٢٤
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-]
[HF] = 7.1 x 10-4
HF (aq) H+ (aq) + F- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.50 0.00
-x +x0.50 - x
0.00
+xx x
Ka =x2
0 50 - x = 7.1 x 10-4 0.50 – x ≈ 0.50Ka << 10.50 x
Ka ≈x2
0.50 = 7.1 x 10-4 x2 = 3.55 x 10-4 x = 0.019 M
[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72[HF] = 0.50 – x = 0.48 M 15.5
When can I use the approximation?
0.50 – x ≈ 0.50Ka << 1
When x is less than 5% of the value from which it is subtracted.L th 5%
x = 0.019 0.019 M0.50 M x 100% = 3.8%
Less than 5%Approximation ok.
What is the pH of a 0.05 M HF solution (at 250C)?
Ka ≈x2
0 05 = 7.1 x 10-4 x = 0.006 M0.05
0.006 M0.05 M x 100% = 12%
More than 5%Approximation not ok.
Must solve for x exactly using quadratic equation or method of successive approximation. 15.5
٠٨/١٢/١٤٣٥
٢٥
Solving weak acid ionization problems:
1. Identify the major species that can affect the pH.
• In most cases, you can ignore the autoionization of water.
Ignore [OH-] because it is determined by [H+]• Ignore [OH-] because it is determined by [H+].
2. Use ICE to express the equilibrium concentrations in terms of single unknown x.
3. Write Ka in terms of equilibrium concentrations. Solve for xby the approximation method. If approximation is not valid, solve for x exactlysolve for x exactly.
4. Calculate concentrations of all species and/or pH of the solution.
15.5
What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?
HA (aq) H+ (aq) + A- (aq)
Initial (M) 0.122 0.00 0.00
Change (M)
Equilibrium (M)
-x +x0.122 - x
+xx x
Ka =x2
0.122 - x= 5.7 x 10-4
x2
0.122 – x ≈ 0.122Ka << 1
Ka ≈x
0.122 = 5.7 x 10-4 x2 = 6.95 x 10-5 x = 0.0083 M
0.0083 M0.122 M x 100% = 6.8%
More than 5%Approximation not ok.
15.5
٠٨/١٢/١٤٣٥
٢٦
Ka =x2
0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0
ax2 + bx + c =0 -b ± b2 – 4ac√2ax =
x = 0 0081 x = 0 0081x = 0.0081 x = - 0.0081
HA (aq) H+ (aq) + A- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122 0.00
-x +x0 122 x
0.00
+xx xEquilibrium (M) 0.122 - x x x
[H+] = x = 0.0081 M pH = -log[H+] = 2.09
15.5
percent ionization = Ionized acid concentration at equilibriumInitial concentration of acid
x 100%
For a monoprotic acid HA
Percent ionization = [H+]
[HA]0x 100% [HA]0 = initial concentration
15.5
٠٨/١٢/١٤٣٥
٢٧
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Weak Bases and Base Ionization Constants
Kb =[NH4
+][OH-][NH ]Kb [NH3]
Kb is the base ionization constant
Kbweak base
strength
15.6
Solve weak base problems like weak acids except solve for [OH-] instead of [H+].
15.6
٠٨/١٢/١٤٣٥
٢٨
Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq)
A- (aq) + H2O (l) OH- (aq) + HA (aq)
Ka
Kb
H2O (l) H+ (aq) + OH- (aq) Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
15.7
Weak Acid and Its Conjugate Base
Ka = KwKb
Kb = KwKa
15.8
٠٨/١٢/١٤٣٥
٢٩
Molecular Structure and Acid Strength
H X H+ + X-
The stronger
The weaker g
the bond the acid
HF << HCl < HBr < HI
15.9
Molecular Structure and Acid Strength
Z O H Z O- + H+δ- δ+
The O-H bond will be more polar and easier to break if:
• Z is very electronegative or
• Z is in a high oxidation state
15.9
٠٨/١٢/١٤٣٥
٣٠
Molecular Structure and Acid Strength
1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number.
Acid strength increases with increasing electronegativity of ZAcid strength increases with increasing electronegativity of Z
H O Cl O
O••
•••••• ••••
••
••••
H O Br O
O••
•••••• ••••
••
••••Cl is more electronegative than Br
HClO3 > HBrO3
15.9
Molecular Structure and Acid Strength
2. Oxoacids having the same central atom (Z) but different numbers of attached groups.
Acid strength increases as the oxidation number of Z increasesAcid strength increases as the oxidation number of Z increases.
HClO4 > HClO3 > HClO2 > HClO4 3 2
15.9
٠٨/١٢/١٤٣٥
٣١
Acid-Base Properties of SaltsNeutral Solutions:
Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strongacid (e.g. Cl-, Br-, and NO3
-).3
NaCl (s) Na+ (aq) + Cl- (aq)H2O
Basic Solutions:
Salts derived from a strong base and a weak acid.
H ONaCH3COOH (s) Na+ (aq) + CH3COO- (aq)H2O
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)
15.10
Acid-Base Properties of SaltsAcid Solutions:
Salts derived from a strong acid and a weak base.
NH Cl ( ) NH + ( ) + Cl ( )H2ONH4Cl (s) NH4+ (aq) + Cl- (aq)2
NH4+ (aq) NH3 (aq) + H+ (aq)
Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid.
Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)3+ 2+
15.10
٠٨/١٢/١٤٣٥
٣٢
Acid Hydrolysis of Al3+
15.10
Acid-Base Properties of SaltsSolutions in which both the cation and the anion hydrolyze:
• Kb for the anion > Ka for the cation, solution will be basic
• Kb for the anion < Ka for the cation, solution will be acidic
• Kb for the anion ≈ Ka for the cation, solution will be neutral
15.10
٠٨/١٢/١٤٣٥
٣٣
Oxides of the Representative ElementsIn Their Highest Oxidation States
15.11
CO2 (g) + H2O (l) H2CO3 (aq)
N2O5 (g) + H2O (l) 2HNO3 (aq)
Arrhenius acid is a substance that produces H+ (H3O+) in water
A Brønsted acid is a proton donor
A L i id i b t th t t i f l t
Definition of An Acid
A Lewis acid is a substance that can accept a pair of electrons
A Lewis base is a substance that can donate a pair of electrons
H+ H O H••••
+ OH-••••••acid base
N H••
H
H
H+ +
acid base 15.12
N H
H
H
H+
٠٨/١٢/١٤٣٥
٣٤
Lewis Acids and Bases
N H••
H
F B
F
+ F B
F
N H
H
N H
H
acid base
F F
N H
H
No protons donated or accepted!No protons donated or accepted!
15.12
Chemistry In Action: Antacids and the Stomach pH Balance
NaHCO3 (aq) + HCl (aq)NaCl (aq) + H2O (l) + CO2 (g)
Mg(OH)2 (s) + 2HCl (aq)MgCl2 (aq) + 2H2O (l)