Wize High School Grade 12 Chemistry Textbook > Acids and Bases

Acid Base Equations Summary

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 Types of Acid or Base Reactions and pH Calculations Summary

1. Strong Acid Alone 
HCl(g)→H(aq)++Cl(aq)−HCl_{(g)}\to H_{(aq)}^++Cl_{(aq)}^-HCl(g)​→H(aq)+​+Cl(aq)−​
[H(aq)+]=[Cl(aq)−]=[HCl](initial)[H_{(aq)}^+]=[Cl_{(aq)}^-]=[HCl]_{(initial)}[H(aq)+​]=[Cl(aq)−​]=[HCl](initial)​
pH=−log⁡[H(aq)+]pH=-\log[H_{(aq)}^+]pH=−log[H(aq)+​]

2. Strong Base Alone 
NaOH→Na(aq)++OH(aq)−NaOH\to Na_{(aq)}^++OH_{(aq)}^-NaOH→Na(aq)+​+OH(aq)−​
[OH(aq)−]=[Na(aq)+]=[NaOH]initial[OH_{(aq)}^-]=[Na_{(aq)}^+]=[NaOH]_{initial}[OH(aq)−​]=[Na(aq)+​]=[NaOH]initial​
pOH=−log⁡[OH(aq)−]pOH=-\log[OH_{(aq)}^-]pOH=−log[OH(aq)−​]
pH=14−pOHpH=14-pOHpH=14−pOH
PAGE BREAK
 3. Weak Acid Alone  
Example: 2.0 mols of weak acid, HA, with a Ka = 2 x 10-4 in 1.0 L of solution. 
Step 1: Make an ICE table
HA(aq)+H2O(l)⇋H3O(aq)++A(aq)−HA_{(aq)}+H_2O_{(l)}\leftrightharpoons H_3O_{(aq)}^++A_{(aq)}^-HA(aq)​+H2​O(l)​⇋H3​O(aq)+​+A(aq)−​
I2.0Mn/a00C−xn/a+x+xE2.0−xn/axx\def\arraystretch{2}\begin{array}{c:c:c:c:c}
I &2.0M &n/a &0 &0\\\hline C &-x &n/a &+x &+x\\ \hline E &2.0-x &n/a &x &x\end{array}ICE​2.0M−x2.0−x​n/an/an/a​0+xx​0+xx​​
 Step 2: Set up Ka equation and solve for [H3O+]  

Ka=[H3O(aq)+][A(aq)−][HA(aq)]=x22.0−x≈x22.0Ka=\cfrac{[H_3O^+_{(aq)}][A^-_{(aq)}]}{[HA_{(aq)}]}=\cfrac{x^2}{2.0-x}\approx \cfrac{x^2}{2.0}Ka=[HA(aq)​][H3​O(aq)+​][A(aq)−​]​=2.0−xx2​≈2.0x2​
x=(2.0)Ka=0.02M=[H3O(aq)+]x=\sqrt{(2.0)K_a}=0.02M=[H_3O^+_{(aq)}]x=(2.0)Ka​​=0.02M=[H3​O(aq)+​]
pH=−log⁡[H3O(aq)+]=1.70pH=-\log [H_3O^+_{(aq)}]=1.70pH=−log[H3​O(aq)+​]=1.70

 4. Weak Base Alone 
Example: 2.0 moles of weak base, B, with a Kb of 7 x 10-7 in 1.0 L of solution. 
Step 1: Make an ICE table  
B(aq)+H2O(l)⇋OH(aq)−+BH(aq)+B_{(aq)}+H_2O_{(l)}\leftrightharpoons OH_{(aq)}^-+BH_{(aq)}^+B(aq)​+H2​O(l)​⇋OH(aq)−​+BH(aq)+​
I2.0Mn/a00C−xn/a+x+xE2.0−xn/axx\def\arraystretch{2}\begin{array}{c:c:c:c:c}
I &2.0M &n/a &0 &0\\\hline C &-x &n/a &+x &+x\\ \hline E &2.0-x &n/a &x &x\end{array}ICE​2.0M−x2.0−x​n/an/an/a​0+xx​0+xx​​

 Step 2: Set up Kb equation and solve for [OH- ] 
  
Kb=[OH(aq)−][BH(aq)+][B(aq)]=x22.0−x≈x22.0K_b=\cfrac{[OH^-_{(aq)}][BH^+_{(aq)}]}{[B_{(aq)}]}=\cfrac{x^2}{2.0-x}\approx \cfrac{x^2}{2.0}Kb​=[B(aq)​][OH(aq)−​][BH(aq)+​]​=2.0−xx2​≈2.0x2​
x=(2.0)Kb=0.00118M=[OH(aq)−]x=\sqrt{(2.0)K_b}=0.00118M=[OH^-_{(aq)}]x=(2.0)Kb​​=0.00118M=[OH(aq)−​]
pOH=−log⁡[OH(aq)−]=2.93pOH=-\log [OH^-_{(aq)}]=2.93pOH=−log[OH(aq)−​]=2.93
pH=14−pOH=11.07pH=14-pOH=11.07pH=14−pOH=11.07

When can we make a simplifying assumption about x? 
In K= x2/(y-x) 
When y/K > 400 you can simplify and ignore the "-x"

 Important Relationships:  
pH=−log⁡⁡[H3O+] and [H3O+]=10−pHpH=-\log⁡[H_3O^+]\ and\ [H_3O^+]=10^{-pH}pH=−log⁡[H3​O+] and [H3​O+]=10−pH
pOH=−log⁡⁡OH−and OH−=10−pOHpOH=-\log⁡OH^-and\ OH^-=10^{-pOH}pOH=−log⁡OH−and OH−=10−pOH

2H2O⇌H3O++OH−2H_2 O⇌H_3 O^++OH^-2H2​O⇌H3​O++OH−

  In pure water, at 25 °C:
[H3O+]=[OH−]=1.0×10−7M[H_3O^+]=[OH^-]=1.0\times10^{-7}M[H3​O+]=[OH−]=1.0×10−7M
Kw=[H3O+][OH−]=1.0×10−14 at 25°CK_w=[H_3O^+][OH^-]=1.0\times10^{-14}\ at\ 25°CKw​=[H3​O+][OH−]=1.0×10−14 at 25°C
pH+pOH=pKw=14.00 at 25°CpH+pOH=pK_w=14.00\ at\ 25°CpH+pOH=pKw​=14.00 at 25°C

PAGE BREAK

Ka is the acid dissociation constant and is defined as the equilibrium constant for the reaction below: 

HA(aq)+H2O(l)⇌A−(aq)+H3O+(aq)HA(aq)+H_2O(l)⇌A^-(aq)+H_3O^+(aq)HA(aq)+H2​O(l)⇌A−(aq)+H3​O+(aq)
Ka=[H3O+][A−][HA]K_a=\cfrac{[H_3 O^+ ][A^- ]}{[HA]}Ka​=[HA][H3​O+][A−]​
pKa=−log⁡⁡Ka and Ka=10−pKapK_a=-\log⁡K_a\ and\ K_a=10^{-pK_a}pKa​=−log⁡Ka​ and Ka​=10−pKa​

Strong acids dissociate completely. 
HNO3(aq)→H+(aq)+NO3−(aq)HNO_3 (aq)→H^+ (aq)+NO_3^- (aq)HNO3​(aq)→H+(aq)+NO3−​(aq)  (not an equilibrium, Ka is very large) 

Weak acids partially dissociate to give H+ or H3O+
CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)CH_3 COOH(aq)⇌CH_3 COO^- (aq)+H^+ (aq)CH3​COOH(aq)⇌CH3​COO−(aq)+H+(aq)  (equilibrium, Ka) 
PAGE BREAK
Kb is the base dissociation constant and is defined as the equilibrium constant for the reaction below:

B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)B(aq)+H_2 O(l)⇌BH^+ (aq)+ OH^- (aq)B(aq)+H2​O(l)⇌BH+(aq)+OH−(aq)
Kb=[BH+][OH−][B]K_b=\cfrac{[BH^+ ][OH^- ]}{[B]}Kb​=[B][BH+][OH−]​
pKb=−log⁡⁡Kb and Kb=10−pKbpK_b=-\log⁡K_b\ and\ K_b=10^{-pK_b}pKb​=−log⁡Kb​ and Kb​=10−pKb​
Strong bases dissociate completely. 
LiOH(s)→Li+(aq)+OH−(aq)LiOH(s)→Li^+(aq)+OH^-(aq)LiOH(s)→Li+(aq)+OH−(aq)  (not an equilibrium, Kb is very large) 

Weak bases partially dissociate (they accept protons to give OH-, but the reaction is not complete)

CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)CH_3NH_2(aq)+H_2O(l)⇌CH_3NH_3^+(aq)+OH^-(aq)CH3​NH2​(aq)+H2​O(l)⇌CH3​NH3+​(aq)+OH−(aq) 

PAGE BREAK

For a given conjugate acid-base pair:

 HA+H2O⇌A−+H3O+                   KaHA+H_2O⇌A^-+H_3O^+\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red}{K_a}HA+H2​O⇌A−+H3​O+                   Ka​
A−+H2O⇌HA+OH−                   Kb‾\underline{A^-+H_2O⇌HA+OH^-\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red}{K_b}}A−+H2​O⇌HA+OH−                   Kb​​
2H2O⇌H3O++OH−                      Kw2H_2 O⇌H_3 O^++OH^- \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red} K_w2H2​O⇌H3​O++OH−                      Kw​
Ka×Kb=Kw=1×10−14 and pKa+pKb=pKw=14.00 (at 25°C)K_a×K_b=K_w=1×10^{-14} \ and \ pK_a+pK_b=pK_w=14.00\  (at\  25 °C)Ka​×Kb​=Kw​=1×10−14 and pKa​+pKb​=pKw​=14.00 (at 25°C)

Acid/Base Important Relationships Cheatsheet

Wize High School Grade 12 Chemistry Textbook > Acids and Bases

Acid Base Equations Summary

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Types of Acid or Base Reactions and pH Calculations Summary

1. Strong Acid Alone

$HCl_{(g)}\to H_{(aq)}^++Cl_{(aq)}^-$

$[H_{(aq)}^+]=[Cl_{(aq)}^-]=[HCl]_{(initial)}$

$pH=-\log[H_{(aq)}^+]$

2. Strong Base Alone

$NaOH\to Na_{(aq)}^++OH_{(aq)}^-$

$[OH_{(aq)}^-]=[Na_{(aq)}^+]=[NaOH]_{initial}$

$pOH=-\log[OH_{(aq)}^-]$

$pH=14-pOH$

3. Weak Acid Alone

$HA_{(aq)}+H_2O_{(l)}\leftrightharpoons H_3O_{(aq)}^++A_{(aq)}^-$

$\def\arraystretch{2}\begin{array}{c:c:c:c:c} I &2.0M &n/a &0 &0\\\hline C &-x &n/a &+x &+x\\ \hline E &2.0-x &n/a &x &x\end{array}$

4. Weak Base Alone

When can we make a simplifying assumption about x?

Important Relationships:

$pH=-\log⁡[H_3O^+]\ and\ [H_3O^+]=10^{-pH}$

$pOH=-\log⁡OH^-and\ OH^-=10^{-pOH}$

$HNO_3 (aq)→H^+ (aq)+NO_3^- (aq)$ (not an equilibrium, Ka is very large)

$CH_3 COOH(aq)⇌CH_3 COO^- (aq)+H^+ (aq)$ (equilibrium, Ka)

$K_b=\cfrac{[BH^+ ][OH^- ]}{[B]}$

$pK_b=-\log⁡K_b\ and\ K_b=10^{-pK_b}$

$LiOH(s)→Li^+(aq)+OH^-(aq)$ (not an equilibrium, Kb is very large)

Acid/Base Important Relationships Cheatsheet

Wize High School Grade 12 Chemistry Textbook > Acids and Bases

Acid Base Equations Summary

Popular Courses

Types of Acid or Base Reactions and pH Calculations Summary

1. Strong Acid Alone

HCl(g)→H(aq)++Cl(aq)−HCl_{(g)}\to H_{(aq)}^++Cl_{(aq)}^-HCl(g)​→H(aq)+​+Cl(aq)−​

[H(aq)+]=[Cl(aq)−]=[HCl](initial)[H_{(aq)}^+]=[Cl_{(aq)}^-]=[HCl]_{(initial)}[H(aq)+​]=[Cl(aq)−​]=[HCl](initial)​

pH=−log⁡[H(aq)+]pH=-\log[H_{(aq)}^+]pH=−log[H(aq)+​]

2. Strong Base Alone

NaOH→Na(aq)++OH(aq)−NaOH\to Na_{(aq)}^++OH_{(aq)}^-NaOH→Na(aq)+​+OH(aq)−​

[OH(aq)−]=[Na(aq)+]=[NaOH]initial[OH_{(aq)}^-]=[Na_{(aq)}^+]=[NaOH]_{initial}[OH(aq)−​]=[Na(aq)+​]=[NaOH]initial​

pOH=−log⁡[OH(aq)−]pOH=-\log[OH_{(aq)}^-]pOH=−log[OH(aq)−​]

pH=14−pOHpH=14-pOHpH=14−pOH

3. Weak Acid Alone

HA(aq)+H2O(l)⇋H3O(aq)++A(aq)−HA_{(aq)}+H_2O_{(l)}\leftrightharpoons H_3O_{(aq)}^++A_{(aq)}^-HA(aq)​+H2​O(l)​⇋H3​O(aq)+​+A(aq)−​

I2.0Mn/a00C−xn/a+x+xE2.0−xn/axx\def\arraystretch{2}\begin{array}{c:c:c:c:c} I &2.0M &n/a &0 &0\\\hline C &-x &n/a &+x &+x\\ \hline E &2.0-x &n/a &x &x\end{array}ICE​2.0M−x2.0−x​n/an/an/a​0+xx​0+xx​​

4. Weak Base Alone

When can we make a simplifying assumption about x?

Important Relationships:

pH=−log⁡⁡[H3O+] and [H3O+]=10−pHpH=-\log⁡[H_3O^+]\ and\ [H_3O^+]=10^{-pH}pH=−log⁡[H3​O+] and [H3​O+]=10−pH

pOH=−log⁡⁡OH−and OH−=10−pOHpOH=-\log⁡OH^-and\ OH^-=10^{-pOH}pOH=−log⁡OH−and OH−=10−pOH

HNO3(aq)→H+(aq)+NO3−(aq)HNO_3 (aq)→H^+ (aq)+NO_3^- (aq)HNO3​(aq)→H+(aq)+NO3−​(aq) (not an equilibrium, Ka is very large)

CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)CH_3 COOH(aq)⇌CH_3 COO^- (aq)+H^+ (aq)CH3​COOH(aq)⇌CH3​COO−(aq)+H+(aq) (equilibrium, Ka)

Kb=[BH+][OH−][B]K_b=\cfrac{[BH^+ ][OH^- ]}{[B]}Kb​=[B][BH+][OH−]​

pKb=−log⁡⁡Kb and Kb=10−pKbpK_b=-\log⁡K_b\ and\ K_b=10^{-pK_b}pKb​=−log⁡Kb​ and Kb​=10−pKb​

LiOH(s)→Li+(aq)+OH−(aq)LiOH(s)→Li^+(aq)+OH^-(aq)LiOH(s)→Li+(aq)+OH−(aq) (not an equilibrium, Kb is very large)

Acid/Base Important Relationships Cheatsheet

$HCl_{(g)}\to H_{(aq)}^++Cl_{(aq)}^-$

$[H_{(aq)}^+]=[Cl_{(aq)}^-]=[HCl]_{(initial)}$

$pH=-\log[H_{(aq)}^+]$

$NaOH\to Na_{(aq)}^++OH_{(aq)}^-$

$[OH_{(aq)}^-]=[Na_{(aq)}^+]=[NaOH]_{initial}$

$pOH=-\log[OH_{(aq)}^-]$

$pH=14-pOH$

$HA_{(aq)}+H_2O_{(l)}\leftrightharpoons H_3O_{(aq)}^++A_{(aq)}^-$

$\def\arraystretch{2}\begin{array}{c:c:c:c:c} I &2.0M &n/a &0 &0\\\hline C &-x &n/a &+x &+x\\ \hline E &2.0-x &n/a &x &x\end{array}$

$pH=-\log⁡[H_3O^+]\ and\ [H_3O^+]=10^{-pH}$

$pOH=-\log⁡OH^-and\ OH^-=10^{-pOH}$

$HNO_3 (aq)→H^+ (aq)+NO_3^- (aq)$ (not an equilibrium, Ka is very large)

$CH_3 COOH(aq)⇌CH_3 COO^- (aq)+H^+ (aq)$ (equilibrium, Ka)

$K_b=\cfrac{[BH^+ ][OH^- ]}{[B]}$

$pK_b=-\log⁡K_b\ and\ K_b=10^{-pK_b}$

$LiOH(s)→Li^+(aq)+OH^-(aq)$ (not an equilibrium, Kb is very large)