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Wize University Chemistry Textbook > Acids and Bases

Applications of Acid and Base Equations

Base Equations (Kb, pKb)Previous Section
Autoionization of Water (Kw)Next Section
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Example: Ka and pKa Values

What is the strongest acid below? What is the weakest acid?

Strongest acid:
Cl3CCOOH

Weakest acid:
HCN

Standard Ka values


The strongest acid has the highest Ka value (and lowest pKa value)
The weakest acid has the lowest Ka value (and the highest pKa value)
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What is the strongest base below? What is the weakest base?

Strongest base:
Dimethylamine
Weakest base:
Aniline

Standard Kb values


The strongest base has the highest Kb value (and the lowest pKb value).
The weakest base has the lowest Kb value (and the highest pKb value).


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Example: pH of Acids and Bases

Part 1: What is the pH of a 0.5 M HCl solution?

HCl is a (strong/weak acid)
strong
It dissociates (completely/incompletely)
completely

Strong acids COMPLETELY dissociate so we can write the reaction: HCl → H+ + Cl-
0.5M
Since the equation is balanced, we can see that there is a 1:1 molar ratio between HCl and H+ so the concentration of H+ will actually be the same!
[H+] =0.5M
plug that into the pH equation: pH=-log[H+]
pH=-log[0.5]
pH=0.30


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Part 2: What is the pH of a 1.0 M CH3COOH solution? Ka = 1.8 x 10-5

This is an (acid/base)
acid
It is (weak/strong)
weak
Does this type of acid dissociate into ions completely/incompletely?
incompletely

CH3COOH is not one of the strong acids we had to memorize, so we automatically know that it is a weak acid! Weak acids have INCOMPLETE dissociation, so we will see an equilibrium in the equation:

CH3COOH(aq) + H2O(l) ⇌ CH3COO- + H3O+(aq)
1.0M ? need this to find pH!

Write out an ICE Table:
I 1M / 0 0
C -x +x +x
E 1-x x x

Write out the Ka expression:

Ka=[CH3COO−][H3O+][CH3COOH]Ka=\frac{\left[CH_3COO^-\right]\left[H_3O^+\right]}{\left[CH_3COOH\right]}

Plug in the Ka value we were given and plug in the equilibrium concentrations from our ICE table:

1.8x10−5=[x][x][1−x]1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[1-x\right]}

Let's check to see if the denomenator has to stay [1-x] (and we would have to use the quadratic equation) OR we might be able to simplify it to ~1.
When the denominator is in the form of y-x, we can check to see if y/K > 400 and if it is, then we can make the assumption and simplify.

Let's see.... 1/1.8x10-5= 55 555 which is much larger than 400! This means that our denominator was [1-x] but we can simplify it to [1] since y is so large and K is so small so the x won't make a difference.

1.8x10−5=x211.8x10^{-5}=\frac{x^2}{1}

Take the square root of both sides:
4.24x10-3=x

If we look back at our ICE table, x=[H+]

so now we can solve for pH:
pH=-log[H+]
pH=-log[4.24x10-3]
pH=2.37 !


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Part 3: What is the pH of a solution prepared by dissolving 0.10 mol of Ba(OH)2 in 1.0 L of pure water?

Is this an acid or base?
base
Is this a strong one or weak one?
Strong
Complete/incomplete dissociation?
Complete

Since we know it's a strong base, that means we know it will COMPLETELY dissociate so we can write out the chemical reaction:
Ba(OH)2 → Ba2+ + 2OH- (Always double check your equation is balanced!!)
Here we are given n=0.10 mol, V=1L. We can use the equation n=cv to solve for c:
c=n/v
c=0.10mol/1L
c=0.1M
0.1M 0.2M
(Since Ba(OH)2 has a concentration of 0.1M, OH- must have a concentration of 0.2M since there is a 1:2 molar ratio!)

[OH-]=0.2M
pOH=-log[OH-]
pOH=-log[0.2]
pOH=0.70
We are asked to solve for pH so use the equation: pH + pOH = 14
pH=14-pOH
pH=14-0.70
pH=13.3 (This logically makes sense since we would expect a base to have a very high pH)

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Part 4: Many liquid household cleaners contain ammonia. The concentration of ammonia (NH3) in these products is usually around 5.0 M. What is the pH of such a cleaning solution? (Kb NH3 = 1.8 x 10-5)

Is this an acid or a base?
Base
Is it strong or weak?
Weak
Complete or incomplete dissociation?
Incomplete

NH3 is a weak base so that means it INCOMPLETELY dissociates:
NH3 + H2O ⇌ NH4+ + OH-
5M
Here we will need to make an ICE table, so we can again plug in the equilibrium concentrations into our Ka expression, solve for [OH-] then pOH, then pH !
I 5M / 0 0
C -x +x +x
E 5-x x x

Now write the Kb expression out:

Kb=[NH4+][OH−][NH3]Kb=\frac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}

Plug in the Kb value we were given and the equilibrium values from our ICE Table:
1.8x10−5=[x][x][5−x]1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[5-x\right]}

Check to see if we can make the simplification of the denominator from [5-x] to ~5.
y/K >400
5/1.8x10-5=277 778. which is much greater than 400 so we can make the simplifcation!

1.8x10−5=[x][x][5]1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[5\right]}
1.8x10-5(5)=x2
9x10-5=x2
Take the square root of both sides:
9.49x10-3=x
x=[OH-] (see the ICE Table)
Solve for pOH:
pOH=-log[OH-]
pOH=-log[9.49x10-3]
pOH=2.02

The question is asking us to solve for pH so use pH + pOH=14
pH=14 - pOH
pH=14-2.02
pH=11.98 !

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)}^-

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

pH=−log⁡[H(aq)+]pH=-\log[H_{(aq)}^+]


2. Strong Base Alone

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

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

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

pH=14−pOHpH=14-pOH

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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)}^-

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}

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}
x=(2.0)Ka=0.02M=[H3O(aq)+]x=\sqrt{(2.0)K_a}=0.02M=[H_3O^+_{(aq)}]
pH=−log⁡[H3O(aq)+]=1.70pH=-\log [H_3O^+_{(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)}^+
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}

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}
x=(2.0)Kb=0.00118M=[OH(aq)−]x=\sqrt{(2.0)K_b}=0.00118M=[OH^-_{(aq)}]
pOH=−log⁡[OH(aq)−]=2.93pOH=-\log [OH^-_{(aq)}]=2.93
pH=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"
Extra Practice
Weak Acids and Bases
Calculate the pH of the following solution.
A solution prepared by dissolving 4.25 x 10-4 g of CH3COONa in 100.0 mL of water.
Note: Acetic Acid Ka = 1.8 x 10-5
Weak Acids and Bases
Calculate the pH and pOH of the following solutions.
a) A 6.30 x 10-2 M solution of CH3COOH in water.
b) A solution prepared by dissolving 4.25 x 10-4 g of CH3COONa in 100.0 mL of water.
Weak Acids and Bases
Calculate the pH of a solution prepared by dissolving 0.500 g of sodium hydrogen phosphate (Na2HPO4, MW = 141.9579 g/mol) in 150.0 mL of water.
Weak Acids and Bases
Calculate the pH and pOH of the following solutions.
a) A 6.30 x 10-2 M solution of CH3COOH in water.
b) A solution prepared by dissolving 4.25 x 10-4 g of CH3COONa in 100.0 mL of water.
Weak Acids and Bases
Will HCO3- act as an acid or as a base?
Weak Acids and Bases
Calculate the pH and pOH of the following solutions, knowing that Ka = 1.8x10^-5.
a) A 5.33 x 10-2 M solution of CH3COOH in water.
b) A solution prepared by dissolving 1.15 x 10-3 g of CH3COONa in 100.0 mL of water.
Weak Acids and Bases
Calculate the pH of a solution prepared by dissolving 0.150 g of sodium hydrogen phosphate (Na2HPO4, MW = 141.9579 g/mol) in 125.0 mL of water.
Ka (HPO42−) = 4.2 x 10−13;  Kb (HPO42−) = 1.6 x 10−7. K_a\ \left(HPO_4^{2-}\right)\ =\ 4.2\ x\ 10^{-13};\ \ K_b\ \left(HPO_4^{2-}\right)\ =\ 1.6\ x\ 10^{-7}.\
Weak Acids and Bases
Calculate the percent dissociation for the following solutions.
a) A 0.444 M aqueous solution of HCN (Ka = 6.2 x 10-10).
b) A 3.50 x 10-2 M aqueous solution of NaCN.
Weak Acids and Bases
Sulfurous acid (H2SO3) is a diprotic weak acid (pKa1 = 1.85, pKa2 = 7.20). Bisulfite (HSO3-) is the conjugate base of sulfurous acid and is an amphoteric species (can act as an acid or a base). Will HSO3- act as an acid or as a base?
Using the data in the table, which of the conjugate bases below is the strongest base?
AcidKaHOAc1.8×10−5HCHO21.8×10−4HClO3.0×10−8HF6.8×10−4\def\arraystretch{1.5} \begin{array}{c} \hline Acid &&&K_a\\ \hline HOAc&&&1.8 \times 10^{-5}\\ HCHO_2&&&1.8 \times 10^{-4}\\ HClO &&& 3.0 \times 10^{-8}\\ HF &&& 6.8 \times 10^{-4}\\ \hline \end{array}
Consider the weak bases below and their Kb values:
C6H7O Kb = 1.3 x 10-10
C2H5NH2 Kb = 5.6 x 10-4
Acids and Bases
Find the pH during titration of 20.00 mL of 0.100 M butanoic acid with 0.100 M NaOH solution after the following additions of base: 0 mL, 10.00 mL, 20.00 mL, 25.00 mL (Ka = 1.54 x 10-5)
Acids and Bases: pH
What is the pH of a 0.10 M solution of NaF? (Ka for HF = 6.8×10-4)?