Weak acid and base ionization reactions and the related equilibrium constants,  Ka and Kb. Relating Ka and Kb to pH, and calculating percent dissociation. 

Key points:

  • For a generic monoprotic weak acid HA\text{HA} with conjugate base A\text{A}^-, the equilibrium constant has the form:
Ka=[H3O+][A][HA]K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}
  • The acid dissociation constant KaK_\text{a} quantifies the extent of dissociation of a weak acid. The larger the value of KaK_\text{a}, the stronger the acid, and vice versa.
  • For a generic weak base B\text{B} with conjugate acid BH+\text{BH}^+, the equilibrium constant has the form:
Kb=[BH+][OH][B]K_\text{b}=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}
  • The base dissociation constant (or base ionization constant) KbK_\text{b} quantifies the extent of ionization of a weak base. The larger the value of KbK_\text{b}, the stronger the base, and vice versa.

Strong vs. weak acids and bases

Strong acids and strong bases refer to species that completely dissociate to form ions in solution. By contrast, weak acids and bases ionize only partially, and the ionization reaction is reversible. Thus, weak acid and base solutions contain multiple charged and uncharged species in dynamic equilibrium.
In this article, we will discuss acid and base dissociation reactions and the related equilibrium constants: KaK_\text{a}, the acid dissociation constant, and KbK_\text{b}, the base dissociation constant.

Warm-up: Comparing acid strength and pH\text{pH}

Problem 1: Weak vs. strong acids at the same concentration

We have two aqueous solutions: a 2.0M2.0\,\text M solution of hydrofluoric acid, HF(aq)\text{HF}(aq), and a 2.0M2.0\,\text{M} solution of hydrobromic acid, HBr(aq)\text{HBr}(aq). Which solution has the lower pH\text{pH}?
Choose 1 answer:
Choose 1 answer:
Answer: HBr(aq)\text{HBr}(aq) has the lower pH\text{pH}
We want to compare the relative amounts of H+\text{H}^+ in solution to compare the pH\text{pH} values. Since the two acids have the same concentration, the acid that dissociates more in solution will have the lower pH\text{pH}.
HBr(aq)\text{HBr}(aq) is a strong acid that dissociates fully in solution:
HBr(aq)Br(aq)+H+(aq)\text{HBr}(aq) \rightarrow \text{Br}^-(aq)+\text{H}^+(aq)
In comparison, HF(aq)\text{HF}(aq) is a weak acid that only partially dissociates in solution:
HF(aq)F(aq)+H+(aq)\text{HF}(aq) \leftrightharpoons \text{F}^-(aq)+\text{H}^+(aq)
Since HBr(aq)\text{HBr}(aq) is a stronger acid compared to HF(aq)\text{HF}(aq), it will have a higher [H+][\text{H}^+] and lower pH\text{pH}.

Problem 2: Weak vs. strong acids at different concentrations

This time we have a 2.0M2.0\,\text M solution of hydrofluoric acid, HF(aq)\text{HF}(aq), and a 1.0M1.0\,\text{M} solution of hydrobromic acid, HBr(aq)\text{HBr}(aq). Which solution has the lower pH\text{pH}?
Assume we don't know the equilibrium constant for the dissociation of hydrofluoric acid.
Choose 1 answer:
Choose 1 answer:
Answer: We need more information!
We want to compare the relative amounts of H+\text{H}^+ (or H3O+\text{H}_3 \text O^+) in solution to compare the pH\text{pH} values. However, this time the two acids do not have the same concentration: HBr(aq)\text{HBr}(aq) is a strong acid that dissociates fully in solution, but we have a lower concentration of HBr(aq)\text{HBr}(aq) compared to the weak acid HF(aq)\text{HF}(aq).
In order to answer this question, we will need to calculate [H+][\text H^+] for both of our solutions. This will require knowing the equilibrium constant for the dissociation of HF(aq)\text{HF}(aq).
HF(aq)F(aq)+H+(aq)\text{HF}(aq) \leftrightharpoons \text{F}^-(aq)+\text{H}^+(aq)
Ka=[F(aq)][H+(aq)][HF(aq)]K_\text a=\dfrac{[\text F^-(aq)][\text H^+(aq)]}{[\text{HF}(aq)]}
Keep reading the article to learn more about solving this problem!

Weak acids and the acid dissociation constant, KaK_\text{a}

Weak acids are acids that don't completely dissociate in solution. In other words, a weak acid is any acid that is not a strong acid.
The strength of a weak acid depends on how much it dissociates: the more it dissociates, the stronger the acid. In order to quantify the relative strengths of weak acids, we can look at the acid dissociation constant KaK_\text{a}, the equilibrium constant for the acid dissociation reaction.
For a generic monoprotic weak acid HA\text{HA}, the dissociation reaction in water can be written as follows:
A monoprotic acid is an acid that can lose one acidic proton per molecule of acid, such as HCl\text{HCl}.
Some acids can lose more than one proton per acid molecule. Sulfuric acid, H2SO4\text H_2 \text{SO}_4, is an example of a diprotic acid, which can lose up to two protons per molecule of H2SO4\text H_2 \text{SO}_4. Triprotic acids have three acidic protons, such as phosphoric acid (H3PO4)(\text H_3 \text{PO}_4).
HA(aq)+H2O(l)H3O+(aq)+A(aq)\text{HA}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{H}_3\text{O}^+(aq)+\text{A}^-(aq)
Based on this reaction, we can write our expression for equilibrium constant KaK_\text{a}:
Ka=[H3O+][A][HA]K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}
Here is a quick review of the rules for writing equilibrium constants:
  • Concentrations of products are multiplied together and placed in the numerator.
  • Concentrations of reactants are multiplied together and placed in the denominator.
  • All concentrations are raised to a power equal to the molar coefficient of each species from the balanced equation. (Therefore, before writing the equilibrium expression, we'll need to have a balanced equation!)
  • Only species in the aqueous and gaseous phases are included in the equilibrium expression. Pure solids and liquids are omitted.
The equilibrium expression is a ratio of products to reactants. The more HA\text{HA} dissociates into H+\text{H}^+ and the conjugate base A\text{A}^-, the stronger the acid, and the larger the value of KaK_\text{a}. Since pH\text{pH} is related to [H3O+][\text H_3 \text O^+], the pH\text{pH} of the solution will be a function of KaK_\text{a} as well as the concentration of the acid: the pH\text{pH} decreases as the concentration of the acid and/or KaK_\text{a} increase.

Common weak acids

Malic acid, C4H6O5\text{C}_4\text{H}_6\text{O}_5, is an organic acid found in apples. Image from Wikimedia Commons, CC BY-SA 3.0.
Carboxylic acids are a common functional group in organic weak acids, and they have the formula COOH-\text{COOH}. Malic acid (C4H6O5)(\text{C}_4\text{H}_6\text{O}_5), an organic acid that contains two carboxylic acid groups, contributes to the tart flavor of apples and some other fruits. Since there are two carboxylic acid groups in the molecule, malic acid can potentially donate up to two protons.
The table below lists some more examples of weak acids and their KaK_\text{a} values.
NameFormulaKa(25C)K_\text{a}(25\,^\circ\text{C})
AmmoniumNH4+\text{NH}_4^+5.6×10105.6\times10^{-10}
Chlorous acidHClO2\text{HClO}_21.2×1021.2\times10^{-2}
Hydrofluoric acidHF\text{HF}7.2×1047.2\times10^{-4}
Acetic acidCH3COOH\text{CH}_3\text{COOH}1.8×1051.8\times10^{-5}
Concept check: Based on the table above, which is a stronger acid-acetic acid or hydrofluoric acid?
The KaK_\text{a} value for HF\text{HF} is greater than the KaK_\text{a} value for CH3COOH\text{CH}_3\text{COOH} (7.2×104>1.8×105)(7.2\times10^{-4}>1.8\times 10^{-5}). The larger KaK_\text{a} value means that hydrofluoric acid will dissociate in solution more than acetic acid. Therefore, hydrofluoric acid is a stronger acid than acetic acid.

Example 1: Calculating % dissociation of a weak acid

One way to quantify how much a weak acid has dissociated in solution is to calculate the percent dissociation. The percent dissociation for weak acid HA\text {HA} can be calculated as follows:
If nitrous acid (HNO2)(\text{HNO}_2) has a KaK_\text{a} of 4.0×1044.0\times10^{-4} at 25C25\,^\circ\text{C}, what is the percent dissociation of nitrous acid in a 0.400 M0.400\text{ M} solution?
Let's go through this example step-by-step!

Step 1: Write the balanced acid dissociation reaction

First, let's write the balanced dissociation reaction of HNO2\text{HNO}_2 in water. Nitrous acid can donate a proton to water to form NO2(aq)\text{NO}_2^-(aq):
HNO2(aq)+H2O(l)H3O+(aq)+NO2(aq)\text{HNO}_2(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{H}_3\text{O}^+(aq)+\text{NO}_2^-(aq)

Step 2: Write the expression for KaK_\text{a}

From the equation in Step 1, we can write the KaK_\text{a} expression for nitrous acid:
Ka=[H3O+][NO2][HNO2]=4.0×104K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{NO}_2^-]}{[\text{HNO}_2]}=4.0\times10^{-4}

Step 3: Find [H+][\text H^+] and [NO2][\text{NO}_2^-] at equilibrium

Next, we can use an ICE\text{ICE} table to determine algebraic expressions for the equilibrium concentrations in our KaK_\text{a} expression:
HNO2(aq)\text{HNO}_2(aq)\rightleftharpoonsH3O+\text{H}_3\text{O}^+NO2\text{NO}_2^-
Initial0.400M0.400\,\text M0000
Changex-x+x+x+x+x
Equilibrium0.400Mx0.400\,\text M-xxxxx
Plugging the equilibrium concentrations into our KaK_\text{a} expression, we get:
Ka=(x)(x)(0.400Mx)=4.0×104K_\text{a}=\dfrac{(x)(x)}{(0.400\,\text M-x)}=4.0\times10^{-4}
Simplifying this expression, we get the following:
x20.400Mx=4.0×104\dfrac{x^2}{0.400\,\text M-x}=4.0\times10^{-4}
This is a quadratic equation that can be solved for xx either by using the quadratic formula or an approximation method.
To learn more about the quadratic formula, check out Sal's videos, Intro to the quadratic formula and Using the quadratic formula.
To learn about using approximation methods, check out the video on the small x approximation for small Kc values.
Either method will give x=0.0126 Mx=0.0126\text{ M}. Therefore, [NO2]=[H3O+]=0.0126 M[\text{NO}_2^-]=[\text{H}_3\text{O}^+]=0.0126\text{ M}.

Step 4: Calculate percent dissociation

To calculate percent dissociation, we can use the equilibrium concentrations we found in Step 3:
Therefore, 3.2%3.2\% of the HNO2\text{HNO}_2 in solution has dissociated into H+\text H^+ and NO2\text{NO}_2^- ions.

Weak bases and KbK_\text{b}

Let's now examine the base dissociation constant (also called the base ionization constant) KbK_\text{b}. We can start by writing the ionization reaction for a generic weak base B\text{B} in water. In this reaction, the base accepts a proton from water to form hydroxide and the conjugate acid, BH+\text{BH}^+:
B(aq)+H2O(l)BH+(aq)+OH(aq)\text{B}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{BH}^+(aq)+\text{OH}^-(aq)
Good question! Perhaps a better name would be the "base ionization constant." But for whatever reason (perhaps to be consistent with the naming of KaK_\text{a}), KbK_\text{b} is called the "base dissociation constant" in most chemistry textbooks, so we will stick with that convention here.
We can write the expression for equilibrium constant KbK_\text{b} as follows:
Kb=[BH+][OH][B]K_\text{b}=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}
From this ratio, we can see that the more the base ionizes to form BH+\text{BH}^+, the stronger the base, and the larger the value of KbK_\text{b}. As such, the pH\text{pH} of the solution will be a function of both the value of KbK_\text{b} as well as the concentration of the base.

Example 2: Calculating the pH\text{pH} of a weak base solution

What is the pH\text{pH} of a 1.50 M1.50\text{ M} solution of ammonia, NH3\text{NH}_3? (Kb=1.8×105)(K_\text{b}=1.8\times10^{-5})
This example is an equilibrium problem with one extra step: finding pH\text{pH} from [OH][\text {OH}^-]. Let's go through the calculation step-by-step.

Step 1: Write the balanced ionization reaction

First, let's write out the base ionization reaction for ammonia. Ammonia will accept a proton from water to form ammonium, NH4+\text{NH}_4^+:
NH3(aq)+H2O(l)NH4+(aq)+OH(aq)\text{NH}_3(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{NH}_4^+(aq)+\text{OH}^-(aq)

Step 2: Write the expression for KbK_\text{b}

From this balanced equation, we can write an expression for KbK_\text{b}:
Kb=[NH4+][OH][NH3]=1.8×105K_\text{b}=\dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}=1.8\times10^{-5}

Step 3: Find [NH4+][\text{NH}_4^+] and [OH][\text{OH}^-] at equilibrium

To determine the equilibrium concentrations, we use an ICE\text{ICE} table:
NH3(aq)\text{NH}_3(aq)\rightleftharpoonsNH4+\text{NH}_4^+OH\text{OH}^-
Initial1.50M1.50\,\text M0000
Changex-x+x+x+x+x
Equilibrium1.50Mx1.50\,\text M-xxxxx
Plugging the equilibrium values into our KbK_\text{b} expression, we get the following:
Kb=(x)(x)1.50Mx=1.8×105K_\text{b}=\dfrac{(x)(x)}{1.50\,\text M-x}=1.8\times10^{-5}
Simplifying, we get:
x21.50Mx=1.8×105\dfrac{x^2}{1.50\,\text M-x}=1.8\times10^{-5}
This is a quadratic equation that can be solved by using the quadratic formula or an approximation method. Either method will yield the solution
x=[OH]=5.2×103 Mx=[\text{OH}^-]=5.2\times10^{-3}\text{ M}

Step 4: Find pH\text{pH} from [OH][\text{OH}^-]

Now that we know the concentration of hydroxide, we can calculate pOH\text{pOH}:
pOH=log[OH]=log(5.2×103)=2.28\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(5.2\times10^{-3})\\ \\ &=2.28\end{aligned}
Recall that at 25C25\,^\circ\text{C}, pH+pOH=14\text{pH}+\text{pOH}=14. Rearranging this equation, we have:
pH=14pOH\text{pH}=14-\text{pOH}
Plugging in our value for pOH\text{pOH}, we get:
pH=14.00(2.28)=11.72\text{pH}=14.00-(2.28)=11.72
Therefore, the pH\text{pH} of the solution is 11.72.

Common weak bases

At left, structure of pyridine. On right, structure of a generic amine: a neutral nitrogen atom with single bonds to R1, R2, and R3.
Pyridine (left) is cyclic nitrogen-containing compound. Amines (right) are organic compounds containing a neutral nitrogen atom with three single bonds to hydrogen or carbon. Both molecules act as weak bases.
From soaps to household cleaners, weak bases are all around us. Amines, a neutral nitrogen with three bonds to other atoms (usually a carbon or hydrogen), are common functional groups in organic weak bases.
Amines act as bases because nitrogen's lone pair of electrons can accept an H+\text H^+. Ammonia, NH3\text{NH}_3 is an example of an amine base. Pyridine, C5H5N\text C_5 \text H_5 \text N, is another example of a nitrogen-containing base.

Summary

  • For a generic monoprotic weak acid HA\text{HA} with conjugate base A\text{A}^-, the equilibrium constant has the form:
Ka=[H3O+][A][HA]K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}
  • The acid dissociation constant KaK_\text{a} quantifies the extent of dissociation of a weak acid. The larger the value of KaK_\text{a}, the stronger the acid, and vice versa.
  • For a generic weak base B\text{B} with conjugate acid BH+\text{BH}^+, the equilibrium constant has the form:
Kb=[BH+][OH][B]K_\text{b}=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}
  • The base dissociation constant (or base ionization constant) KbK_\text{b} quantifies the extent of ionization of a weak base. The larger the value of KbK_\text{b}, the stronger the base, and vice versa.

Attributions

  1. Weak Acids and Bases” from UC Davis ChemWiki, CC BY-NC-SA 3.0
The modified article is licensed under a CC-BY-NC-SA 4.0 license.

Additional References

Zumdahl, S.S., and Zumdahl S.A. (2003). Atomic Structure and Periodicity. In Chemistry (6th ed., pp. 290-94), Boston, MA: Houghton Mifflin Company.

Try it!

Problem 1: Finding KbK_\text{b} from pH\text{pH}

A 1.50 M1.50\text{ M} solution of pyridine, C5H5N\text{C}_5\text{H}_5\text{N}, has a pH\text{pH} of 9.709.70 at 25C25\,^\circ\text{C}. What is the KbK_\text{b} of pyridine?
Choose 1 answer:
Choose 1 answer:
Let's first write the base ionization reaction for pyridine:
C5H5N(aq)+H2O(l)C5H5NH+(aq)+OH(aq)\text{C}_5\text{H}_5\text{N}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{C}_5\text{H}_5\text{NH}^+(aq)+\text{OH}^-(aq)
The corresponding KbK_\text{b} expression is:
Kb=[C5H5NH+][OH][C5H5N]K_\text{b}=\dfrac{[\text{C}_5\text{H}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]}
To calculate KbK_\text{b}, we'll need to determine the concentrations of the species in the equilibrium expression.
We can make an ICE\text{ICE} table for the ionization reaction to find the concentrations at equilibrium:
C5H5N(aq)\text{C}_5\text{H}_5\text{N}(aq)\rightleftharpoonsC5H5NH+\text{C}_5\text{H}_5\text{NH}^+OH(aq)\text{OH}^-(aq)
Initial1.50 M1.50\text{ M}0 M0\text{ M}0 M0\text{ M}
Changex-x+x+x+x+x
Equilibrium1.50 Mx1.50\text{ M}-xxxxx
Plugging our expressions for the equilibrium concentrations of each species into the KbK_\text{b} equation, we have:
Kb=(x)(x)(1.50 Mx)=x21.50 MxK_\text{b}=\dfrac{(x)(x)}{(1.50\text{ M}-x)}=\dfrac{x^2}{1.50\text{ M}-x}
We can use pH\text{pH} to solve for [OH][\text{OH}^-], which will give us xx. First we can use pH\text{pH} to calculate the pOH\text{pOH} of the solution:
pOH=149.70=4.30\text{pOH}=14-9.70=4.30
We can then use pOH\text{pOH} to solve for [OH][\text{OH}^-], which will give us xx.
[OH]=10pOH=104.30=5.05×105 M\begin{aligned}[\text{OH}^-]&=10^{-\text{pOH}}\\ \\ &=10^{-4.30}\\ \\ &=5.05\times10^{-5}\text{ M}\end{aligned}
Therefore, x=5.05×105 Mx=5.05\times10^{-5}\text{ M}.
Finally, we can plug this value for xx into our KbK_\text{b} expression to solve for KbK_\text{b}:
Kb=x21.50x=(5.05×105)2(1.505.05×105)=1.7×109\begin{aligned}K_\text{b}&=\dfrac{x^2}{1.50-x}\\ \\ &=\dfrac{(5.05\times10^{-5})^2}{(1.50-5.05\times10^{-5})}\\ \\ &=1.7\times10^{-9}\end{aligned}
Therefore, the KbK_\text{b} of pyridine is 1.7×1091.7\times10^{-9}.
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