Understanding Acid Dissociation Constant (Ka) and pKa

The Acid Dissociation Constant (Ka) and its related concept, pKa, are critical in understanding the strength of acids in chemistry. These terms provide deeper insights into the dissociation of weak acids in aqueous solutions, and they help us predict the behavior of acids in different environments.

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1. What is Acid Dissociation Constant (Ka)?

The Acid Dissociation Constant (Ka) is a measure of the strength of an acid in solution. Specifically, it quantifies the extent to which an acid dissociates (breaks apart) into its ions in water. Strong acids dissociate completely, while weak acids only partially dissociate.

The dissociation of a weak acid (HA) in water can be represented as:

$$\text{HA}+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}+{\text{A}}^{-}$$

Where:

• HA is the weak acid.
• H₃O⁺ is the hydronium ion.
• A⁻ is the conjugate base.

The Ka expression is derived from this equilibrium reaction and is given by:

$$K_a=\frac{[{\text{H}}_3{\text{O}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$$

Where:

• $[\text{H}_3\text{O}^+]$ is the concentration of hydronium ions.
• $[\text{A}^-]$ is the concentration of the conjugate base.
• $[\text{HA}]$ is the concentration of the undissociated acid.

A larger Ka value indicates a stronger acid (greater dissociation), while a smaller Ka value indicates a weaker acid.

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2. What is pKa?

pKa is the negative logarithm of the Ka value:

$$pK_a = -\log(K_a)$$

pKa provides a more convenient way to express the strength of an acid. The lower the pKa value, the stronger the acid, because a lower pKa corresponds to a larger Ka (greater dissociation).

• Low pKa (e.g., pKa = 1) → Strong acid
• High pKa (e.g., pKa = 9) → Weak acid

By using pKa, we can easily compare the strength of different acids. For example, an acid with a pKa of 3 is stronger than an acid with a pKa of 5.

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3. Relationship Between Ka and pKa

The relationship between Ka and pKa is straightforward:

$$p{K}_a=-\log (K_a)$$

A large Ka (strong acid) corresponds to a small pKa, while a small Ka (weak acid) corresponds to a large pKa.

For example:

• If $K_a = 10^{-2}$, then $p{K}_a=2.$
• If $K_a = 10^{-5}$, then $p{K}_a=5.$

This inverse relationship helps students quickly determine the strength of an acid by examining its pKa value.

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4. How to Calculate Ka and pKa

Example: Suppose you have a weak acid with a pH of 4.75 and a concentration of 0.1 M. To find Ka and pKa:

1. Step 1: Calculate the concentration of $[\text{H}^+]$ from the pH:

   $$[{\text{H}}^{+}]=10^{-\text{pH}}=10^{-4.75}=1.78\times 10^{-5}\text{M}$$

2. Step 2: Use the concentration of $[\text{H}^+]$ to find the dissociation of the acid. In this case, since it’s a weak acid, we assume that the concentration of $[\text{A}^-] = [\text{H}^+]$.

3. Step 3: Plug the values into the Ka expression:

   $$K_a=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]-[{\text{H}}^{+}]}=\frac{(1.78\times 10^{-5}{)}^2}{0.1-1.78\times 10^{-5}}=3.16\times 10^{-5}$$

4. Step 4: Calculate pKa:

   $$p{K}_a=-\log (3.16\times 10^{-5})=4.5$$

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Acid Dissociation Constant Practice Problems with Solutions

Question 1

You are given a weak monobasic acid, HA.

(i) Write an equilibrium equation to represent its dissociation in aqueous solution.

(ii) Derive an expression for the dissociation constant, $K_a$, in terms of the degree of dissociation $x$ of the acid.

(iii) Given that the $K_a$ is equal to $3.5 \times 10^{-5} \, \text{mol/dm}^3$, calculate the pH value of an aqueous solution of concentration $1.0 \times 10^{-3} \, \text{mol/dm}^3$.

(iv) What would be the respective pH values of aqueous solutions of HCl and $H_2SO_4$, each of concentration $1.0 \times 10^{-3} \, \text{mol/dm}^3$?

Solution

This question relates to weak acid dissociation and pH calculations. Let's address each part:

(i) Equilibrium Equation: For a weak monobasic acid, HA, the dissociation in aqueous solution is given by:

$$HA (aq) \rightleftharpoons H^+ (aq) + A^- (aq)$$

(ii) Expression for Dissociation Constant (Ka): The dissociation constant $K_a$ is expressed as:

$$K_a = \frac{[H^+][A^-]}{[HA]}$$

If $x$ is the degree of dissociation, then:

• $[H^+] = xC$
  
• $[A^-] = xC$
  
• $[HA] = C(1 - x)$
  

Thus, the expression for $K_a$ becomes:

$$K_a = \frac{(xC)(xC)}{C(1-x)} = \frac{x^2C}{1-x}$$

Where $C$ is the initial concentration of the acid.

(iii) pH Calculation: Given that $K_a = 3.5 \times 10^{-5} \, \text{mol/dm}^3$ and the concentration of the acid is $1.0 \times 10^{-3} \, \text{mol/dm}^3$:

Using the approximation that $1 - x \approx 1$ for weak acids, the equation simplifies to:

$$K_a = x^2C$$

Substitute the values:

$$3.5 \times 10^{-5} = x^2 \times 1.0 \times 10^{-3}$$

Solve for $x^2$:

$$x^2 = \frac{3.5 \times 10^{-5}}{1.0 \times 10^{-3}} = 3.5 \times 10^{-2}$$
$$x = \sqrt{3.5 \times 10^{-2}} \approx 0.187$$

Thus, the concentration of $H^+$ is:

$$[H^+] = xC = 0.187 \times 1.0 \times 10^{-3} = 1.87 \times 10^{-4} \, \text{mol/dm}^3$$

Finally, the pH is:

$$pH = -\log[H^+] = -\log(1.87 \times 10^{-4}) \approx 3.73$$

(iv) pH of HCl and H2SO4 Solutions:

For HCl (a strong acid), it dissociates completely, so:

$$[H^+] = 1.0 \times 10^{-3} \, \text{mol/dm}^3$$

Thus:

$$pH = -\log(1.0 \times 10^{-3}) = 3$$

For H2SO4, each molecule dissociates to give two protons, so:

$$[H^+] = 2 \times 1.0 \times 10^{-3} = 2.0 \times 10^{-3} \, \text{mol/dm}^3$$

Thus:

$$pH=-\log (2.0\times 10^{-3})\approx 2.70\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}$$

Question 2

The compound HA is a very weak acid, and in aqueous solution, the following equilibrium is established:

$$HA + H_2O \rightleftharpoons H_3O^+ + A^-$$

(i) Deduce the relationship between the pH of the aqueous solution, the acid dissociation constant $K_a$, and the initial acid concentration $HA$.

(ii) Show that for every four-fold increase in the concentration of $HA$, the concentration of $H_3O^+$ in the solution is doubled.

Solution 

This question involves understanding the relationship between pH, the dissociation constant $K_a$, and the concentration of a weak acid $HA$.

(i) For the weak acid dissociation:

$$HA + H_2O \rightleftharpoons H_3O^+ + A^-$$

The dissociation constant $K_a$ is:

$$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$

For a weak acid, $[H_3{O}^{+}]=[A^{-}]=x\mathrm{C,\; where }$$x$ is the degree of dissociation and $C$ is the initial concentration of the acid. Assuming $[HA] \approx C$ (since the acid is weak and dissociation is minimal), we get:

$$K_a = \frac{x^2C}{C} = x^2$$

Thus, $x = \sqrt{K_a}$, and since $[H_3{O}^{+}]=x\mathrm{C,\; we\; have:}$

$$[H_3O^+] = \sqrt{K_a C}$$

The pH is then related to $[H_3O^+]$ by:

$$pH = -\log [H_3O^+] = -\log (\sqrt{K_a C})$$

This simplifies to:

$$pH = \frac{1}{2}(-\log K_a) - \frac{1}{2} \log C$$

(ii) Show that for every four-fold increase in H$A$, the concentration of  $H_3{O}^{+}<b\; style="display:\; inline;">is\; doubled:</b>$

From the expression $[H_3O^+] = \sqrt{K_a C}$, if the concentration $C$ is increased four-fold, the new concentration $C'$ is $4C$. The new $[H_3O^+]$ is:

$$[H_3O^+]_{new} = \sqrt{K_a \times 4C} = 2 \times \sqrt{K_a C}$$

Thus, the concentration of $H_3O^+$ is doubled when the concentration of $HA$ is increased four-fold.

This completes the solutions for both parts of the question.

Question 3

$H_2X$ is an inorganic acid that is fully dissociated in diluted solutions. In solutions greater than 0.5M, however, only the first dissociation is complete, while the second dissociation is incomplete with a $K_a$ value of $1.3 \times 10^{-2}$. Show by calculation that the concentration of $[H_3O^+]$ in a 1M solution of $H_2X$ is less than 2M.

Solution:

Given that:

1. The first dissociation of $H_2X$ is complete:

   $$H_2X \rightleftharpoons H^+ + HX^-$$
   

   Since this dissociation is complete, for a 1M solution of $H_2X$, the concentration of $[H^+]$ and $[HX^-]$ will each be 1M.

2. The second dissociation of $HX^-$ is incomplete with a $K_a$ value of $1.3 \times 10^{-2}$:

   $$HX^- + H_2O \rightleftharpoons H_3O^+ + X^{2-}$$
   

   Let the concentration of $X^{2-}$ (and additional $H_3O^+$) formed by the second dissociation be $x$.

   The expression for the $K_a$ of the second dissociation is:

   $$K_a = \frac{[H_3O^+][X^{2-}]}{[HX^-]}$$

   Substituting the known values:

   $$1.3 \times 10^{-2} = \frac{x^2}{1 - x}$$

   Since $x$ is relatively small compared to 1M, we can approximate $1 - x \approx 1$. This simplifies the equation to:

   $$1.3\times 10^{-2}=x^2$$

   Solving for $x$:

   $$x = \sqrt{1.3 \times 10^{-2}} = 0.114$$
   

Thus, the additional $[H_3O^+]$ contributed by the second dissociation is $0.114\mathrm{M.}$

Now, the total concentration of $[H_3O^+]$ will be the sum of the contribution from both dissociations:

$$[H_3O^+] = 1 + 0.114 = 1.114M$$

Since $1.114M < 2M$, we have shown that the concentration of $[H_3O^+]$ for a 1M solution of $H_2X$ is less than 2M.

Final Answer:

The concentration of $[H_3O^+]$ for a 1M solution of $H_2X$ is approximately 1.114M, which is less than 2M as required.

Question 4

A given sample of ethanoic acid solution has a pH = 2.86. 1.08 g of NaOH is dissolved in distilled water and made up to 250 cm³. 25 cm³ of this NaOH solution is required to neutralize 23.8 cm³ of the ethanoic acid solution. Calculate the $pK_a$ of the ethanoic acid.

Solution:

1. Find the moles of NaOH:

   Moles of NaOH = $\frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{1.08}{40} = 0.027 \text{ mol}$

2. Find concentration of NaOH:

   $$\text{Concentration of NaOH} = \frac{\text{moles of NaOH}}{\text{volume of solution in dm³}} = \frac{0.027}{0.250} = 0.108 \text{ mol/dm³}$$
   

3. Determine the moles of NaOH used in titration:

   For 25 cm³ of NaOH:

   $$\text{Moles of NaOH in 25 cm³} = \frac{25}{1000} \times 0.108 = 0.0027 \text{ mol}$$
   

4. Moles of ethanoic acid:

   The moles of NaOH used is equal to the moles of ethanoic acid neutralized:

   $$\text{Moles of ethanoic acid in 23.8 cm³} = 0.0027 \text{ mol}$$
   

   To find the concentration of ethanoic acid in the original solution:

   $$\text{Concentration of ethanoic acid} = \frac{0.0027}{\frac{23.8}{1000}} = 0.113 \text{ mol/dm³}$$
   

5. Find $[H^+]$ from pH:

   $$[H^+] = 10^{-\text{pH}} = 10^{-2.86} = 1.38 \times 10^{-3} \text{ mol/dm³}$$
   

6. Use the expression for $K_a$:

   $$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$$

   Since $[H^+] = [CH_3COO^-]$, the equation simplifies to:

   $$K_a = \frac{(1.38 \times 10^{-3})^2}{0.113 - 1.38 \times 10^{-3}} \approx \frac{1.9 \times 10^{-6}}{0.111} = 1.71 \times 10^{-5}$$
   

7. Calculate $pK_a$:

   $$p{K}_a=-\log K_a=-\log (1.71\times 10^{-5})\approx 4.77$$

Question 5

Explain why 0.1M HCl has a pH = 1 but 0.1M $CH_3COOH$ has an approximate pH value of 3. ($K_a$ of $C{H}_{3}COO\mathrm{H\; = }$$1.85 \times 10^{-5}$).

Solution:

1. 0.1M HCl:

   HCl is a strong acid, meaning it fully dissociates in solution. Therefore:

   $$[H^+] = 0.1 \text{ mol/dm³}$$
   

   The pH is given by:

   $$pH = -\log[H^+] = -\log(0.1) = 1$$
   

2. 0.1M $CH_3COOH$:

   Acetic acid is a weak acid, which only partially dissociates. To calculate the pH, we use the $K_a$ expression:

   $$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$$

   Let $x$ be the concentration of $[H^+]$ formed. Initially, $[CH_3COOH] = 0.1 \text{ M}$, and we assume $x\ll \mathrm{0.1.\; Thus:}$

   $$K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1}$$

   Substituting the given $K_a$:

   $$1.85 \times 10^{-5} = \frac{x^2}{0.1}$$

   Solving for $x$:

   $$x^2 = 1.85 \times 10^{-6} \quad \Rightarrow \quad x = 1.36 \times 10^{-3} \text{ M}$$
   

   Therefore, the pH is:

   $$pH = -\log(1.36 \times 10^{-3}) \approx 2.87 \quad (\text{approximately } 3)$$
   

Question 6

Calculate the pH of a 0.01M aqueous solution of dichloroethanoic acid, given that its $K_a$ is $5.0 \times 10^{-2}$. 

Solution:

For a weak acid, the dissociation is given by:

$$K_a = \frac{[H^+][A^-]}{[HA]}$$

Let $x$ be the concentration of $H^+$ ions:

$$K_a = \frac{x^2}{0.01 - x} \approx \frac{x^2}{0.01}$$

Substitute $K_a = 5.0 \times 10^{-12}$:

$$5.0\times 10^{-12}=\frac{x^2}{\mathrm{0.01}}$$

Solving for $x$:

$$x^2 = 5.0 \times 10^{-14} \quad \Rightarrow \quad x = \sqrt{5.0 \times 10^{-14}} = 7.07 \times 10^{-7}$$

Thus, the pH is:

$$pH = -\log(7.07 \times 10^{-7}) \approx 6.65$$

Question 7

(i) The $H_3O^+$ in an aqueous solution of a strong acid is doubled if the concentration of the acid is doubled. However, in a weak acid, $H_3O^+$ increases by only a factor of $\sqrt{2}$.

(ii) The addition of sodium ethanoate to an aqueous solution of ethanoic acid raises the pH, but the addition of sodium chloride to an aqueous solution of HCl has practically no effect on the pH.

Solution:

(i) For a strong acid, it completely dissociates, so $[H_3O^+]$ is directly proportional to the acid concentration. For a weak acid, the dissociation follows an equilibrium governed by $K_a$, so the increase in $[H_3O^+]$ is less pronounced and proportional to the square root of the concentration.

(ii) Sodium ethanoate acts as a conjugate base to ethanoic acid, creating a buffer system and increasing the pH. In the case of HCl, sodium chloride does not affect the pH since chloride ions do not hydrolyze in water.

Question 8

The pH of 0.1M methanoic acid is 2.40 at 300K. Calculate the acid dissociation constant, $K_a$, assuming the degree of dissociation $x$ of the acid is negligible.

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Solution:

From the pH:

$$[H^{+}]=10^{-2.40}=3.98\times 10^{-3}\text{ M}$$

For a weak acid:

$$K_a = \frac{[H^+]^2}{[HA] - [H^+]}$$

Substituting values:

$$K_a = \frac{(3.98 \times 10^{-3})^2}{0.1 - 3.98 \times 10^{-3}} \approx \frac{1.58 \times 10^{-5}}{0.096} = 1.65 \times 10^{-4} \text{ mol/dm³}$$

This is approximately $1.58 \times 10^{-4} \text{ mol/dm³}$, as given.

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5. Comparing Strong Acids vs. Weak Acids

• Strong Acids (e.g., HCl, HNO₃) have very high Ka values and low pKa values, indicating that they dissociate almost completely in water.

• Weak Acids (e.g., acetic acid, citric acid) have lower Ka values and higher pKa values, meaning they only partially dissociate.

Acid	Ka	pKa
Hydrochloric acid (HCl)	~$10^7$	~ −7
Acetic acid (CH₃COOH)	$1.8\times 10^{-5}$	4.75
Citric acid	$7.4\times 10^{-4}$	3.14

The table shows that the stronger the acid, the larger the Ka and the smaller the pKa.

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6. Applications of Ka and pKa

Understanding Ka and pKa is crucial in:

• Buffer solutions: pKa values help design buffer systems for laboratory experiments, especially in biological systems where pH must be tightly regulated.
• Drug design: The pKa of drugs determines their absorption in the body and how they interact with biological molecules.

For example, this guide on buffer solutions explains how pKa is essential in preparing stable buffers in labs.

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7. Frequently Asked Questions (FAQs)

Q: Why is pKa important?
A: pKa helps predict how an acid behaves in different environments. It determines the degree of dissociation, which is vital in chemical reactions, biological systems, and pharmaceutical applications.

Q: Can a strong acid have a Ka less than 1?
A: No, strong acids have Ka values much greater than 1, indicating nearly complete dissociation.

Q: What is the relationship between pKa and pH?
A: pKa is the pH at which half of the acid is dissociated. When the pH is equal to the pKa, the concentrations of the acid and its conjugate base are equal.

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Conclusion

The Acid Dissociation Constant (Ka) and pKa are key concepts in understanding the strength and behavior of acids in solution. By mastering these ideas, you can predict acid-base reactions, design buffer solutions, and gain a deeper understanding of chemical equilibrium.

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