Question

$\Lambda_m$ for 0.01 M nicotinic acid (weak monoprotic acid) is 10 times smaller than that of 0.1 M formic acid. If

$\lambda_{nicotinate ion}^{\circ} = \lambda_{formate ion}^{\circ}$ then, $(\alpha << 1 \text{ for both acids})$

• A. $|(pK_a)_{nicotinic \text{ }acid} - (pK_a)_{formic \text{ }acid}| = 3$
• B. $\frac{(K_a)_{nicotinic \text{ }acid}}{(K_a)_{formic \text{ }acid}} = 10^2$
• C. $\frac{(K_a)_{formic \text{ }acid}}{(K_a)_{nicotinic \text{ }acid}} = 10^3$
• D. $(pK_a)_{formic \text{ }acid} - (pK_a)_{nicotinic \text{ }acid} = -2$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The problem involves comparing the dissociation constants ($K_a$) and $pK_a$ values of two weak monoprotic acids: nicotinic acid (HA1) and formic acid (HA2), based on their molar conductivities.

1. Given Information and Relationships:

• Nicotinic Acid (HA1):

  • Concentration, $C_1 = 0.01$ M
  • Molar conductivity, $\Lambda_{m1}$
  • Degree of dissociation, $\alpha_1$
  • Dissociation constant, $(K_a)_1$

• Formic Acid (HA2):

  • Concentration, $C_2 = 0.1$ M
  • Molar conductivity, $\Lambda_{m2}$
  • Degree of dissociation, $\alpha_2$
  • Dissociation constant, $(K_a)_2$

• Conductivity Relationship: $\Lambda_{m1} = \frac{1}{10} \Lambda_{m2}$

• Limiting Molar Conductivity of Ions: $\lambda_{nicotinate ion}^{\circ} = \lambda_{formate ion}^{\circ}$

• Approximation: $\alpha << 1$ for both acids.

2. Limiting Molar Conductivities of Acids ($\Lambda_m^{\circ}$):

For a monoprotic acid HA, $\Lambda_m^{\circ} = \lambda_{H^+}^{\circ} + \lambda_{A^-}^{\circ}$.

• For nicotinic acid: $\Lambda_{m1}^{\circ} = \lambda_{H^+}^{\circ} + \lambda_{nicotinate ion}^{\circ}$
• For formic acid: $\Lambda_{m2}^{\circ} = \lambda_{H^+}^{\circ} + \lambda_{formate ion}^{\circ}$

Since $\lambda_{nicotinate ion}^{\circ} = \lambda_{formate ion}^{\circ}$, it implies that $\Lambda_{m1}^{\circ} = \Lambda_{m2}^{\circ}$. Let's denote this common value as $\Lambda^{\circ}$.

3. Relation between Molar Conductivity and Degree of Dissociation:

The degree of dissociation ($\alpha$) for a weak electrolyte is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$. So, $\Lambda_m = \alpha \Lambda_m^{\circ}$.

• For nicotinic acid: $\Lambda_{m1} = \alpha_1 \Lambda^{\circ}$
• For formic acid: $\Lambda_{m2} = \alpha_2 \Lambda^{\circ}$

4. Using the Given Conductivity Relationship:

Substitute the expressions for $\Lambda_{m1}$ and $\Lambda_{m2}$ into the given relation $\Lambda_{m1} = \frac{1}{10} \Lambda_{m2}$: $\alpha_1 \Lambda^{\circ} = \frac{1}{10} (\alpha_2 \Lambda^{\circ})$ Since $\Lambda^{\circ} \neq 0$, we can cancel it out: $\alpha_1 = \frac{1}{10} \alpha_2 \implies \alpha_2 = 10 \alpha_1$

5. Relation between Degree of Dissociation and Dissociation Constant ($K_a$):

For a weak acid HA, $HA \rightleftharpoons H^+ + A^-$.

At equilibrium, concentrations are $C(1-\alpha)$, $C\alpha$, $C\alpha$. The dissociation constant is $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$. Given that $\alpha << 1$, we can approximate $1-\alpha \approx 1$. So, $K_a = C\alpha^2$. This implies $\alpha = \sqrt{\frac{K_a}{C}}$.

6. Applying to Both Acids:

• For nicotinic acid: $(K_a)_1 = C_1 \alpha_1^2 = 0.01 \alpha_1^2$
• For formic acid: $(K_a)_2 = C_2 \alpha_2^2 = 0.1 \alpha_2^2$

7. Substituting $\alpha_2 = 10 \alpha_1$ into the $K_a$ expression for formic acid:

$(K_a)_2 = 0.1 (10 \alpha_1)^2 = 0.1 (100 \alpha_1^2) = 10 \alpha_1^2$

8. Finding the Ratio of $K_a$ Values:

Now, let's find the ratio of the dissociation constants: $\frac{(K_a)_2}{(K_a)_1} = \frac{10 \alpha_1^2}{0.01 \alpha_1^2} = \frac{10}{0.01} = \frac{10}{1/100} = 10 \times 100 = 1000$ So, $\frac{(K_a)_{formic \text{ }acid}}{(K_a)_{nicotinic \text{ }acid}} = 10^3$. This matches option C.

9. Finding the Relationship between $pK_a$ Values:

We know that $pK_a = -\log K_a$. From $\frac{(K_a)_{formic \text{ }acid}}{(K_a)_{nicotinic \text{ }acid}} = 10^3$: Take logarithm on both sides: $\log \left( \frac{(K_a)_{formic \text{ }acid}}{(K_a)_{nicotinic \text{ }acid}} \right) = \log(10^3)$ $\log (K_a)_{formic \text{ }acid} - \log (K_a)_{nicotinic \text{ }acid} = 3$ Multiply by -1: $-(\log (K_a)_{formic \text{ }acid} - \log (K_a)_{nicotinic \text{ }acid}) = -3$ $(-\log (K_a)_{formic \text{ }acid}) - (-\log (K_a)_{nicotinic \text{ }acid}) = -3$ $(pK_a)_{formic \text{ }acid} - (pK_a)_{nicotinic \text{ }acid} = -3$

Now, let's check the options:

• A. $|(pK_a)_{nicotinic \text{ }acid} - (pK_a)_{formic \text{ }acid}| = 3$ From our result, $(pK_a)_{nicotinic \text{ }acid} - (pK_a)_{formic \text{ }acid} = 3$. So, $|(pK_a)_{nicotinic \text{ }acid} - (pK_a)_{formic \text{ }acid}| = |3| = 3$. This option is correct.
• B. $\frac{(K_a)_{nicotinic \text{ }acid}}{(K_a)_{formic \text{ }acid}} = 10^2$ From our result, $\frac{(K_a)_{nicotinic \text{ }acid}}{(K_a)_{formic \text{ }acid}} = \frac{1}{10^3} = 10^{-3}$. This option is incorrect.
• C. $\frac{(K_a)_{formic \text{ }acid}}{(K_a)_{nicotinic \text{ }acid}} = 10^3$ This matches our derived result. This option is correct.
• D. $(pK_a)_{formic \text{ }acid} - (pK_a)_{nicotinic \text{ }acid} = -2$ From our result, $(pK_a)_{formic \text{ }acid} - (pK_a)_{nicotinic \text{ }acid} = -3$. This option is incorrect.

Both options A and C are correct based on the calculations.

The final answer is $\boxed{A, C}$

Explanation of the solution:

1. Equate limiting molar conductivities: Since the limiting molar conductivities of the anions are equal ($\lambda_{nicotinate ion}^{\circ} = \lambda_{formate ion}^{\circ}$) and both acids share the $\lambda_{H^+}^{\circ}$ component, their limiting molar conductivities ($\Lambda_m^{\circ}$) are equal.
2. Relate $\Lambda_m$ and $\alpha$: Use the formula $\Lambda_m = \alpha \Lambda_m^{\circ}$. Given $\Lambda_{m,nic} = \frac{1}{10} \Lambda_{m,form}$ and $\Lambda_{m,nic}^{\circ} = \Lambda_{m,form}^{\circ}$, it leads to $\alpha_{nic} = \frac{1}{10} \alpha_{form}$, or $\alpha_{form} = 10 \alpha_{nic}$.
3. Relate $K_a$ and $\alpha$: For weak acids with $\alpha << 1$, $K_a = C\alpha^2$.
4. Calculate $K_a$ ratio: Substitute the concentrations ($C_{nic} = 0.01$ M, $C_{form} = 0.1$ M) and the relationship between $\alpha$ values into the $K_a$ expressions. $(K_a)_{nic} = 0.01 \alpha_{nic}^2$ $(K_a)_{form} = 0.1 \alpha_{form}^2 = 0.1 (10 \alpha_{nic})^2 = 0.1 (100 \alpha_{nic}^2) = 10 \alpha_{nic}^2$ The ratio $\frac{(K_a)_{form}}{(K_a)_{nic}} = \frac{10 \alpha_{nic}^2}{0.01 \alpha_{nic}^2} = \frac{10}{0.01} = 1000 = 10^3$. This confirms option C.
5. Calculate $pK_a$ difference: Use the relation $pK_a = -\log K_a$. From $\frac{(K_a)_{form}}{(K_a)_{nic}} = 10^3$, taking negative logarithm yields $(pK_a)_{form} - (pK_a)_{nic} = -3$. Therefore, $(pK_a)_{nic} - (pK_a)_{form} = 3$. So, $|(pK_a)_{nic} - (pK_a)_{form}| = |3| = 3$. This confirms option A.