In a solvent 50% of an acid HA dimerises and the rest dissociates. The van't Hoff factor of the acid

Q: Why did we assume 1 mole initially? Can I assume something else?

You can assume any amount, say 'x' moles. The van't Hoff factor is a ratio, so the initial amount 'x' will eventually cancel out in the numerator and denominator. Assuming 1 makes the calculations much faster and more intuitive.

Q: How do we know the reactions go to 100% completion?

The problem states '50% dimerises and the rest dissociates' without providing any equilibrium constants or degree of dissociation ($\alpha$). In such standard JEE problems, unless $\alpha$ is given or asked for, we assume the respective processes go to 100% completion for their allocated fractions.

Q: Why is the moles of dimer 0.25 and not 0.5?

Dimerisation means two molecules combine to form one ($2HA \rightarrow H_2A_2$). Therefore, the number of moles of the product is exactly half the number of moles of the reactant consumed. Since 0.5 moles reacted, $0.5 / 2 = 0.25$ moles of dimer are formed.

Q: What if the acid was trimerising instead?

If it were trimerising ($3HA \rightarrow H_3A_3$), you would divide the reacting moles by 3. So 0.5 moles would yield $0.5 / 3$ moles of the trimer.

Q: Why do we add all the final moles together? They are different chemicals!

The van't Hoff factor depends on the *total* number of solute particles in the solution, regardless of their chemical identity. Colligative properties are 'democratic'—they depend only on the number of particles, not what those particles are.

Question

In a solvent 50% of an acid HA dimerises and the rest dissociates. The van't Hoff factor of the acid is ………… $	imes 10^{-2}$. (Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### The Split Personality of Acid HA: Dimerisation and Dissociation

Imagine you are observing a beaker filled with a solvent. You drop exactly $1$ mole of a weak acid, $HA$, into it. But this acid has a fascinating split personality! Half of it wants to pair up and form a larger molecule, while the other half wants to break apart into smaller ions.

Our goal is to find the **van't Hoff factor ($i$)**, which is a measure of the effect of a solute on colligative properties. It is defined simply as the ratio of the total number of moles of all particles present in the solution at equilibrium to the initial number of moles of the solute.

$$i = \frac{	ext{Total final moles}}{	ext{Initial moles}}$$

### The Dimerisation Phase

Let's look at the first half of our acid. The problem states that $50\%$ of the acid undergoes dimerisation. Since we started with $1$ mole, this means $0.5$ moles of $HA$ will dimerise.

In dimerisation, two molecules of $HA$ join hands to form one single molecule of $H_2A_2$.

$$2HA ightleftharpoons H_2A_2$$

Because two moles of reactant yield only one mole of product, the number of moles of the dimer formed will be exactly half of the $HA$ consumed.

$$n_{H_2A_2} = \frac{0.5}{2} = 0.25 	ext{ moles}$$

### The Dissociation Phase

Now for the remaining $50\%$. This $0.5$ moles of $HA$ dissociates completely into $H^+$ and $A^-$ ions.

$$HA ightleftharpoons H^+ + A^-$$

One mole of $HA$ gives one mole of each ion. Therefore, $0.5$ moles of $HA$ will yield $0.5$ moles of $H^+$ and $0.5$ moles of $A^-$.

### The Master Equation

Let's count the total particles swimming in our beaker now. We have the dimer molecules and the dissociated ions all mixed together.

$$n_{	ext{final}} = n_{H_2A_2} + n_{H^+} + n_{A^-}$$

$$n_{	ext{final}} = 0.25 + 0.5 + 0.5 = 1.25 	ext{ moles}$$

### Final Calculation

Plugging this into our van't Hoff factor formula, we divide the final $1.25$ moles by the initial $1$ mole.

$$i = \frac{1.25}{1} = 1.25$$

There is a slight catch here. The question asks for the answer in the specific format of $x 	imes 10^{-2}$. To match this, we simply rewrite our decimal as a whole number multiplied by the required power of ten.

$$i = 125 	imes 10^{-2}$$

Our final integer value is **$125$**! Always remember to read the question carefully to see if the reaction goes to $100\%$ completion. If it doesn't, you would need to incorporate the degree of dissociation or association ($\alpha$) into your calculations.

Animated Solution for Chemistry - Solutions: In a solvent 50% of an acid HA dimerises and the rest dissociates. The van't Hoff factor of the acid is ………… $\times 1 0^{- 2}$ . (Round off to the nearest integer)

Enter Numerical Value:

Visualized Solution

$n_{initial} = 1$

$i = \frac{n _{final}}{n _{initial}}$

$2 H A ⇌ H_{2} A_{2}$

$H A ⇌ H^{+} + A^{-}$

$n_{final} = 1.25$

$i = 125 \times 1 0^{- 2}$

$α < 1$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Split Personality of Acid HA

The Dimerisation Phase

The Dissociation Phase

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=ninitial​nfinal​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2HA⇌H2​A2​

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nfinal​=1.25

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=125×10−2

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