2 molal solution of a weak acid HA has a freezing point of 3.885°C. The degree of dissociation of th

Q: Why is the freezing point given as a positive value in the question?

It's a common typo in such problems. Since water freezes at $0^\circ\text{C}$, the freezing point of an aqueous solution must be negative. The value $3.885^\circ\text{C}$ actually represents the magnitude of the depression in freezing point ($\Delta T_f$).

Q: How do we know the acid dissociates into exactly two ions?

The chemical formula $\text{HA}$ implies a monoprotic acid. When it dissociates in water, it yields one proton ($\text{H}^+$) and one conjugate base anion ($\text{A}^-$), making a total of two ions per molecule.

Q: Why is the van't Hoff factor $i = 1 + \alpha$?

Initially, assume we have $1 \text{ mole}$ of $\text{HA}$. At equilibrium, $1-\alpha \text{ moles}$ of $\text{HA}$ remain, and $\alpha \text{ moles}$ of $\text{H}^+$ and $\alpha \text{ moles}$ of $\text{A}^-$ are formed. The total number of moles of particles is $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$.

Q: What if the acid was a strong acid instead of a weak acid?

For a strong acid, it would dissociate almost completely, meaning $\alpha \approx 1$. Consequently, the van't Hoff factor $i$ would be approximately $2$, leading to a much larger freezing point depression.

Q: Why do we multiply by $10^{-3}$ at the end?

The question specifically asks for the answer in the format of $x \times 10^{-3}$ to extract an integer value for the numerical response. Since $\alpha = 0.05$, we rewrite it as $50 \times 10^{-3}$ to find that $x = 50$.

Question

2 molal solution of a weak acid HA has a freezing point of $3.885^\circ	ext{C}$. The degree of dissociation of this acid is ......... $	imes 10^{-3}$. (Round off to the nearest integer).
[Given : Molal depression constant of water = $1.85	ext{ K kg mol}^{-1}$, freezing point of pure water = $0^\circ	ext{C}$]

Sigma Percentile · Accepted Answer

Welcome! Let's dive into this interesting physical chemistry problem. We have a $2 	ext{ molal}$ solution of a weak acid, $	ext{HA}$. The problem states its freezing point is $3.885^\circ	ext{C}$.

Now, wait a minute! Water freezes at $0^\circ	ext{C}$, so an aqueous solution must freeze below zero. This means the text implies the **depression in freezing point**, $\Delta T_f$, is exactly $3.885^\circ	ext{C}$. Let's visualize our beaker with the acid solution and break down the math.

### The Master Equation

To connect the freezing point depression with the properties of our solute, we need our master equation. The depression in freezing point is given by:

$$\Delta T_f = i \cdot K_f \cdot m$$

Here, $\Delta T_f$ equals the van't Hoff factor $i$, times the molal depression constant $K_f$, times the molality $m$. The van't Hoff factor is crucial here because our weak acid $	ext{HA}$ will partially dissociate into ions, increasing the total number of particles in the solution.

### Calculating the van't Hoff Factor

Let's substitute the values we know into our formula. We plug in $3.885$ for $\Delta T_f$. We leave $i$ as our unknown. For $K_f$, we put $1.85$, and for molality, we put $2$. Notice how we are just setting up the raw structure before doing any heavy lifting:

$$3.885 = i \cdot 1.85 \cdot 2$$

Now, let's do the math. $1.85$ times $2$ gives us exactly $3.7$. So, we have $3.885 = i \cdot 3.7$. Dividing both sides by $3.7$, we find that the van't Hoff factor $i$ is exactly $1.05$. This value tells us the effective number of particles per molecule of $	ext{HA}$.

### Degree of Dissociation

Now, let's look at what happens inside the solution. The weak acid $	ext{HA}$ dissociates into $	ext{H}^+$ and $	ext{A}^-$ ions:

$$	ext{HA} ightleftharpoons 	ext{H}^+ + 	ext{A}^-$$

If $\alpha$ is the degree of dissociation, for every $1 	ext{ mole}$ of $	ext{HA}$, we get $\alpha 	ext{ moles}$ of $	ext{H}^+$ and $\alpha 	ext{ moles}$ of $	ext{A}^-$, while $1 - \alpha 	ext{ moles}$ of $	ext{HA}$ remain undissociated. The total number of particles is $(1 - \alpha) + \alpha + \alpha$, which simplifies to $1 + \alpha$. So, $i = 1 + \alpha$.

We just found that $i$ is $1.05$. Let's equate the two expressions for $i$:

$$1.05 = 1 + \alpha$$

Subtracting $1$ from both sides, we get $\alpha = 0.05$. This means $5\%$ of the acid molecules have dissociated.

### Formatting the Final Answer

We are almost there! The question asks for the degree of dissociation in the format of some number times $10^{-3}$. Let's rewrite $0.05$. We can write it as $50 	imes 10^{-3}$.

So, the integer value we are looking for is **$50$**. And that is our final answer! What a beautiful problem that seamlessly connected colligative properties with ionic equilibrium.

Enter Numerical Value:

Visualized Solution

$A na l y z in g t h e S e t u p$

$D e p r ess i o nin F r eez in g P o in t$

$S u b s t i t u t in g t h e V a l u es$

$C a l c u l a t in g v a n^{'} t H o f f F a c t or$

$D e g r eeo f D i ssoc ia t i o n$

$F in d in g α$

$F or ma tt in g t h e A n s w er$

$C o n c l u s i o n$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Master Equation

Calculating the van't Hoff Factor

Degree of Dissociation

Formatting the Final Answer

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Enter Numerical Value:

Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculatingvan′tHoffFactor

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

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

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

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