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Solutions - Van't Hoff factor

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Van't Hoff factor (ii) is a measure of the effect of a solute on colligative properties such as osmotic pressure, relative lowering of vapor pressure, elevation of boiling point, and depression of freezing point.

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It accounts for the dissociation or association of solute particles in a solution. For non-electrolytes that do not associate or dissociate, i=1i = 1.

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Dissociation: When a solute molecule breaks into multiple ions (e.g., NaClNaCl, K2SO4K_2SO_4), the number of particles increases, making i>1i > 1.

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Association: When solute molecules join together to form a larger molecule (e.g., dimerization of acetic acid in benzene), the number of particles decreases, making i<1i < 1.

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The factor is calculated as the ratio of the observed colligative property to the theoretical (calculated) colligative property, or the ratio of normal molar mass to the abnormal (observed) molar mass.

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Degree of dissociation (α\alpha) and degree of association (α\alpha) are quantitatively linked to ii and the number of ions/molecules (nn) involved in the process.

📐Formulae

i=Observed Colligative PropertyCalculated Colligative Propertyi = \frac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}}

i=Normal Molar MassAbnormal Molar Massi = \frac{\text{Normal Molar Mass}}{\text{Abnormal Molar Mass}}

i=Total moles of particles after association/dissociationTotal moles of particles before association/dissociationi = \frac{\text{Total moles of particles after association/dissociation}}{\text{Total moles of particles before association/dissociation}}

For Dissociation: α=i−1n−1\text{For Dissociation: } \alpha = \frac{i - 1}{n - 1}

For Association: α=i−1(1/n)−1\text{For Association: } \alpha = \frac{i - 1}{(1/n) - 1}

ΔP/P0=i⋅Xsolute\Delta P / P^0 = i \cdot X_{solute}

ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot m

ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m

π=i⋅C⋅R⋅T\pi = i \cdot C \cdot R \cdot T

💡Examples

Problem 1:

Determine the Van't Hoff factor (ii) for BaCl2BaCl_2 if it is 80%80\% dissociated in an aqueous solution.

Solution:

For BaCl2BaCl_2, the dissociation equation is: BaCl2→Ba2++2Cl−BaCl_2 \rightarrow Ba^{2+} + 2Cl^-. Thus, the number of particles n=1+2=3n = 1 + 2 = 3. Given α=80%=0.80\alpha = 80\% = 0.80. Using the formula: i=1+(n−1)αi = 1 + (n - 1)\alpha, we get i=1+(3−1)×0.80=1+2×0.80=1+1.60=2.60i = 1 + (3 - 1) \times 0.80 = 1 + 2 \times 0.80 = 1 + 1.60 = 2.60.

Explanation:

Since BaCl2BaCl_2 dissociates into 3 ions, if it were 100%100\% dissociated, ii would be 33. At 80%80\% dissociation, the value is 2.602.60.

Problem 2:

Benzoic acid (C6H5COOHC_6H_5COOH) dimerizes in benzene. If the observed molar mass is 244 g/mol244\, g/mol and the normal molar mass is 122 g/mol122\, g/mol, calculate the Van't Hoff factor (ii).

Solution:

i=Normal Molar MassAbnormal Molar Mass=122244=0.5i = \frac{\text{Normal Molar Mass}}{\text{Abnormal Molar Mass}} = \frac{122}{244} = 0.5

Explanation:

In cases of association, the observed molar mass is higher than the calculated one because the number of particles decreases. A value of 0.50.5 indicates complete dimerization.

Problem 3:

Calculate the osmotic pressure of a 0.1 M0.1\, M solution of K2SO4K_2SO_4 at 27∘C27^\circ C, assuming complete dissociation. (R=0.0821 L⋅atm⋅K−1⋅mol−1R = 0.0821\, L\cdot atm\cdot K^{-1}\cdot mol^{-1})

Solution:

K2SO4K_2SO_4 dissociates as K2SO4→2K++SO42−K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}, so n=3n = 3. For complete dissociation, i=3i = 3. T=27+273=300 KT = 27 + 273 = 300\, K. Using π=iCRT\pi = iCRT: π=3×0.1×0.0821×300=7.389 atm\pi = 3 \times 0.1 \times 0.0821 \times 300 = 7.389\, atm.

Explanation:

The Van't Hoff factor multiplies the concentration to account for the triple number of particles resulting from the salt's dissociation.

Van't Hoff factor - Revision Notes & Key Formulas | ICSE Class 12 Chemistry