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What would be the value of Van ’t Hoff factor for a dilute solution of $ {K_2}S{O_4} $ in water?

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Last updated date: 13th Jul 2024
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Answer
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Hint: The Van ‘t Hoff factor for an ionic compound that is soluble in water is the ratio of the amount of discrete ions that are produced when the compound dissolves in water to the amount of substance. It is shown by the symbol $ i $ .

Complete answer:
The Van 't Hoff factor, $ i $ , is a measure of the effect of a solute on colligative properties such as osmotic pressure, relative lowering in vapor pressure, boiling-point elevation and freezing-point depression. For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance. This is true for ideal solutions only, as occasionally ion pairing occurs in solution. When solute particles neither dissociate nor associate in solution, $ i $ equals $ 1 $ .
The value of $ i $ is the actual number of particles in solution after dissociation divided by the number of formula units initially dissolved in solution and means the number of particles per formula unit of the solute when a solution is dilute.
 $ {K_2}S{O_4} $ dissociates into its constituents as follows:
 $ {K_2}S{O_4} \to 2{K^ + } + S{O_4}^{2 - } $
Here, the total ions produced are $ 3 $ , therefore, $ i = 3 $ .
Since, $ {K_2}S{O_4} $ completely dissociates therefore its Van ‘t Hoff factor $ = \dfrac{3}{1} = 3 $

Note:
In the actual solution of $ {K_2}S{O_4} $ in water, the value of Van ‘t Hoff factor is usually less than $ 3 $ . This happens because in dilute solution, the salts do not ionize completely. That is why the value of Van t’ Hoff factor remains less than the theoretically calculated value.