Problem 169
Question
The degree of dissociation \((\alpha)\) of a weak electrolyte, \(\mathrm{A}_{\mathrm{x}} \mathrm{B}_{y}\) is related to van't Hoff factor (i) by the expression: (a) \(\alpha=\frac{x+y-1}{i-1}\) (b) \(\alpha=\frac{x+y+1}{i-1}\) (c) \(\alpha=\frac{i-1}{(x+y-1)}\) (d) \(\alpha=\frac{i-1}{x+y+1}\)
Step-by-Step Solution
Verified Answer
The correct expression is (c) \(\alpha=\frac{i-1}{(x+y-1)}\).
1Step 1: Understanding the Problem
The problem asks us to find the correct expression that relates the degree of dissociation \((\alpha)\) of a weak electrolyte \(\mathrm{A}_{\mathrm{x}} \mathrm{B}_{y}\) with the van't Hoff factor \(i\). We have four options to determine which one is correct.
2Step 2: Conceptual Background
The van't Hoff factor \(i\) accounts for the number of particles in solution after dissociation compared to the number initially. For a weak electrolyte \(\mathrm{A}_{\mathrm{x}} \mathrm{B}_{y}\), the dissociation can be represented as: \[ \mathrm{A}_{\mathrm{x}} \mathrm{B}_{y} \rightarrow x \mathrm{A} + y \mathrm{B} \] The total number of particles after dissociation is \(1 + \alpha(x + y - 1)\). Thus, the van't Hoff factor is \(i = 1 + \alpha(x + y - 1)\).
3Step 3: Relating \(\alpha\) and \(i\)
We have derived the relationship \(i = 1 + \alpha(x + y - 1)\). To find \(\alpha\), rearrange this equation: \[ \alpha(x + y - 1) = i - 1 \] Thus, \[ \alpha = \frac{i - 1}{x + y - 1} \]
4Step 4: Comparing with Options
Now, compare the derived expression \(\alpha = \frac{i - 1}{x + y - 1}\) with the options provided in the question: (a) \(\alpha=\frac{x+y-1}{i-1}\), (b) \(\alpha=\frac{x+y+1}{i-1}\), (c) \(\alpha=\frac{i-1}{(x+y-1)}\), (d) \(\alpha=\frac{i-1}{x+y+1}\). The correct option is (c) as it matches our derived expression.
Key Concepts
Van't Hoff FactorWeak Electrolyte DissociationChemical Equilibrium
Van't Hoff Factor
The van't Hoff factor \(i\) is a crucial concept in understanding how solutes behave in a solution, especially in the case of electrolytes. It signifies the number of particles present in the solution following the dissociation of the solute.
For non-electrolytes, which do not dissociate in solution, \(i\) is typically 1, as there are no additional particles formed. However, for electrolytes, it becomes important to account for the breaking down into ions.
For non-electrolytes, which do not dissociate in solution, \(i\) is typically 1, as there are no additional particles formed. However, for electrolytes, it becomes important to account for the breaking down into ions.
- In the context of weak electrolytes like \(A_xB_y\), the dissociation is partial, hence the van't Hoff factor may not be an integer.
- The van't Hoff factor can be expressed as \(i = 1 + \alpha(x + y - 1)\), where \(\alpha\) is the degree of dissociation.
Weak Electrolyte Dissociation
Weak electrolytes are compounds that only partially dissociate into ions when dissolved in water. Unlike strong electrolytes, which fully dissociate, weak electrolytes establish a dynamic equilibrium between the dissolved ions and undissociated molecules.
When a weak electrolyte such as \(A_xB_y\) dissociates, it results in:
When a weak electrolyte such as \(A_xB_y\) dissociates, it results in:
- \(A_xB_y \) \(\rightarrow\) \(xA^+ + yB^- \).
- Instead of completely turning into ions, only a fraction represented by the degree of dissociation \(\alpha\) converts, affecting the total number of particles in solution.
Chemical Equilibrium
Chemical equilibrium is a key concept in chemistry, pertinent to reactions involving weak electrolytes. It describes a state where the rate of the forward reaction (dissociation into ions) equals the rate of the reverse reaction (recombination into molecules).
At equilibrium in a solution of a weak electrolyte like \(A_xB_y\), the concentration of reactants and products remains constant over time. This balance allows for:
At equilibrium in a solution of a weak electrolyte like \(A_xB_y\), the concentration of reactants and products remains constant over time. This balance allows for:
- Predicting ion concentrations using equilibrium constants.
- Deriving relationships between measurable properties, such as in the case of the van't Hoff factor and degree of dissociation.
Other exercises in this chapter
Problem 167
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water \(\left(\Delta
View solution Problem 168
On mixing, heptane and octane form an ideal solution. At \(373 \mathrm{~K}\), the vapour pressures of the two liquid components (heptane and octane) are \(105 \
View solution Problem 170
Ethylene glycol is used as antifreeze in a cold climate. Mass of ethylene glycol which should be added to \(4 \mathrm{~kg}\) of water to prevent it form freezin
View solution Problem 171
\(\mathrm{K}_{\mathrm{f}}\) for water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\). If your automobile radiator holds \(1.0 \mathrm{~kg}\) of water,
View solution