Consider the dissociation of the weak acid HX as given below:
\[\text{HX(aq)} \rightleftharpoons \text{H}^+(\text{aq}) + \text{X}^-(\text{aq}), \, K_a = 1.2 \times 10^{-5}\]
\([K_a: \text{dissociation constant}]\)
The osmotic pressure of \(0.03 \, \text{M}\) aqueous solution of HX at 300 K is ______ \( \times 10^{-2} \, \text{bar} \) (nearest integer).
Given: \(R = 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}\)
\[\text{HX(aq)} \rightleftharpoons \text{H}^+(\text{aq}) + \text{X}^-(\text{aq}), \, K_a = 1.2 \times 10^{-5}\]
\([K_a: \text{dissociation constant}]\)
The osmotic pressure of \(0.03 \, \text{M}\) aqueous solution of HX at 300 K is ______ \( \times 10^{-2} \, \text{bar} \) (nearest integer).
Given: \(R = 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}\)
Correct Answer: 76
Approach Solution - 1
The dissociation of HX is represented as:
\[\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-\]
Initial concentration of HX: 0.03 M
At equilibrium:
\[[\text{HX}] = 0.03 - x, \quad [\text{H}^+] = x, \quad [\text{X}^-] = x\]
Using the dissociation constant $K_a$:
\[K_a = \frac{x^2}{0.03 - x}\]
For $K_a = 1.2 \times 10^{-5}$ and $0.03 - x \approx 0.03$ (since $K_a$ is very small):
\[1.2 \times 10^{-5} = \frac{x^2}{0.03}\]
\[x^2 = 1.2 \times 10^{-5} \times 0.03 = 3.6 \times 10^{-7}\]
\[x = \sqrt{3.6 \times 10^{-7}} = 6 \times 10^{-4}\]
Total solute concentration:
\[C_{\text{total}} = [\text{HX}] + [\text{H}^+] + [\text{X}^-] = 0.03 - x + x + x = 0.03 + x\]
\[C_{\text{total}} = 0.03 + 6 \times 10^{-4} = 0.0306 \, \text{M}\]
Osmotic pressure $\Pi$ is calculated using:
\[\Pi = C_{\text{total}}RT\]
\[\Pi = (0.0306) \times (0.083) \times (300)\]
\[\Pi = 76.19 \, \text{bar}\]
Nearest integer:
\[\Pi = 76 \times 10^{-2} \, \text{bar}\]
Approach Solution -2
The problem requires the calculation of the osmotic pressure of a weak acid solution, given its concentration, dissociation constant, and temperature.
Concept Used:
The osmotic pressure (\(\Pi\)) of a solution containing a dissociating solute is given by the van't Hoff equation:
\[ \Pi = i \cdot C \cdot R \cdot T \]where:
- \(i\) is the van't Hoff factor, which accounts for the number of particles in the solution after dissociation.
- \(C\) is the molar concentration of the solution.
- \(R\) is the ideal gas constant.
- \(T\) is the absolute temperature in Kelvin.
For a weak acid HX that dissociates as \(\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-\), the van't Hoff factor is related to the degree of dissociation (\(\alpha\)) by:
\[ i = 1 + \alpha \]The degree of dissociation (\(\alpha\)) can be found from the acid dissociation constant (\(K_a\)) and the concentration (\(C\)) using the Ostwald's dilution law:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]For a weak acid where \(\alpha \ll 1\), this can be approximated as \(K_a \approx C\alpha^2\).
Step-by-Step Solution:
Step 1: List the given values.
- Concentration, \(C = 0.03 \, \text{M}\)
- Acid dissociation constant, \(K_a = 1.2 \times 10^{-5}\)
- Temperature, \(T = 300 \, \text{K}\)
- Gas constant, \(R = 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}\)
Step 2: Calculate the degree of dissociation (\(\alpha\)).
Using the approximate formula for a weak acid, \(K_a \approx C\alpha^2\):
\[ \alpha = \sqrt{\frac{K_a}{C}} \]Substituting the given values:
\[ \alpha = \sqrt{\frac{1.2 \times 10^{-5}}{0.03}} = \sqrt{\frac{1.2 \times 10^{-5}}{3 \times 10^{-2}}} = \sqrt{0.4 \times 10^{-3}} = \sqrt{4 \times 10^{-4}} \] \[ \alpha = 2 \times 10^{-2} = 0.02 \]Since \(\alpha = 0.02\) is much less than 1, our approximation is valid.
Step 3: Calculate the van't Hoff factor (\(i\)).
The dissociation is \(\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-\), so one molecule produces two ions. The van't Hoff factor is:
\[ i = 1 + \alpha = 1 + 0.02 = 1.02 \]Step 4: Calculate the osmotic pressure (\(\Pi\)).
Using the van't Hoff equation for osmotic pressure:
\[ \Pi = i \cdot C \cdot R \cdot T \]Substitute all the calculated and given values:
\[ \Pi = (1.02) \times (0.03 \, \text{mol L}^{-1}) \times (0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}) \times (300 \, \text{K}) \] \[ \Pi = 1.02 \times 0.03 \times 0.083 \times 300 \, \text{bar} \] \[ \Pi = 1.02 \times (0.03 \times 300) \times 0.083 \, \text{bar} \] \[ \Pi = 1.02 \times 9 \times 0.083 \, \text{bar} \] \[ \Pi = 9.18 \times 0.083 \, \text{bar} \] \[ \Pi = 0.76194 \, \text{bar} \]Final Computation & Result:
The problem asks for the answer in the format ______ \(\times 10^{-2}\) bar (nearest integer).
We need to convert our result to this format:
\[ \Pi = 0.76194 \, \text{bar} = 76.194 \times 10^{-2} \, \text{bar} \]Rounding to the nearest integer, we get 76.
The osmotic pressure of the solution is 76 \( \times 10^{-2} \, \text{bar} \).
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