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Chemistry Question on Colligative Properties

Consider the dissociation of the weak acid HX as given below:
HX(aq)⇌H+(aq)+X−(aq), Ka=1.2×10−5\text{HX(aq)} \rightleftharpoons \text{H}^+(\text{aq}) + \text{X}^-(\text{aq}), \, K_a = 1.2 \times 10^{-5}
[Ka:dissociation constant][K_a: \text{dissociation constant}]
The osmotic pressure of 0.03 M0.03 \, \text{M} aqueous solution of HX at 300 K is ______ ×10−2 bar\times 10^{-2} \, \text{bar} (nearest integer).
Given: R=0.083 L bar mol−1K−1R = 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}

Answer

The dissociation of HX is represented as:
HX⇌H++X−\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-
Initial concentration of HX: 0.03 M
At equilibrium:
[HX]=0.03−x,[H+]=x,[X−]=x[\text{HX}] = 0.03 - x, \quad [\text{H}^+] = x, \quad [\text{X}^-] = x
Using the dissociation constant KaK_a:
Ka=x20.03−xK_a = \frac{x^2}{0.03 - x}
For Ka=1.2×10−5K_a = 1.2 \times 10^{-5} and 0.03−x≈0.030.03 - x \approx 0.03 (since KaK_a is very small):
1.2×10−5=x20.031.2 \times 10^{-5} = \frac{x^2}{0.03}
x2=1.2×10−5×0.03=3.6×10−7x^2 = 1.2 \times 10^{-5} \times 0.03 = 3.6 \times 10^{-7}
x=3.6×10−7=6×10−4x = \sqrt{3.6 \times 10^{-7}} = 6 \times 10^{-4}
Total solute concentration:
Ctotal=[HX]+[H+]+[X−]=0.03−x+x+x=0.03+xC_{\text{total}} = [\text{HX}] + [\text{H}^+] + [\text{X}^-] = 0.03 - x + x + x = 0.03 + x
Ctotal=0.03+6×10−4=0.0306 MC_{\text{total}} = 0.03 + 6 \times 10^{-4} = 0.0306 \, \text{M}
Osmotic pressure Π\Pi is calculated using:
Π=CtotalRT\Pi = C_{\text{total}}RT
Π=(0.0306)×(0.083)×(300)\Pi = (0.0306) \times (0.083) \times (300)
Π=76.19 bar\Pi = 76.19 \, \text{bar}
Nearest integer:
Π=76×10−2 bar\Pi = 76 \times 10^{-2} \, \text{bar}

Explanation

Solution

The dissociation of HX is represented as:
HX⇌H++X−\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-
Initial concentration of HX: 0.03 M
At equilibrium:
[HX]=0.03−x,[H+]=x,[X−]=x[\text{HX}] = 0.03 - x, \quad [\text{H}^+] = x, \quad [\text{X}^-] = x
Using the dissociation constant KaK_a:
Ka=x20.03−xK_a = \frac{x^2}{0.03 - x}
For Ka=1.2×10−5K_a = 1.2 \times 10^{-5} and 0.03−x≈0.030.03 - x \approx 0.03 (since KaK_a is very small):
1.2×10−5=x20.031.2 \times 10^{-5} = \frac{x^2}{0.03}
x2=1.2×10−5×0.03=3.6×10−7x^2 = 1.2 \times 10^{-5} \times 0.03 = 3.6 \times 10^{-7}
x=3.6×10−7=6×10−4x = \sqrt{3.6 \times 10^{-7}} = 6 \times 10^{-4}
Total solute concentration:
Ctotal=[HX]+[H+]+[X−]=0.03−x+x+x=0.03+xC_{\text{total}} = [\text{HX}] + [\text{H}^+] + [\text{X}^-] = 0.03 - x + x + x = 0.03 + x
Ctotal=0.03+6×10−4=0.0306 MC_{\text{total}} = 0.03 + 6 \times 10^{-4} = 0.0306 \, \text{M}
Osmotic pressure Π\Pi is calculated using:
Π=CtotalRT\Pi = C_{\text{total}}RT
Π=(0.0306)×(0.083)×(300)\Pi = (0.0306) \times (0.083) \times (300)
Π=76.19 bar\Pi = 76.19 \, \text{bar}
Nearest integer:
Π=76×10−2 bar\Pi = 76 \times 10^{-2} \, \text{bar}