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Chapter 13: Problem 185

A \(0.100-\mathrm{g}\) sample of the weak acid HA (molar mass \(=\) \(100.0 \mathrm{g} / \mathrm{mol}\) ) is dissolved in \(500.0 \mathrm{g}\) water. The freezing point of the resulting solution is \(-0.0056^{\circ} \mathrm{C}\). Calculate the value of \(K_{\mathrm{a}}\) for this acid. Assume molality equals molarity in this solution.

Short Answer

Expert verified
The value of the acid dissociation constant (\(K_{a}\)) for the weak acid HA is approximately \(1.125\times 10^{-3}\).

Step by step solution

01

Calculate the molality of the solution

First, we need to find the molality of the solution (i.e., the number of moles of solute per kilogram of solvent). We can determine the molality using the molar mass of the acid and the mass of the solute and solvent. Given mass of acid (HA): \(0.100\,\text{g}\) Molar mass of acid (HA): \(100.0\,\text{g/mol}\) Mass of solution: \(0.100\,\text{g}\) + \(500.0\,\text{g}\) = \(500.100\,\text{g}\) Moles of HA: \(\frac{0.100\,\text{g}}{100.0\,\text{g/mol}}=0.001\,\mathrm{mol}\) Mass of water: \(500.0\,\text{g}\) = \(0.500\,\mathrm{kg}\) Molality, \(m=\frac{\mathrm{moles\,of\,solute}}{\mathrm{mass\,of\,solvent\,(kg)}}=\frac{0.001\,\mathrm{mol}}{0.500\,\mathrm{kg}}=0.002\,\mathrm{mol/kg}\)
02

Determine the ions in the solution

Next, we'll use the freezing point depression information to determine the number of ions that the acid produces when dissolved. The freezing point depression is given by the equation: \(ΔT_{f}=K_{f} \times m \times i\) ΔTₓ=f is the freezing point depression, Kₓ is the freezing point depression constant for water (1.86 °C/molal), m is the molality, and i is the van't Hoff factor (the number of particles into which the solute dissociates when dissolved). We need to solve for i. First, let's plug in the known values: \(-0.0056^{\circ} \mathrm{C} = 1.86\, °C \mathrm{molal/kg} \times 0.002\,\mathrm{mol/kg} \times i\)
03

Calculate the van't Hoff factor (i)

Now, let's solve for i: \(i=\frac{-0.0056^{\circ} \mathrm{C}}{1.86\, °C \mathrm{molal/kg} \times 0.002\,\mathrm{mol/kg}} = 1.5\) Since the value of i is close to 2, the acid HA dissociates into two ions, H⁺ and A⁻.
04

Calculate the dissociation

Now, let's calculate the dissociation of HA into ions: HA \(\rightleftharpoons\) H⁺ + A⁻ Initial concentration of HA: 0.002 M (as molality and molarity are nearly equal) \[ \begin{array}{c|c|c|c} & HA & H+ & A^- \\ \hline Initial & 0.002 & 0 & 0 \\ Change & -x & +x & +x \\ Equilibrium & 0.002-x & x & x \end{array} \]
05

Apply the equilibrium constant expression

Now, we'll use the acid dissociation constant expression: \(K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\) As the value of i is close to 2, we can assume that the equilibrium concentration of H⁺ and A⁻ is equal (x). Thus, \(K_{a} = \frac{x^2}{0.002 - x}\) Since i is close to 2, we know that a significant amount of the weak acid HA has dissociated into H⁺ and A⁻. So, we can make an approximation: \(x<<0.002\) Therefore, \(K_{a} \approx \frac{x^2}{0.002}\)
06

Calculate the concentration of H⁺

At this point, we have enough information to determine the concentration of H⁺ ions in the solution. Since we know they are equal to the concentration of A⁻ ions, we can calculate x using the van't Hoff factor from step 3: \(x = \frac{1.5}{2}[HA]\) \(x = 0.0015\,\text{M}\)
07

Calculate the value of \(K_{a}\)

Finally, we can use the calculated value of x to find the value of \(K_{a}\): \(K_{a} = \frac{x^2}{0.002} = \frac{(0.0015\,M)^2}{0.002\,M} = \frac{2.25 \times 10^{-6} \,M^2}{0.002\,M}= 1.125 \times 10^{-3}\) Thus, the value of \(K_{a}\) for the weak acid HA is approximately \(1.125\times 10^{-3}\).

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