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Chapter 14: Problem 191

A \(0.100-\mathrm{g}\) sample of the weak acid \(\mathrm{HA}\) (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 \(K_a\) for the weak acid is approximately 1.1 × 10^-4.

Step by step solution

01

Calculate molality of the solution

First, we need to determine the number of moles of the weak acid (HA) in the solution by using the given mass and molar mass: n(HA) = mass of HA / molar mass of HA = \( 0.100\mathrm{g} / 100.0\mathrm{g}·\mathrm{mol}^{-1} \) = \( 0.00100\mathrm {mol} \) Now we will determine the molality of the solution, using the mass of water, which assuming molality equals molarity: molality(HA) = moles of HA / mass of water (in kg) = \( 0.00100\mathrm{mol} / 0.500\mathrm{kg} \) = \( 0.00200\mathrm{mol}·\mathrm{kg}^{-1} \)
02

Find the molality of the solute

Next, we will use the freezing point depression formula to find the molality of the solute: ΔT = Kf × m × i where ΔT is the freezing point depression, Kf is the cryoscopic constant for water (1.86 °C·kg/mol), m is molality, and i is the van't Hoff factor. In this problem, i = 2, because the weak acid dissociates into two ions: H+ and A-. Rearrange the formula to find the molality of the solute: m = ΔT / (Kf × i) = \( -0.0056\mathrm{°C} / (1.86\mathrm{°C}·\mathrm{kg}·\mathrm{mol}^{-1} × 2) \) = \( 0.001505\mathrm{mol}·\mathrm{kg}^{-1} \)
03

Set up the ion concentration ratios

Now that we have the molality of the solute, we can set up equilibrium expressions for the dissociation of the weak acid into ions: \( HA \rightleftharpoons H^+ + A^- \) The equilibrium concentrations for the ions should be the same, assuming complete dissociation of the weak acid: \[ H^+ = A^- = x \] While the concentration of the weak acid is: \[ HA = 0.00200 - x \]
04

Calculate Ka using equilibrium concentrations

Finally, we will set up an equilibrium expression for the acid dissociation reaction and solve for Ka: \[ K_a = \frac{[H^+][A^-]}{[HA]} \] Substitute the equilibrium concentrations into the equation: \[ K_a = \frac{x^2}{0.00200-x} \] Since the value of x (concentration of the ions) is relatively small and can be neglected when compared to 0.00200, we can simplify the equation: \[ K_a \approx \frac{x^2}{0.00200} \] Now, substitute the molality of the solute (0.001505 mol·kg^-1) for x in the equation: \[ K_a \approx \frac{(0.001505)^2}{0.00200} = 1.1 \times 10^{-4} \] Thus, the value of Ka for this weak acid is approximately 1.1 × 10^-4.

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Most popular questions from this chapter

Place the species in each of the following groups in order of increasing acid strength. a. $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}\( (bond energies: \)\mathrm{H}-\mathrm{O}, 467 \mathrm{kJ} / \mathrm{mol}$ $\mathrm{H}-\mathrm{S}, 363 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{kJ} / \mathrm{mol} )$ b. $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}$ c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}^{+}\) (bond energies: $\mathrm{N}-\mathrm{H}, 391 \mathrm{kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H},$ 322 \(\mathrm{kJ} / \mathrm{mol} )\) Give reasons for the orders you chose.

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{Li}_{2} \mathrm{O}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{SrO}\)

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$

What are the major species present in the following mixtures of bases? a. 0.050 M NaOH and 0.050 M LiOH b. 0.0010\(M \mathrm{Ca}(\mathrm{OH})_{2}\) and 0.020\(M \mathrm{RbOH}\) What is [OH \(^{-} ]\) and the pH of each of these solutions?

Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. $C_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}$ \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)
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