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Chapter 16: Problem 109

Caproic acid \(\left(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COOH}\right)\) is found in small amounts in coconut and palm oils and is used in making artificial flavors. A saturated solution of the acid contains \(11 \mathrm{~g} / \mathrm{L}\) and has a \(\mathrm{pH}\) of 2.94. Calculate \(K_{a}\) for the acid.

Short Answer

Expert verified
The acid dissociation constant (\(K_{a}\)) for caproic acid is approximately \(1.35 \times 10^{-5}\).

Step by step solution

01

Determine the concentration of H+ ions using the pH value

To determine the concentration of H+ ions, we can use the formula: pH = -log10[H+] Given pH = 2.94, we can calculate the H+ concentration: 2.94 = -log10[H+] Rearranging the equation to solve for [H+]: [H+] = 10^(-2.94)
02

Write the balanced dissociation equation of caproic acid

The balanced dissociation equation for caproic acid (HA) would be: HA <-> H+ + A-
03

Set up an ICE table to determine the equilibrium concentrations

Set up the ICE (Initial, Change, Equilibrium) table for concentration changes: | [HA] | [H+] | [A-] ------+--------+--------+-------- I | Co | 0 | 0 C | -x | +x | +x E | Co - x | x | x We are given that the concentration of caproic acid is \(\frac{11 g}{L}\) in the saturated solution. Let's find the moles of the acid: \(Co = \frac{11 g}{(C_{5}H_{11}COOH)g/mol}\) \(Co = \frac{11}{116.15}\) mol/L \(Co ≈ 0.095 M\) Since it's a saturated solution, most of the caproic acid will remain undissociated. Hence, we can assume \(Co - x ≈ Co = 0.095M\). Also, the concentration of H+ is calculated from Step 1, which is our x value: [H+] = x = \(10^{-2.94}\)
04

Calculate the Ka using the equilibrium concentrations

Now, let's calculate the Ka: \(K_{a} = \frac{[H+][A^-]}{[HA]}\) Substitute the equilibrium concentration values: \(K_{a} = \frac{x \cdot x}{Co - x}\) \(K_{a} = \frac{(10^{-2.94})^2}{0.095}\) Calculate Ka: \(K_{a} ≈ 1.35 \times 10^{-5}\) So, the calculated \(K_{a}\) for caproic acid is \(1.35 \times 10^{-5}\).

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Most popular questions from this chapter

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in \(500.0 \mathrm{~mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL}\) (d) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}\left(\mathrm{OH}_{2}\right),\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, (c) \(1.00 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},\) (d) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \quad M \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

(a) Which of the following is the stronger Bronsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2} ?\) (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{HS}^{-}\) or \(\mathrm{HSO}_{4}^{-}\) ? Briefly explain your choices.

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Calculate the percent ionization of propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\) in solutions of each of the following concentrations \(\left(K_{a}\right.\) is given in Appendix \(\left.D\right):\) (a) \(0.250 \mathrm{M}\), (b) \(0.0800 \mathrm{M}\), (c) \(0.0200 \mathrm{M}\).
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