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Chapter 17: Problem 17

(a) Calculate the percent ionization of \(0.0085 \mathrm{Mbutanoic}\) acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.0085 \mathrm{M}\) butanoic acid in a solution containing \(0.075 M\) sodium butanoate.

Short Answer

Expert verified
In the absence of sodium butanoate, the percent ionization of 0.0085 M butanoic acid is approximately 1.41%. However, in the presence of 0.075 M sodium butanoate, the percent ionization of 0.0085 M butanoic acid significantly decreases, to approximately 0.02%.

Step by step solution

01

Write the Ka expression

The Ka expression for butanoic acid (\(CH_{3}CH_{2}CH_{2}COOH\)) dissociating into protons and butanoate (\(CH_{3}CH_{2}CH_{2}COO^-\)) can be written as follows: \[K_a = \frac{[H^+][CH_{3}CH_{2}CH_{2}COO^{-}]}{[CH_{3}CH_{2}CH_{2}COOH]}\]
02

Set up the ICE table

To find the concentration of the ionized species, we can use the Initial, Change, and Equilibrium (ICE) table method. Initially, we have 0.0085 M butanoic acid, and no dissociation. So the initial concentrations are: [H+] = 0 [Butanoate] = 0 [Butanoic acid] = 0.0085 M As the acid dissociates, we get the following changes in concentrations: [H+] = +x [Butanoate] = +x [Butanoic acid] = -x At equilibrium, the concentrations are: [H+] = x [Butanoate] = x [Butanoic acid] = 0.0085 - x
03

Calculate the proton concentration ([H+])

We can now substitute these values into the Ka expression: \[1.5\times10^{-5} = \frac{x\cdot x}{0.0085 - x}\] It is reasonable to assume that x << 0.0085, so we can simplify the equation as follows: \[1.5\times10^{-5} \approx \frac{x^2}{0.0085}\] Now we can solve for x (the proton concentration): \[x = \sqrt{1.5\times10^{-5} \times 0.0085} \approx 1.20\times10^{-4} M\]
04

Calculate the percent ionization

Now that we have the proton concentration, we can find the percent ionization of butanoic acid by dividing the proton concentration by the initial concentration of butanoic acid and multiplying by 100: Percent ionization = \(\frac{[H^+]}{[Initial\:butanoic\:acid]}\) × 100 Percent ionization = \(\frac{1.20\times10^{-4}}{0.0085}\) × 100 ≈ 1.41 % b) Percent ionization of 0.0085 M butanoic acid in a 0.075 M sodium butanoate solution
05

Consider the effect of sodium butanoate

Sodium butanoate (\(CH_{3}CH_{2}CH_{2}COONa\)) is the salt of butanoic acid, which means it can affect the ionization of butanoic acid. Its presence results in a common ion effect, as sodium butanoate dissociates into butanoate and sodium ions when dissolved in water. This will change the equilibrium concentration of the butanoate ion, and thus affect the percent ionization. Initial concentration of butanoate ions = 0.075 M. The new ICE table Equilibrium concentrations are: [H+] = x [Butanoate] = 0.075 + x [Butanoic acid] = 0.0085 - x
06

Calculate the new proton concentration ([H+])

Using the new equilibrium concentrations, we can rewrite the Ka expression: \[1.5\times10^{-5} = \frac{x(0.075 + x)}{0.0085 - x}\] We can again assume that x << 0.0085 and x << 0.075. Therefore, we can simplify the equation as follows: \[1.5\times10^{-5} \approx \frac{x\cdot 0.075}{0.0085}\] Now solving for x: \[x \approx \frac{1.5\times10^{-5} \times 0.0085}{0.075} \approx 1.70\times10^{-6} M\]
07

Calculate the new percent ionization

Now that we have the new proton concentration, we can find the percent ionization of butanoic acid in the presence of sodium butanoate: Percent ionization = \(\frac{[H^+]}{[Initial\:butanoic\:acid]}\) × 100 Percent ionization = \(\frac{1.70\times10^{-6}}{0.0085}\) × 100 ≈ 0.02 % In conclusion, the percent ionization of 0.0085 M butanoic acid is: a) ≈ 1.41 % in the absence of sodium butanoate b) ≈ 0.02 % in the presence of 0.075 M sodium butanoate

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Most popular questions from this chapter

(a) If the molar solubility of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(1.24 \times 10^{-3} \mathrm{~mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that $1.1 \times 10^{-2} \mathrm{~g} \mathrm{SrF}_{2}\( dissolves per \)100 \mathrm{~mL}$ of aqueous solution at \(25^{\circ} \mathrm{C} .\) Calculate the solubility product for \(\mathrm{SrF}_{2} .(\mathbf{c})\) The \(K_{s p}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10} .\) What is the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) ?

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Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathbf{a})\) a solution formed by adding \(30.0 \mathrm{~g}\) of furoic acid and \(25.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.300 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) $\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\( and \)30.0 \mathrm{~mL}\( of \)0.250 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$ and diluting the total volume to \(125 \mathrm{~mL},(\mathbf{c})\) a solution prepared by adding $25.0 \mathrm{~mL}\( of \)1.00 \mathrm{M} \mathrm{NaOH}\( solution to \)100.0 \mathrm{~mL}\( of \)0.100 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$
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