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Chapter 14: Problem 151

Acrylic acid \(\left(\mathrm{CH}_{2}=\mathrm{CHCO}_{2} \mathrm{H}\right)\) is a precursor for many important plastics. \(K_{\mathrm{a}}\) for acrylic acid is \(5.6 \times 10^{-5}\) a. Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of acrylic acid. b. Calculate the percent dissociation of a \(0.10-M\) solution of acrylic acid. c. Calculate the \(\mathrm{pH}\) of a \(0.050-M\) solution of sodium acrylate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Short Answer

Expert verified
a. The pH of a 0.10-M solution of acrylic acid is approximately 2.13. b. The percent dissociation of a 0.10-M solution of acrylic acid is approximately 7.48%. c. The pH of a 0.050-M solution of sodium acrylate is approximately 12.30.

Step by step solution

01

Write down the expression for \(K_{a}\)#

We will first write the expression for \(K_{a}\) of acrylic acid: \(K_{a} = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}\)
02

Assume the concentration changes#

We will assume small amounts of \(H^+\) and \(A^-\) formed from the initial concentration of acrylic acid (\(0.10M\)): \([\mathrm{H}^+] = [\mathrm{A}^-] = x\) \([\mathrm{HA}] = 0.10 - x\)
03

Replace concentrations with values in \(K_{a}\) equation#

Now we can plug these concentrations back into the expression for \(K_{a}\) and solve for x: \(5.6 \times 10^{-5} = \frac{x^2}{0.10 - x}\)
04

Solve for x (concentration of \(H^+\))#

Since this is a weak acid, we can assume that x is small compared to \(0.10\), so we can approximate the equation as: \(5.6 \times 10^{-5} \approx \frac{x^2}{0.10}\) x = \(7.48 \times 10^{-3}\)
05

Calculate the pH#

Now that we have the concentration of the \(H^+\) ions, we can calculate the pH using the formula: pH = \(-\log_{10}(H^+)\) pH = \(-\log_{10}(7.48 \times 10^{-3})\)
06

Final answer for \(\mathrm{pH}\) of \(0.10-M\) acrylic acid solution#

Calculating the pH of our \(0.10-M\) acrylic acid solution, we get: pH \(\approx 2.13\) #b. Calculate the percent dissociation of a \(0.10-M\) solution of acrylic acid.#
07

Calculate the percent dissociation#

Use the formula for percent dissociation: Percent dissociation = \(\frac{\text{moles of undissociated acid}}{\text{initial moles of acid}} \times 100\%\) Percent dissociation = \(\frac{x}{0.10} \times 100\%\)
08

Final answer for percent dissociation#

Plug the previously calculated x into the formula and find the percent dissociation: Percent dissociation \(\approx 7.48\%\) #c. Calculate the \(\mathrm{pH}\) of a \(0.050-M\) solution of sodium acrylate#
09

Calculate the concentration of \(OH^-\) ions#

Sodium acrylate fully dissociates in water, so the concentration of \(OH^-\) ions will be equal to the initial concentration of sodium acrylate: \([\mathrm{OH}^-] = 0.050\,\mathrm{M}\)
10

Calculate the pOH#

Now we calculate the pOH using the concentration of \(OH^-\) ions: pOH = \(-\log_{10}(OH^-)\) pOH = \(-\log_{10}(0.050)\)
11

Calculate the pH from the pOH#

We can calculate the pH using the relation between pH and pOH: pH + pOH = 14 pH = 14 - pOH
12

Final answer for \(\mathrm{pH}\) of \(0.050-M\) sodium acrylate solution#

Calculating the pH of our \(0.050-M\) sodium acrylate solution, we get: pH \(\approx 12.30\)

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times\right.\) \(10^{-5}\) ) has a pH of \(2.80 .\) Calculate the water solubility of benzoic acid in moles per liter.

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right]\), \(\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Consider a \(0.60-M\) solution of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), lactic acid \(\left(K_{\mathrm{a}}=\right.\) \(\left.1.4 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) ii. \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) iii. \(\mathrm{H}^{+}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{OH}^{-}\) b. Complete the following ICE table in terms of \(x\), the amount \((\mathrm{mol} / \mathrm{L})\) of lactic acid that dissociates to reach equilibrium. c. What is the equilibrium concentration for \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) ? d. Calculate the \(\mathrm{pH}\) of the solution.

Trichloroacetic acid \(\left(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\) is a corrosive acid that is used to precipitate proteins. The \(\mathrm{pH}\) of a \(0.050-M\) solution of trichloroacetic acid is the same as the \(\mathrm{pH}\) of a \(0.040-\mathrm{M} \mathrm{HClO}_{4}\) solution. Calculate \(K_{\mathrm{a}}\) for trichloroacetic acid.

Consider \(50.0 \mathrm{~mL}\) of a solution of weak acid \(\mathrm{HA}\left(K_{\mathrm{a}}=\right.\) \(1.00 \times 10^{-6}\) ), which has a pH of \(4.000\). What volume of water must be added to make the \(\mathrm{pH}=5.000 ?\)
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