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 \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

In order to find the pH, we need to first establish the equilibrium equation and determine the concentration of hydrogen ions. We can then use the pH formula to find the pH. Acrylic acid dissociates as follows: \[\mathrm{CH}_{2}=\mathrm{CHCO}_{2}\mathrm{H} \rightleftharpoons \mathrm{H}^{+} + \mathrm{CH}_{2}=\mathrm{CHCO}_{2}^{-}\] The equilibrium constant, Ka, can be expressed as follows: \[K_\mathrm{a} = \frac{[\mathrm{H}^{+}][\mathrm{CH}_{2}=\mathrm{CHCO}_{2}^{-}]}{[\mathrm{CH}_{2}=\mathrm{CHCO}_{2}\mathrm{H}]}\] Initially, let's assume that x moles per liter of acrylic acid will dissociate. Therefore, at equilibrium, the concentrations of the components will be as follows: [CH₂=CHCO₂H] = (0.10 - x) M [H⁺] = [CH₂=CHCO₂⁻] = x M Plugging these values into the equilibrium equation: \[K_\mathrm{a} = \frac{x^2}{(0.10 - x)}\] Now, substitute the given Ka value and solve for x, which corresponds to [H⁺]: \[5.6 \times 10^{-5} = \frac{x^2}{(0.10 - x)}\] We can assume that x is much smaller than 0.10 since acrylic acid is a weak acid: \[x^2 \approx 5.6 \times 10^{-5}\times 0.10\] \[x \approx \sqrt{5.6 \times 10^{-6}}\] Now, compute the value of x: \[x \approx 2.37 \times 10^{-3}\] Finally, we can find the pH using the formula pH = -log[H⁺]: \[pH = -\log(2.37 \times 10^{-3}) \approx 2.63\]

The percent dissociation can be calculated using the formula: \[\%\text{ dissociation } = \frac{[\text{Dissociated Acid}] }{[\text{Initial Acid}]} \times 100\] In this case, [Dissociated Acid] is equal to the [H⁺] concentration, which is \(2.37 \times 10^{-3} \,\text{M}\). The initial concentration of the acid was 0.10 M. Substitute these values into the percent dissociation formula: \[\%\text{ dissociation } = \frac{2.37 \times 10^{-3}}{0.10} \times 100\] Now, calculate the percent dissociation: \[\%\text{ dissociation } \approx 2.37\%\]

Sodium acrylate is the sodium salt of the conjugate base of acrylic acid, which means it will cause the pH to increase. To find the pH of this solution, we first need to find the pOH and then use the relationship: pH + pOH = 14. The dissociation of sodium acrylate in water produces hydroxide ions (OH⁻): \[\mathrm{NaC}_{3}\mathrm{H}_{3}\mathrm{O}_{2} \rightarrow \mathrm{Na}^{+} + \mathrm{CH}_{2}=\mathrm{CHCO}_{2}^{-} + \mathrm{H}_{2}O \rightarrow \mathrm{CH}_{2}=\mathrm{CHCO}_{2}^{-} + \mathrm{OH}^{-}\] Acrylic acid's pKa can be calculated using the relationship pKa = -log(Ka): \[pK_\mathrm{a} = -\log(5.6 \times 10^{-5}) \approx 4.25 \] The pKb of the conjugate base (acrylate ion) can be calculated using the relationship pKa + pKb = 14: \[pK_\mathrm{b} = 14 - 4.25 = 9.75\] Now, we can find Kb using the relationship Kb = 10^(−pKb): \[K_\mathrm{b} = 10^{-9.75} \approx 1.78 \times 10^{-10}\] Assume y moles per liter of sodium acrylate generate hydroxide ions: [OH⁻] = [CH₂=CHCO₂⁻] = y M The Kb expression is: \[K_\mathrm{b} = \frac{[\mathrm{OH}^{-}][\mathrm{CH}_{2}=\mathrm{CHCO}_{2}^{-}] }{[\mathrm{CH}_{2}=\mathrm{CHCO}_{2}\mathrm{H}]}\] Substitute the initial concentrations: \[1.78 \times 10^{-10} = \frac{y^2}{(0.050-y)}\] Assuming y is much smaller than 0.050: \[y^2 \approx 1.78 \times 10^{-10}\times 0.050\] \[y \approx \sqrt{8.90 \times 10^{-12}}\] Now, compute the value of y, which corresponds to [OH⁻]: \[y \approx 9.43 \times 10^{-6}\] Next, we need to find the pOH using the formula pOH = -log[OH⁻]: \[pOH = -\log(9.43 \times 10^{-6}) \approx 5.03\] Finally, we can find the pH using the relationship pH = 14 - pOH: \[pH = 14 - 5.03 \approx 8.97\] To summarize: a) The pH of a 0.10-M solution of acrylic acid is approximately 2.63. b) The percent dissociation of a 0.10-M solution of acrylic acid is approximately 2.37%. c) The pH of a 0.050-M solution of sodium acrylate is approximately 8.97.