Question

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

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.

Step-by-step solution

a. Calculate the pH of a 0.10-M solution of acrylic acid.

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⁺]: \[pH = -\log(2.37 \times 10^{-3}) \approx 2.63\]

b. Calculate the percent dissociation of a 0.10-M solution of acrylic acid.

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: \[\%\text{ dissociation } \approx 2.37\%\]

c. Calculate the pH of a 0.050-M solution of sodium acrylate.

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.

Key concepts

Weak Acid Dissociation

Weak acids only partially dissociate in solution, releasing some of their hydrogen ions into the water. This partial dissociation is characterized by the acid dissociation constant, known as the equilibrium constant or Ka. For acrylic acid, the dissociation is represented as follows: \[ \mathrm{CH}_{2} = \mathrm{CHCO}_{2} \mathrm{H} \rightleftharpoons \mathrm{H}^{+} + \mathrm{CH}_{2} = \mathrm{CHCO}_{2}^{-} \]As acrylic acid is a weak acid, its equilibrium heavily favors the reactants, meaning not all of the acid molecules dissociate. Ka is a measure of the strength of the weak acid, and for acrylic acid, it is given as \(5.6 \times 10^{-5}\). The smaller this value, the weaker the acid is, indicating a lesser tendency to dissociate.

pH Calculation

Calculating the pH of a solution involves finding the concentration of hydrogen ions present when a weak acid is dissolved in water. The relationship between hydrogen ion concentration and pH is given by the formula:\[ \text{pH} = -\log[\mathrm{H}^{+}] \]When dealing with weak acids like acrylic acid, we start by setting up the equilibrium expression: \[K_{\mathrm{a}} = \frac{[\mathrm{H}^{+}][\mathrm{CH}_{2} = \mathrm{CHCO}_{2}^{-}]}{[\mathrm{CH}_{2} = \mathrm{CHCO}_{2} \mathrm{H}]}\]Assuming the degree of dissociation is small, we approximate that the initial concentration remains nearly constant, simplifying the calculations. By substituting and solving for \([\mathrm{H}^{+}]\), we can then use the pH formula to find the solution's acidity level.

Percent Dissociation

Percent dissociation quantifies how much of the weak acid dissociates into ions in solution. For weak acids, this is typically a small percentage. It is calculated using the formula:\[\% \text{ dissociation} = \frac{[\text{Dissociated Acid}]}{[\text{Initial Acid}]} \times 100\]In this context, dissociated acid refers to the concentration of hydrogen ions produced from the weak acid's partial dissociation. For example, in a 0.10 M solution of acrylic acid with \([\mathrm{H}^{+}] = 2.37 \times 10^{-3} \ \text{M}\), the percent dissociation is approximately 2.37%. This illustrates the weak nature of acrylic acid as only a small fraction of molecules dissociate fully.

Sodium Acrylate

Sodium acrylate is the sodium salt of the conjugate base of acrylic acid. When it dissolves in water, it increases the solution's pH by generating hydroxide ions \((\mathrm{OH}^{-})\). This results in a basic solution because the conjugate base creates more \(\mathrm{OH}^{-}\) ions, making the solution less acidic.The dissociation in water is as follows:\[\mathrm{NaC}_{3}\mathrm{H}_{3}\mathrm{O}_{2} \rightarrow \mathrm{Na}^{+} + \mathrm{CH}_{2} = \mathrm{CHCO}_{2}^{−} + \mathrm{OH}^{-}\]To calculate the pH of a sodium acrylate solution, we first determine its pOH and then convert this value to pH using the relationship:\[\text{pH} + \text{pOH} = 14\]Understanding that sodium acrylate creates a basic environment helps predict how solutions containing this compound will react, particularly in buffer systems or when modifying the acidity of a mixture.