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Chapter 14: Problem 153

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 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 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
The pH of a 0.10-M solution of acrylic acid is approximately 2.63, and its percent dissociation is approximately 2.37%. The pH of a 0.050-M solution of sodium acrylate (NaC3H3O2) is approximately 8.78.

Step by step solution

01

Write the dissociation equilibrium

The dissociation of acrylic acid, CH2=CHCO2H, can be represented as: \( CH_{2}=CHCO_{2}H \rightleftharpoons CH_{2}=CHCO_{2}^- + H^+ \)
02

Write the expression for Ka

We can now write the expression for the given Ka value: \[ K_{a} = \frac{[CH_{2}=CHCO_{2}^-][H^+]}{[CH_{2}=CHCO_{2}H]} \]
03

Set up the ICE table to find [H+]

To find [H+], we will use the reaction and set up an ICE (Initial, Change, Equilibrium) table: | | CH2=CHCO2H | -> | CH2=CHCO2- | + | H+ | |------------|------------|------------|------------|-----|----| | Initial | 0.10 M | | 0 | | 0 | | Change | -x | | +x | | +x | | Equilibrium| 0.10-x | | x | | x | where x represents the change in concentration.
04

Substitute the equilibrium concentrations into the Ka expression

Now we can substitute the equilibrium concentrations into the Ka expression: \( 5.6\times10^{-5} = \frac{x^2}{0.10-x} \)
05

Solve for x (which represents [H+])

Since Ka is small, we can assume that x is very small compared to 0.10 M, and we can simplify the equation: \( 5.6\times10^{-5} \approx \frac{x^2}{0.10} \) Solving for x, we find that x = [H+] ≈ 2.37 x 10^(-3) M.
06

Calculate the pH

Now that we have the [H+] concentration, we can find the pH using the pH formula: \( pH = -\log{[H^+]} \) pH = -log(2.37 x 10^(-3)) ≈ 2.63 Answer: The pH of a 0.10 M solution of acrylic acid is ≈ 2.63. #b. Calculate the percent dissociation of a 0.10-M solution of acrylic acid.#
07

Calculate the percent dissociation using the formula

The percent dissociation can be calculated using the formula: \( \% dissociation = \frac{[H^+]_{equilibrium}}{[CH_2=CHCO_2H]_{initial}} \times 100 \)
08

Substitute the values in the formula to find the percent dissociation

Inputting the values we found earlier, we obtain: \(\% dissociation = \frac{2.37\times10^{-​3}}{0.10} \times 100 \approx 2.37\% \) Answer: The percent dissociation of a 0.10-M solution of acrylic acid is ≈ 2.37%. #c. calculate the pH of a 0.050-M solution of sodium acrylate \( NaC_3H_3O_2 \)#
09

Write the dissociation equilibrium of sodium acrylate

The dissociation of sodium acrylate can be represented as: \( NaC_{3}H_{3}O_{2} \rightleftharpoons Na^+ + C_{3}H_{3}O_{2}^- \)
10

Calculate the Kb for the conjugate base using Ka

We know the relationship between Ka, Kb and Kw: \( Ka \times Kb = Kw \) Where Kw is the ion product of water, which is equal to \( 1 \times 10^{-14} \). \( Kb = \frac{Kw}{Ka} = \frac{1 \times 10^{-14}}{5.6 \times 10^{-5}} \) Kb ≈ 1.79 x 10^(-10)
11

Write the dissociation equilibrium of the conjugate base

The dissociation equilibrium of the conjugate base can be represented as: \( C_{3}H_{3}O_{2}^- (aq) + H_2O (l) \rightleftharpoons C_{3}H_{3}O_{2}H (aq) + OH^- (aq) \)
12

Write the expression for Kb and set up ICE table

We can now write the expression for the given Kb value: \[ Kb = \frac{[OH^-][C_{3}H_{3}O_{2}H]}{[C_{3}H_{3}O_{2}^-]} \] We will use the reaction and set up an ICE (Initial, Change, Equilibrium) table: | | C3H3O2- | + | H2O | -> | C3H3O2H | + | OH- | |------------|---------|-------|------|-----|---------|---|------| | Initial | 0.050 | | | | 0 | | 0 | | Change | -x | | | | +x | | +x | | Equilibrium| 0.050-x | | | | x | | x | where x represents the change in concentration.
13

Substitute the equilibrium concentrations into the Kb expression

Now we can substitute the equilibrium concentrations into the Kb expression: \( 1.79\times10^{-10} = \frac{x^2}{0.050-x} \)
14

Solve for x (which represents [OH-])

Since Kb is small, we can assume that x is very small compared to 0.050 M, and we can simplify the equation: \( 1.79\times10^{-10} \approx \frac{x^2}{0.050} \) Solving for x, we find that x = [OH-] ≈ 5.98 x 10^(-6) M.
15

Calculate the pOH and pH

Now we have the [OH-] concentration, we can find the pOH using the pOH formula: \( pOH = -\log{[OH^-]} \) pOH = -log(5.98 x 10^(-6)) ≈ 5.22 Next, we can find the pH using the relationship between pH and pOH: \( pH + pOH = 14 \) pH = 14 - pOH = 14 - 5.22 ≈ 8.78 Answer: The pH of a 0.050-M solution of sodium acrylate (NaC3H3O2) is ≈ 8.78.

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Most popular questions from this chapter

A solution of formic acid (HCOOH, \(K_{\mathrm{a}}=1.8 \times 10^{-4} )\) has a \(\mathrm{pH}\) of 2.70 . Calculate the initial concentration of formic acid in this solution.

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