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Chapter 16: Problem 23

For the ionization of phenylacetic acid, $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \\\ K_{\mathrm{a}}=4.9 \times 10^{-5} \end{array}$$ (a) What is \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right]\) in \(0.186 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) (b) What is the \(\mathrm{pH}\) of \(0.121 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\)

Short Answer

Expert verified
The concentration of \([C_6H_5CH_2CO_{2}^-]\) ion in \(0.186M\, C_6H_5CH_2CO_2H\) solution and the pH of \(0.121M\, C_6H_5CH_2CO_2H\) solution can be calculated by following the above steps. Since the solutions for 'x' provide both the ion concentration and \([H_3O^+]\) needed for pH calculation, a clear understanding of weak acid ionization and solving equations is critical for solving this exercise.

Step by step solution

01

Set up ICE table

To solve this problem, we use an ICE (Initial, Change, Equilibrium) table. The initial concentration of the acid is given as 0.186 M and the ion and hydronium ion have concentrations of 0 initially. During the reaction the acid loses some amount of H+, which will be represented as ‘x’, hence decreasing by ‘x’. Hence adding ‘x’ to the ion (product) and hydronium ion (product).
02

Write the equation for Ka and solve for ‘x’

The ionization constant or acid dissociation constant Ka is given by the equation: \[ Ka = \frac{[H_3O^{+}][C_6H_5CH_2CO_{2}^{-}]}{[C_6H_5CH_2CO_2H]} \]Substitute the equilibrium concentrations in terms of ‘x’ into the equation and solve for ‘x’, which represents the concentration of the ion. We now have: \[4.9 * 10 ^{-5} = \frac{x * x}{0.186 - x}\]. Solve this quadratic equation to get the value of 'x'.
03

Finding ion concentration of the phenylacetic acid

After solving the quadratic equation we get two solutions for 'x' \([-H_3O^{+}]\) and \([C_6H_5CH_2CO_{2}^-]\). Of the two solutions, the negative one must be discarded (as concentration cannot be negative). Hence, the positive solution for 'x' represents the equilibrium concentration of \([C_6H_5CH_2CO_{2}^-]\) ion.
04

Calculation of pH

pH is calculated using the equation: \[ pH = -log[H_3O^{+}] \]Since \([H_3O^{+}]\) is equal to 'x', substitute the positive ‘x’ value obtained in step 3 and compute to find the pH.

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Most popular questions from this chapter

What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH},\) and \(\mathrm{pOH}\) of \(0.55 \mathrm{M}\) \(\mathrm{M} \mathrm{HClO}_{2} ?\)

\(\operatorname{In} 0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) is equal to \((\mathrm{a}) 0.050 \mathrm{M}\) (b) \(0.10 \mathrm{M} ;\) (c) \(0.11 \mathrm{M} ;\) (d) \(0.20 \mathrm{M}\).

Indicate whether each of the following is a Lewis acid or base. (a) \(\mathrm{OH}^{-} ;\) (b) \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{B} ;\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

One handbook lists a value of 9.5 for \(\mathrm{p} \mathrm{K}_{\mathrm{b}}\) of quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{N},\) a weak base used as a preservative for anatomical specimens and to make dyes. Another handbook lists the solubility of quinoline in water at \(25^{\circ} \mathrm{C}\) as \(0.6 \mathrm{g} / 100 \mathrm{mL} .\) Use this information to calculate the \(\mathrm{pH}\) of a saturated solution of quinoline in water.

In your own words, define or explain the following terms or symbols: (a) \(K_{\mathrm{w}} ;\) (b) \(\mathrm{pH} ;\) (c) \(\mathrm{p} K_{\mathrm{a}} ;\) (d) hydrolysis; (e) Lewis acid.
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