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Chapter 16: Problem 25

Fluoroacetic acid occurs in gifblaar, one of the most poisonous of all plants. A 0.318 M solution of the acid is found to have a \(\mathrm{pH}=1.56 .\) Calculate \(K_{\mathrm{a}}\) of fluoroacetic acid. $$\mathrm{CH}_{2} \mathrm{FCOOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \quad\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CH}_{2} \mathrm{FCOO}^{-}(\mathrm{aq}) \quad K_{\mathrm{a}}=?$$

Short Answer

Expert verified
To find the \(K_a\) value of the fluoroacetic acid, first calculate the hydronium ion concentration from the given pH, then find the equilibrium concentrations of the reactants and products, and finally substitute these values into the \(K_a\) expression.

Step by step solution

01

Calculating Hydronium Ion Concentration

First, determine the concentration of the hydronium ion (\(\mathrm{H}_{3}\mathrm{O}^{+}\)) from the pH using the equation: \[\mathrm{[H_3 O^{+}] = 10^{-pH}}\] thus \[\mathrm{[H_3 O^{+}] = 10^{-1.56}}\]
02

Finding Equilibrium Concentrations

Since the degree of ionization of weak acids like fluoroacetic acid is very small, the concentration of the acid at equilibrium is almost equal to the initial concentration. So we can assume: \[\mathrm{[CH_{2} FCOOH]_{equilibrium} = 0.318 M}\] The acid dissociates to form \(\mathrm{H_{3}O^{+}}\) and \(\mathrm{CH_{2}FCOO^{-}}\) in equal amounts, so their concentrations at equilibrium are also equal to \(\mathrm{[H_{3}O^{+}]}\], which was calculated in Step 1.
03

Calculating the Ka Value

Lastly, we calculate \(K_a\) using the equilibrium concentrations in the \(K_a\) expression: \[K_a = \frac{{[\mathrm{H}_3 \mathrm{O}^{+}][\mathrm{CH}_2 \mathrm{FCOO^-}]}}{{[\mathrm{CH}_2\mathrm{ FCOOH}]} }\] Substitute the values we have found into the \(K_a\) expression to get the final \(K_a\) value of the acid.

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Most popular questions from this chapter

The three following reactions are acid-base reactions according to the Lewis theory. Draw Lewis structures, and identify the Lewis acid and Lewis base in each reaction. (a) \(\mathrm{B}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{BF}_{3} \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{OBF}_{3}\)

Explain why \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in a strong acid solution doubles as the total acid concentration doubles, whereas in a weak acid solution, \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) increases only by about a factor of \(\sqrt{2}\)

Suppose you wanted to produce an aqueous solution of \(\mathrm{pH}=8.65\) by dissolving one of the following salts in water. Which salt would you use, and at what molarity? (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{KHSO}_{4} ;\) (c) \(\mathrm{KNO}_{2}\); (d) \(\mathrm{NaNO}_{3}\).

\(50.00 \mathrm{mL}\) of \(0.0155 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) is mixed with \(75.00 \mathrm{mL}\) of 0.0106 M KOH(aq). What is the pH of the final solution?

What mass of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), would you dissolve in \(350.0 \mathrm{mL}\) of water to produce a solution with a \(\mathrm{pH}=2.85 ?\) $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} & \\ K_{\mathrm{a}}=6.3 \times 10^{-5} \end{aligned}$$
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