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Chapter 1: Problem 24

A 1 -molar solution of a very weak monoprotic acid has a pH of 5. What is the value of \(K_{a}\) for the acid? (A) \(K_{a}=1 \times 10^{-10}\) (B) \(K_{a}=1 \times 10^{-7}\) (C) \(K_{a}=1 \times 10^{-5}\) (D) \(K_{a}=1 \times 10^{-2}\)

Short Answer

Expert verified
The value of \(K_{a}\) for the acid is \(1 \times 10^{-10}\).

Step by step solution

01

Calculate the Hydronium Ion concentration

The relationship between pH and \([H^{+}]\) can be expressed as follows: \(pH = -log_{10} [H^{+}]\). Since the pH = 5, the hydronium ion concentration (\([H^{+}]\)) can be calculated as \([H^{+}] = 10^{-pH} = 10^{-5} M\).
02

Considering the Dissociation of the Acid

A weak monoprotic acid dissociates in water as follows: \(HA ⇌ H^{+} + A^{-}\). In this problem, one logical assumption is that \([H^{+}]\) equals \([A^{-}]\), because every molecule of HA that dissociates produces one hydrogen ion and one A- ion. Therefore, \([H^{+}]\) equals \([A^{-}]\) and equals \(10^{-5} M\). Also use the given information that the initial concentration of HA which is a 1 M solution.
03

Calculate \(K_{a}\) of the Acid

The acid dissociation constant (\(K_{a}\)) can be calculated by the formula: \(K_{a} = [H^{+}][A^{-}]/[HA]\). Substitute the values calculated from previous steps to find the \(K_{a}\) value: \(K_{a} = (10^{-5})(10^{-5})/(1) = 10^{-10}\).

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Most popular questions from this chapter

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ When the cell is connected, which of the following reactions takes place at the anode? (A) \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) (B) \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}+2 e^{-}\) (C) \(2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2}(g)\) (D) \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}+2 e^{-}\)

Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation? (A) \(\operatorname{NaF}(s)\) (B) \(\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{NaOH}(\mathrm{s})\) (D) \(\mathrm{NH}_{3}(a q)\)

Which compound, \(\mathrm{CaCl}_{2}\) or \(\mathrm{CaO}\) , would you expect to have a high melting point? Why? (A) \(\mathrm{CaCl}_{2}\) because there are more ions per lattice unit (B) \(\mathrm{CaCl}_{2}\) because a chlorine ion is smaller than an oxygen ion (C) Cao, because the charge of oxygen ion exceeds that of chlorine ion (D) CaO, because the common charges of calcium and oxygen ions are identical in magnitude

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) Based on the information given in the table below, what is \(\Delta H^{\circ}\) for the above reaction? \(\begin{array}{cc}{\text { Bond }} & {\text { Average bond energy }(\mathrm{kJ} / \mathrm{mol})} \\ {\mathrm{H}-\mathrm{H}} & {500} \\\ {\mathrm{O}=\mathrm{O}} & {500} \\ {\mathrm{O}-\mathrm{H}} & {500}\end{array}\) (A) \(-2,000 \mathrm{kJ}\) (B) \(-500 \mathrm{kJ}\) (C) \(+1,000 \mathrm{kJ}\) (D) \(+2,000 \mathrm{kJ}\)

$$\mathrm{Br}_{2}(g)+\mathrm{I}_{2}(g) \leftrightarrow 2 \mathrm{IBr}(g)$$ At \(150^{\circ} \mathrm{C},\) the equilibrium constant, \(K_{c},\) for the reaction shown above has a value of \(300 .\) This reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr(g) was found to be 3 atm. Which of the following could be the partial pressures due to \(\operatorname{Br}_{2}(g)\) and \(I_{2}(g)\) in the container? \(\begin{array}{lll}{} & {\operatorname{Br}_{2}(g)} & {\mathrm{I}_{2}(g)} \\\ {\text { (A) }} & {0.1 \mathrm{atm}} & {0.3 \mathrm{atm}} \\ {\text { (B) }} & {0.3 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (C) }} & {1 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (D) }} & {1 \mathrm{atm}} & {3 \mathrm{atm}}\end{array}\)
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