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Chapter 14: Problem 81

One mole of a weak acid HA was dissolved in \(2.0 \mathrm{~L}\) of solution. After the system had come to equilibrium, the concentration of HA was found to be \(0.45 M\). Calculate \(K_{\mathrm{a}}\) for HA.

Short Answer

Expert verified
The acidity constant, \(K_{\mathrm{a}}\), for the weak acid HA is approximately \(5.56 \times 10^{-3}\).

Step by step solution

01

Initial Concentration of HA

Initially, one mole of HA is dissolved in 2.0 L of solution, so its initial concentration can be calculated by: \[ \text{Initial concentration of HA} = \frac{\text{moles of HA}}{\text{volume of solution in L}} \] \[ [\text{HA}]_{initial} = \frac{1 \ \text{mol}}{2.0 \ \text{L}} = 0.50 \ \text{M} \] Step 2: Calculate change in concentration
02

Change in Concentration

From the information given, we know that the equilibrium concentration of HA is 0.45 M. We can calculate the change in concentration of HA during this process. \[ \text{Change in concentration of HA} = [\text{HA}]_{initial} - [\text{HA}]_{equilibrium} \] \[ \text{Change in concentration of HA} = 0.50 \ \text{M} - 0.45 \ \text{M} = 0.05 \ \text{M} \] Step 3: Find the concentration of \(H_3O^+\) and \(A^-\) at equilibrium
03

Equilibrium Concentrations

Since the initial concentration of \(H_3O^+\) and \(A^-\) is 0, the change in concentration of HA will also be the concentration of \(H_3O^+\) and \(A^-\) at equilibrium: \[ [\text{H}_3\text{O}^+]_{equilibrium} = [\text{A}^-]_{equilibrium} = 0.05 \ \text{M} \] Step 4: Calculate \(K_{\mathrm{a}}\) using the equilibrium expression
04

\(K_{\mathrm{a}}\) Calculation

The equilibrium expression for the reaction is given by: \[ K_{\mathrm{a}} = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \] Substitute the equilibrium concentrations of \(H_3O^+\), \(A^-\) and HA into the expression: \[ K_{\mathrm{a}} = \frac{(0.05 \ \text{M})(0.05 \ \text{M})}{(0.45 \ \text{M})} \] \[ K_{\mathrm{a}} = \frac{0.0025 \ \text{M}^2}{0.45 \ \text{M}} = 5.56 \times 10^{-3} \] So the acidity constant, \(K_{\mathrm{a}}\), for the weak acid HA is approximately \(5.56 \times 10^{-3}\).

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Most popular questions from this chapter

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).

The \(\mathrm{pH}\) of human blood is steady at a value of approximately \(7.4\) owing to the following equilibrium reactions: \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)\) Acids formed during normal cellular respiration react with the \(\mathrm{HCO}_{3}^{-}\) to form carbonic acid, which is in equilibrium with \(\mathrm{CO}_{2}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) During vigorous exercise, a person's \(\mathrm{H}_{2} \mathrm{CO}_{3}\) blood levels were \(26.3 \mathrm{~m} M\), whereas his \(\mathrm{CO}_{2}\) levels were \(1.63 \mathrm{~m} M .\) On resting, the \(\mathrm{H}_{2} \mathrm{CO}_{3}\) levels declined to \(24.9 \mathrm{~m} M\). What was the \(\mathrm{CO}_{2}\) blood level at rest?

Calculate \(\left[\mathrm{OH}^{-}\right], \mathrm{pOH}\), and \(\mathrm{pH}\) for each of the following. a. \(0.00040 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) b. a solution containing \(25 \mathrm{~g}\) KOH per liter c. a solution containing \(150.0 \mathrm{~g} \mathrm{NaOH}\) per liter

Calculate the \(\mathrm{pH}\) of the following solutions. a. \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(1.0 \times 10^{-10} M \mathrm{NaOH}\) c. \(2.0 \mathrm{M} \mathrm{NaOH}\)

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10-M\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\).
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