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Show that for a weak acid, the percent ionization should vary as the inverse square root of the acid concentration.

Short Answer

Expert verified
The percent ionization of a weak acid varies as the inverse square root of the acid concentration, as demonstrated by the relationship Percent ionization = \(\cfrac{\sqrt{K_a}}{\sqrt{[\text{initial concentration}]}} × 100\% = \cfrac{\sqrt{K_a}}{\sqrt{C}} × 100\%\), where K_a is the ionization constant and C is the initial concentration of the acid. This relationship is derived by substituting the equilibrium concentrations in terms of the ionization extent (x) into the K_a expression and making an approximation for weak acids.

Step by step solution

01

Define percent ionization and ionization constant

The percent ionization of a weak acid is the ratio of the concentration of ionized acid to the initial concentration of acid, multiplied by 100. This can be represented as: Percent ionization = \(\cfrac{[\text{Ionized acid}]}{[\text{Initial acid}]} × 100\%\) A weak acid will partially ionize in water, and this process is represented by the ionization constant, denoted as K_a, for a given weak acid (HA): \(HA \rightleftharpoons H^+ + A^-\) K_a is defined as the ratio of the product of the concentrations of the ions and the concentration of the undissociated acid: \(K_a = \cfrac{[H^+][A^-]}{[HA]}\)
02

Assume equilibrium concentrations

At equilibrium, assume that x moles of HA dissociate. As a result, H+ and A− concentrations will each increase by x while [HA] will decrease by x. Therefore, we can represent the equilibrium concentrations as follows: [H+] = [A−] = x [HA] = initial concentration - x
03

Write K_a expression in terms of x

Substitute the equilibrium concentrations in terms of x into the K_a expression: \(K_a = \cfrac{x \cdot x}{\text{initial concentration} - x} = \cfrac{x^2}{\text{initial concentration} - x}\)
04

Apply the approximation of weak acid concentrations

Since HA is a weak acid, it is reasonable to assume that x is much smaller than the initial concentration of the acid. This simplifies our equation to: \(K_a \approx \cfrac{x^2}{[\text{initial concentration}]}\) Now, we can solve for x: \(x = \sqrt{K_a \times [\text{initial concentration}]}\)
05

Calculate percent ionization and simplify

Percent ionization is given as the ratio of ionized acid to the initial concentration of the acid, multiplied by 100: Percent ionization = \(\cfrac{x}{[\text{initial concentration}]} × 100\%\) We substitute our expression for x (from step 4) into the percent ionization formula: Percent ionization = \(\cfrac{\sqrt{K_a \times [\text{initial concentration}]}}{[\text{initial concentration}]} × 100\%\) Finally, the percent ionization in terms of the inverse square root of the acid concentration, C, is: Percent ionization = \(\cfrac{\sqrt{K_a}}{\sqrt{[\text{initial concentration}]}} × 100\% = \cfrac{\sqrt{K_a}}{\sqrt{C}} × 100\%\) This demonstrates that the percent ionization of a weak acid varies as the inverse square root of the acid concentration, as required.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weak Acid
A weak acid is an acid that only partially ionizes in a solution. Unlike strong acids, which completely dissociate, weak acids are very much reversible and only a small fraction of the acid molecules lose protons to form ions.
This means that in a solution of a weak acid, most of the acid molecules remain intact.
  • The key factor is the partial ionization.
  • This results in a dynamic balance, or equilibrium, between the ionized and unionized forms.
The extent to which a weak acid ionizes is determined by its ionization constant, or \(K_a\). Understanding weak acids is crucial for calculating equilibrium conditions and predicting the behavior of the acid in various solutions.
Ionization Constant
The ionization constant, denoted as \(K_a\), quantifies the degree of ionization of a weak acid in water. It is a specific type of equilibrium constant and gives us insight into the strength of the acid.
  • The formula for \(K_a\) involves the concentrations of hydrogen ions \([H^+]\), the conjugate base \([A^-]\), and the undissociated acid \([HA]\):
\[K_a = \cfrac{[H^+][A^-]}{[HA]}\]This formula demonstrates the proportionality between the products and reactants in equilibrium. A larger \(K_a\) indicates a stronger weak acid because more ions are produced. Conversely, a smaller \(K_a\) suggests fewer ions and a weaker acid. Calculating \(K_a\) helps in predicting how the acid will behave under different conditions, especially when calculating percent ionization.
Equilibrium Concentrations
In the context of weak acids, equilibrium concentrations refer to the balanced amounts of ionized and unionized species in the solution.
When a weak acid is in water, it partially dissociates until it reaches equilibrium.
  • At equilibrium, concentrations do not change, although the reactions continue to occur.
  • These equilibrium concentrations can be shown using algebraic terms like \(x\) to represent changes due to dissociation.
For instance, if \(x\) moles of the acid dissociate, the concentrations of \([H^+]\) and \([A^-]\) both increase by \(x\), while \([HA]\) decreases by \(x\). This helps in plugging values into the \(K_a\) expression, enabling the calculation of percent ionization and providing a clearer picture of the weak acid's behavior in a solution.
Acid Concentration
Acid concentration is a starting point in understanding the behavior of a weak acid in a solution. It's the initial amount of acid present before any dissociation occurs.
  • It's commonly represented as \([HA]\), the initial concentration of the acid.
  • Knowing the initial concentration helps in determining how much of the acid will ionize.
When calculating percent ionization, you divide the concentration of ionized acid by the initial concentration and multiply by 100. The relationship between percent ionization and acid concentration shows that the ionization inversely correlates with the square root of the initial concentration. This means as the initial concentration increases, the ionization percentage decreases, illustrating the inverse relationship in weak acid solutions.

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Most popular questions from this chapter

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\). Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (b) \(\mathrm{HCO}_{3}^{-}\)

Designate the Brønsted-Lowry acid and the BronstedLowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base on the right side: (a) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \rightleftharpoons\) (c) \(\mathrm{HCHO}_{2}(a q)+\mathrm{PO}_{4}{ }^{3-}(a q) \stackrel{\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)}{\mathrm{CHO}_{2}^{-}(a q)+\mathrm{HPO}_{4}{ }^{2-}(a q)}\)

(a) Using dissociation constants from Appendix \(D\), determine the value for the equilibrium constant for each of the following reactions. (Remember that when reactions are added, the corresponding equilibrium constants are multiplied.) (i) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (ii) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) (b) We usually use single arrows for reactions when the forward reaction is appreciable (K much greater than 1) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\), received is name because it is present in sour milk as a product of bacterial action. It is also responsible for the soreness in muscles after vigorous exercise. (a) The \(\mathrm{pK}_{a}\) of lactic acid is \(3.85\). Compare this with the value for propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}, \mathrm{p} K_{a}=4.89\right)\), and explain the differ- ence. (b) Calculate the lactate ion concentration in a \(0.050 \mathrm{M}\) solution of lactic acid. (c) When a solution of sodium lactate, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COONa}\), is mixed with an aqueous copper(II) solution, it is possible to obtain a solid salt of copper(II) lactate as a blue-green hydrate, \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COO}\right)_{2} \mathrm{Cu} \cdot x \mathrm{H}_{2} \mathrm{O} .\) Elemental analysis of the solid tells us that the solid is \(22.9 \% \mathrm{Cu}\) and \(26.0 \% \mathrm{C}\) by mass. What is the value for \(x\) in the formula for the hydrate? (d) The acid-dissociation constant for the \(\mathrm{Cu}^{2+}(a q)\) ion is \(1.0 \times 10^{-8} .\) Based on this value and the acid-dissociation constant of lactic acid, predict whether a solution of copper(II) lactate will be acidic, basic, or neutral. Explain your answer.

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