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(a) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid in a solution containing \(0.050 \mathrm{M}\) sodium lactate.

Short Answer

Expert verified
(a) Lactic acid ionizes 2.36%. (b) With sodium lactate, it ionizes 0.28%.

Step by step solution

01

Write the Ionization Reaction

Lactic acid, \ \( CH_3CHOHCOOH \, \, \) ionizes in water to form hydrogen ions and lactate ions. The ionization reaction is: \[ CH_3CHOHCOOH_{(aq)} \rightleftharpoons H^+_{(aq)} + CH_3CHOHCOO^-_{(aq)} \]
02

Set Up the Expression for Ka

The acid dissociation constant is given by: \[ K_a = \frac{{[H^+][CH_3CHOHCOO^-]}}{{[CH_3CHOHCOOH]}} \]Where \( K_a = 1.4 \times 10^{-4} \) for lactic acid.
03

Setting Initial Concentrations and Changes

Initial concentration of lactic acid is 0.250 M, and it partially ionizes. Let \( x \) be the concentration of \( H^+ \) ions at equilibrium. Initially: \[ [CH_3CHOHCOOH] = 0.250 \, M \] \[ [H^+] = 0 \, M \] \[ [CH_3CHOHCOO^-] = 0 \, M \] Changes at equilibrium: \[ [CH_3CHOHCOOH] = 0.250 - x \] \[ [H^+] = x \] \[ [CH_3CHOHCOO^-] = x \]
04

Substitute into Ka Expression for Case (a)

Substitute the equilibrium concentrations into the expression for \( K_a \): \[ K_a = \frac{{x^2}}{{0.250 - x}} = 1.4 \times 10^{-4} \] Assume \( x \ll 0.250 \), so \( 0.250 - x \approx 0.250 \). \[ \frac{{x^2}}{{0.250}} = 1.4 \times 10^{-4} \] \[ x^2 = 1.4 \times 10^{-4} \times 0.250 \] Calculate \( x \): \[ x = \sqrt{3.5 \times 10^{-5}} \approx 5.9 \times 10^{-3} \]
05

Calculate Percent Ionization for Case (a)

Percent ionization is \[ \frac{x}{\text{initial concentration}} \times 100 = \frac{5.9 \times 10^{-3}}{0.250} \times 100 = 2.36\% \]
06

Set Up ICE Table for Case (b)

For case (b), the solution consists of 0.250 M lactic acid and 0.050 M sodium lactate. Initial:\[ [CH_3CHOHCOOH] = 0.250 \, M \]\[ [H^+] = 0 \, M \]\[ [CH_3CHOHCOO^-] = 0.050 \, M \]Equilibrium changes:\[ [CH_3CHOHCOOH] = 0.250 - x \]\[ [H^+] = x \]\[ [CH_3CHOHCOO^-] = 0.050 + x \]
07

Substitute into Ka Expression for Case (b)

Substitute into the \( K_a \) expression: \[ K_a = \frac{x(0.050 + x)}{0.250 - x} = 1.4 \times 10^{-4} \] Assume \( x \ll 0.050 \), hence \( 0.050 + x \approx 0.050 \) and \( 0.250 - x \approx 0.250 \). Substitute in: \[ \frac{x \cdot 0.050}{0.250} = 1.4 \times 10^{-4} \]Solving for \( x \): \[ x = \frac{1.4 \times 10^{-4} \times 0.250}{0.050} = 7.0 \times 10^{-4} \]
08

Calculate Percent Ionization for Case (b)

Percent ionization is: \[ \frac{x}{0.250} \times 100 = \frac{7.0 \times 10^{-4}}{0.250} \times 100 \approx 0.28\% \]

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Key Concepts

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

Acid Dissociation Constant
The acid dissociation constant, represented as \( K_a \), is a vital parameter in understanding the strength of an acid. It measures the extent to which an acid can donate protons in a solution.
In mathematical terms, it is given by the expression:
  • \( K_a = \frac{{[H^+][A^-]}}{{[HA]}} \)
Here, \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of anions formed, and \([HA]\) is the concentration of the non-dissociated acid.
A larger \( K_a \) value indicates a stronger acid because it shows a higher degree of ionization.
In the case of lactic acid with \( K_a = 1.4 \times 10^{-4} \), this small value shows that lactic acid is a weak acid, not fully ionizing in solution.
Ionization Equilibrium
Ionization equilibrium refers to the state where the rate of ionization of a weak acid equals the rate of recombination of ions in the solution.
At this stage, the concentrations of ions and the non-dissociated acid remain constant. This balance can be disrupted by changes in concentration, temperature, or the presence of other substances.
For lactic acid, the ionization equilibrium is expressed as:\[ CH_3CHOHCOOH_{(aq)} \rightleftharpoons H^+_{(aq)} + CH_3CHOHCOO^-_{(aq)} \]This equilibrium equation shows how lactic acid partially dissociates into hydrogen ions \( H^+ \) and lactate ions \( CH_3CHOHCOO^- \).
As conditions change, such as adding more acid or a common ion, the equilibrium can shift according to Le Châtelier's principle, affecting percent ionization.
Weak Acids
Weak acids, like lactic acid, do not completely dissociate in water. Instead, they partially ionize, leading to a dynamic balance between ionized and non-ionized forms.
The degree of ionization is small and characterized by a low \( K_a \) value, reflective of its minor dissociation tendency.
Key characteristics of weak acids include:
  • They have a \( K_a \) much less than 1.
  • Only a small fraction of their molecules release \( H^+ \) into the solution.
  • The rate of recombination of \( H^+ \) and \( A^- \) is significant, leading to equilibrium.
In calculating percent ionization, the weak nature of acids is crucial, dictating assumptions like \( x \ll 0.250 \) in problem-solving, where \( x \) represents equilibrium concentrations.
Common Ion Effect
The common ion effect occurs when a salt containing an ion in common with the weak acid is added to the solution. This effect reduces the ionization of the acid.
For example, when adding sodium lactate to lactic acid, the concentration of the lactate ion \( CH_3CHOHCOO^- \) increases.
This stresses the equilibrium by increasing the concentration of ions formed by dissociation, so the solution compensates by decreasing the ionization of the acid.
  • This is seen in the shift of equilibrium towards the non-ionized acid, reducing the percent ionization.
  • Calculated using \( K_a = \frac{x(0.050 + x)}{0.250 - x} = 1.4 \times 10^{-4} \) where \( x \) represents \([H^+]\) in the presence of sodium lactate.
The effect is crucial in buffering solutions and controlling pH, showing the practical impacts of interaction between weak acids and their salts.

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Most popular questions from this chapter

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(P \mathrm{bF}_{2}\) (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (e) CuBr.

In nonaqueous solvents, it is possible to react HF to create \(\mathrm{H}_{2} \mathrm{~F}^{+} .\) Which of these statements follows from this observation? (a) HF can act like a strong acid in nonaqueous solvents, (b) HF can act like a base in nonaqueous solvents, (c) HF is thermodynamically unstable, \((\mathbf{d})\) There is an acid in the nonaqueous medium that is a stronger acid than HE.

Which of the following solutions is a buffer? (a) A solution made by mixing \(50 \mathrm{~mL}\), of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(250 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH},(\mathbf{b})\) A solution made by \(\mathrm{mix}\) ing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(25 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) nitric acid \(\left(\mathrm{HNO}_{3}\right),(\mathbf{c})\) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) potassium formate \((\mathrm{HCOOK})\) and \(25 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KNO}_{3},\) (d) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\), and \(25 \mathrm{~mL}\). of \(0.200 \mathrm{MKOH}\).

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid HA and a base B are mixed. The pH of the resulting solution is 9.2 . (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{h}\) for B?

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

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