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Lactic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right)\) has one acidic hydrogen. A \(0.10 \mathrm{M}\) solution of lactic acid has a \(\mathrm{pH}\) of \(2.44\). Calculate \(K_{a}\).

Short Answer

Expert verified
The \(K_{a}\) for lactic acid is \(1.38 \times 10^{-4}\).

Step by step solution

01

Convert pH to [H+] concentration

To find the concentration of \(H^+\) ions in the solution, we can use the formula: \[ [H^+] = 10^{-\mathrm{pH}}\] We are given the pH of the solution, so we can plug it into the formula: \[ [H^+] = 10^{-2.44} \] Calculate the concentration: \[ [H^+] = 3.63 \times 10^{-3} \mathrm{M} \]
02

Set up the ICE table

We will set up an ICE (Initial, Change, Equilibrium) table to help us keep track of the concentrations of all species involved in the dissociation. | | CH\(_3\)CH(OH)COOH | H\(^+\) | CH\(_3\)CH(OH)COO\(^-\) | |-------------------|-------------------|-------|--------------------| | Initial(moles/L) | 0.10 | 0 | 0 | | Change(moles/L) | -x | +x | +x | | Equilibrium | 0.10 - x | x | x |
03

Relate [H+] and [CH\(_3\)CH(OH)COO\(^-\)] concentrations

From the ice table, we can see that the equilibrium concentration of both \(H^+\) ions and CH\(_3\)CH(OH)COO\(^-\) ions are equal to x moles/L: \[ [\mathrm{H}^+] = [\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COO}^-] = x \] We calculated the concentration of \(H^+\) ions in Step 1: \[ x = 3.63 \times 10^{-3} \mathrm{M} \]
04

Calculate the equilibrium concentration of CH\(_3\)CH(OH)COOH

Now, we will find the equilibrium concentration of lactic acid (CH\(_3\)CH(OH)COOH). From the ICE table, the concentration at equilibrium is (0.10 - x) moles/L: \[ [\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}]_{eq} = 0.10 - x \] Substitute the known value of x: \[ [\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}]_{eq} = 0.10 - 3.63 \times 10^{-3} \] Calculate the equilibrium concentration of lactic acid: \[ [\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}]_{eq} = 0.09637 \mathrm{M} \]
05

Calculate \(K_a\)

Now, we have equilibrium concentrations for all the species involved in the dissociation. We can use the formula for \(K_a\): \[ K_a = \frac{[\mathrm{H}^+][\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}]} \] Plug in the equilibrium concentrations: \[ K_a = \frac{(3.63 \times 10^{-3})(3.63 \times 10^{-3})}{0.09637} \] Calculate \(K_a\): \[ K_a = 1.38 \times 10^{-4} \] Thus, the \(K_{a}\) for lactic acid is \(1.38 \times 10^{-4}\).

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Key Concepts

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

Lactic Acid
Lactic acid is a fascinating organic compound that plays a key role in various biochemical processes. Its chemical formula is \( ext{CH}_{3} ext{CH}( ext{OH}) ext{COOH}\), and it possesses one acidic hydrogen, which means it can donate a proton in a chemical reaction. This ability to donate a proton classifies lactic acid as a weak acid. In many physiological scenarios, such as during intense exercise, lactic acid is produced in muscles, where it can cause temporary discomfort and fatigue.

In terms of its chemical behavior in solutions, lactic acid partially dissociates into hydrogen ions \( ext{H}^+\) and the lactate ion \( ext{CH}_3 ext{CH}( ext{OH}) ext{COO}^-\). This partial dissociation is quantitatively measured by the acid dissociation constant, \(K_a\), which reveals the strength of the acid in aqueous solutions. Understanding the behavior of lactic acid in solutions is crucial in settings ranging from biological to industrial.
pH Calculation
The pH of a solution is a measure of its acidity or basicity, often serving as a crucial parameter in chemical studies. It is calculated using the hydrogen ion concentration \([H^+]\). The formula to find pH is: \[ \text{pH} = -\log_{10}[H^+] \]In the given problem, the pH is used to find \([H^+]\) concentration for a 0.10 M solution of lactic acid with a pH of 2.44. This is done using the inverse logarithmic relationship:\[ [H^+] = 10^{-\text{pH}} \]Calculating this yields \([H^+] \approx 3.63 \times 10^{-3} \text{ M}\).

The pH gives insight into the concentration of hydrogen ions present in the solution, which is important for calculating the \(K_a\) of weak acids like lactic acid.
ICE Table
An ICE table is a helpful visual tool used in chemistry to calculate the changes in concentrations of substances during a chemical reaction. It stands for Initial, Change, and Equilibrium - three stages that a solution goes through in a reaction process.

Setting up an ICE table involves listing the initial concentrations of all reactants and products. If the initial concentration of the hydrogen ion \([H^+]\) is zero, this entry in the table would reflect that. The change line in the table accounts for the amount the species concentration increases or decreases, represented as \(x\) for a typical acid dissociation scenario. Finally, the equilibrium line in the table shows concentrations at complete equilibrium based on initial and change values.

For the dissociation of lactic acid, the ICE table helps calculate the equilibrium concentrations of \([H^+]\), \([ ext{CH}_3 ext{CH}( ext{OH}) ext{COO}^-]\), and \([ ext{CH}_3 ext{CH}( ext{OH}) ext{COOH}]\). Using these, you can easily find the acid dissociation constant \(K_a\) to analyze the acidity of the lactic acid solution in a more structured and simple way.

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Most popular questions from this chapter

The hypochlorite ion, \(\mathrm{ClO}^{-}\), acts as a weak base. (a) Is \(\mathrm{ClO}^{-}\)a stronger or weaker base than hydroxylamine? (b) When \(\mathrm{ClO}^{-}\) acts as a base, which atom, \(\mathrm{Cl}\) or \(\mathrm{O}\), acts as the proton acceptor? (c) Can you use formal charges to rationalize your answer to part (b)?

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$ \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ A \(0.035 \mathrm{M}\) solution of ephedrine has a \(\mathrm{pH}\) of 11.33. (a) What are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}\), and \(\mathrm{OH}\) ? (b) Calculate \(K_{b}\) for ephedrine.

Predict whether aqueous solutions of the following substances are acidic, basic, or neutral: (a) \(\mathrm{AlCl}_{3}\), (b) \(\mathrm{NaBr}\), (c) \(\mathrm{NaClO}\), (d) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}\right] \mathrm{NO}_{3}\), (e) \(\mathrm{Na}_{2} \mathrm{SO}_{3}\).

Hemoglobin plays a part in a series of equilibria involving protonation- deprotonation and oxygenation-deoxygenation. The overall reaction is approximately as follows: $$ \mathrm{HbH}^{+}(a q)+\mathrm{O}_{2}(a q) \rightleftharpoons \mathrm{HbO}_{2}(a q)+\mathrm{H}^{+}(a q) $$ where \(\mathrm{Hb}\) stands for hemoglobin and \(\mathrm{HbO}_{2}\) for oxyhemoglobin. (a) The concentration of \(\mathrm{O}_{2}\) is higher in the lungs and lower in the tissues. What effect does high \(\left[\mathrm{O}_{2}\right]\) have on the position of this equilibrium? (b) The normal \(\mathrm{pH}\) of blood is 7.4. Is the blood acidic, basic, or neutral? (c) If the blood \(\mathrm{pH}\) is lowered by the presence of large amounts of acidic metabolism products, a condition known as acidosis results. What effect does lowering blood pH have on the ability of hemoglobin to transport \(\mathrm{O}_{2}\) ?

Explain the following observations: (a) \(\mathrm{HCl}\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S}\); (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is a stronger acid than \(\mathrm{H}_{3} \mathrm{AsO}_{4}\); (c) \(\mathrm{HBrO}_{3}\) is a stronger acid than \(\mathrm{HBrO}_{2}\); (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is a stronger acid than \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\); (e) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is a stronger acid than phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\).

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