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Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH},\) is found in sour milk. A solution containing \(1.00 \mathrm{g} \mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) in 100.0 \(\mathrm{mL}\) of \(0.0500 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) has a \(\mathrm{pH}=4.11 .\) What is \(K_{\mathrm{a}}\) of lactic acid?

Short Answer

Expert verified
The Ka of lactic acid is \(1.413x10^{-5}\).

Step by step solution

01

Find the concentration of sodium lactate

First, you need to find the concentration of sodium lactate in the solution. The molar mass of \(\mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) is about 112 g/mol. Given that there is 1g of sodium lactate in the solution, the molarity, M, of \(\mathrm{NaCH}_{3} \mathrm{CH}_{2}\mathrm{COO}\) is \(1g/(112 g/mol) =0.00893 mol\). Since it's in 100 mL of solution, the concentration is \(0.00893 mol/0.1 L = 0.0893 M\).
02

Find the concentration of acid and hydronium ion

You are told that there is 0.050 M of lactic acid in the solution, so that is your \(\[HA\]\). The pH of the solution is given as 4.11, so to find the \([H_3O^+]\), use the formula \([H_3O^+] =10^{-pH} = 10^{-4.11}=7.94x10^{-5} M\).
03

Substitute into Ka expression

Finally, substitute these values into the Ka expression: \(Ka = [H_3O^+][A^-]/[HA] = ( 7.94x10^{-5} M)(0.0893 M)/(0.0500 M) = 1.413x10^{-5}\).

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Key Concepts

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

Lactic Acid
Lactic acid, chemically represented as \(\text{CH}_{3}\text{CH}_{2}\text{COOH}\), is an organic compound commonly found in sour milk and other fermented products. It's a simple alpha-hydroxy acid, which means it has both a carboxylic group \(-COOH\) and a hydroxyl group \(-OH\) on an adjacent carbon. This structure allows lactic acid to participate in various biochemical reactions, especially in metabolic pathways.A key property of lactic acid is its ability to dissociate in aqueous solutions, releasing a hydrogen ion \(H^+\) and forming the lactate ion \(CH_3CH_2COO^-\). This process is crucial in determining the acid's strength and behavior in solutions. The strength of lactic acid, like other acids, can be measured by its **acid dissociation constant** \(K_a\). The larger the \(K_a\), the stronger the acid and the more readily it donates a proton to water to produce \(H_3O^+\).
pH Calculation
Understanding pH calculation is essential for measuring the acidity of a solution, such as one containing lactic acid. The **pH** is a numeric scale used to specify the acidity or basicity of an aqueous solution, calculated by the formula:\[pH = -\log[H_3O^+]\]In the provided solution, a pH of 4.11 indicates a certain concentration of hydronium ions \([H_3O^+]\). We calculate this by rearranging the pH formula to find \([H_3O^+]\):\[[H_3O^+] = 10^{-4.11} \approx 7.94 \times 10^{-5} \text{ M}\]This calculation helps signify how acidic or basic a solution is, where lower pH values represent higher acidity and higher concentrations of \(H_3O^+\) ions. It's important for tasks such as determining the acid dissociation constant \(K_a\), which provides deeper insight into the acid's dissociation in its environment and is a crucial step in many chemical analysis processes.
Concentration of Solutions
In chemistry, the concentration of solutions is a fundamental concept representing the amount of solute dissolved in a given volume of solvent. For this exercise, calculating the concentration is essential to determining the acid dissociation constant \(K_a\).When you're given the mass of a compound such as sodium lactate, \(\text{NaCH}_3\text{CH}_2\text{COO}\), converting this mass to moles is the first step. Utilizing the molar mass, you can calculate the moles:
  • Molar mass of sodium lactate = 112 g/mol
  • Mass = 1 g
  • Moles = \( \frac{1 \, \text{g}}{112 \, \text{g/mol}} = 0.00893 \, \text{mol} \)
Since the solution's volume is 100 mL (or 0.1 L), the molarity \(M\) of sodium lactate is then calculated as:\[M = \frac{0.00893 \, \text{mol}}{0.1 \, \text{L}} = 0.0893 \, \text{M}\]These concentration values are essential for plugging into the \(K_a\) expression \(K_a = \frac{[H_3O^+][A^-]}{[HA]}\), where \([A^-]\) and \([HA]\) reflect the concentrations of the lactate ion and lactic acid respectively. By understanding and calculating these concentrations, one can further analyze and understand the behaviors of acidic and basic solutions.

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Most popular questions from this chapter

Write equations to show how each of the following buffer solutions reacts with a small added amount of a strong acid or a strong base: (a) HCOOH-KHCOO; (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}-\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\) (c) \(\mathrm{KH}_{2} \mathrm{PO}_{4}-\mathrm{Na}_{2} \mathrm{HPO}_{4}\)

A solution is prepared that is \(0.150 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.250 \mathrm{M} \mathrm{NaHCOO}\) (a) Show that this is a buffer solution. (b) Calculate the pH of this buffer solution. (c) What is the final pH if 1.00 L of 0.100 M HCl is added to \(1.00 \mathrm{L}\) of this buffer solution?

Is a solution of sodium dihydrogen citrate, \(\mathrm{NaH}_{2} \mathrm{Cit}\) likely to be acidic, basic, or neutral? Explain. Citric \(\mathrm{acid}, \mathrm{H}_{3} \mathrm{Cit}, \mathrm{is} \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\)

You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all \(0.100 \mathrm{M},\) are available to you: HCOOH, CH \(_{3} \mathrm{COOH}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{NaCHOO}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) and \(\mathrm{NaH}_{2} \mathrm{PO}_{4} . \quad\) Describe how you would prepare this buffer solution. [Hint: What volumes of which solutions would you use?

In 1922 Donald D. van Slyke ( J. Biol. Chem., 52, 525) defined a quantity known as the buffer index: \(\beta=\mathrm{d} C_{\mathrm{b}} / \mathrm{d}(\mathrm{pH}),\) where \(\mathrm{d} C_{\mathrm{b}}\) represents the increment of moles of strong base to one liter of the buffer. For the addition of a strong acid, he wrote \(\beta=-\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH})\) By applying this idea to a monoprotic acid and its conjugate base, we can derive the following expression: \(\beta=2.303\left(\frac{K_{w}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\frac{\mathrm{CK}_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left(\mathrm{K}_{\mathrm{a}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\right)^{2}}\right)\) where \(C\) is the total concentration of monoprotic acid and conjugate base. (a) Use the above expression to calculate the buffer index for the acetic acid buffer with a total acetic acid and acetate ion concentration of \(2.0 \times 10^{-2}\) and a \(\mathrm{pH}=5.0\) (b) Use the buffer index from part (a) and calculate the \(\mathrm{pH}\) of the buffer after the addition of of a strong acid. (Hint: Let \(\left.\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH}) \approx \Delta C_{\mathrm{a}} / \Delta \mathrm{pH} .\right)\) (c) Make a plot of \(\beta\) versus \(\mathrm{pH}\) for a \(0.1 \mathrm{M}\) acetic acid buffer system. Locate the maximum buffer index as well as the minimum buffer indices.

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