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Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\) ) reacts with acids in foods to form carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(g)\) forms bubbles and causes the cake to rise. \((\mathbf{a})\) A rule of thumb in baking is that \(1 / 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\). Write the chemical equation for this neutralization reaction. (b) The density of baking soda is \(2.16 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of mol/L. (One cup \(=236.6 \mathrm{~mL}=48\) teaspoons \() .(\mathbf{c})\) If \(1 / 2\) teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at a pressure of \(101.3 \mathrm{kPa}\), in an oven set to \(177^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The balanced chemical equation for the neutralization reaction between sodium bicarbonate and lactic acid is: \( \mathrm{NaHCO}_3 + \mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{COOH} \rightarrow \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}_{2} + \mathrm{CH}_{3}\mathrm{CH}(\mathrm{OH})\mathrm{COONa} \) Using the given rule of thumb and the provided information, the concentration of lactic acid in one cup of sour milk is calculated to be approximately 0.2679 mol/L. At a pressure of 101.3 kPa and a temperature of 177°C, the volume of carbon dioxide gas produced during this reaction is approximately 0.1052 L.

Step by step solution

01

Write the balanced chemical equation for the neutralization reaction

Sodium bicarbonate reacts with lactic acid to form water, carbon dioxide, and sodium lactate. The unbalanced chemical equation for this reaction is: NaHCO3 + CH3CH(OH)COOH -> H2O + CO2 + CH3CH(OH)COONa Since there's an equal amount of atoms for each element on both sides, the equation is already balanced.
02

Calculate the concentration of lactic acid in one cup of sour milk

Firstly, convert the density of baking soda to mass. According to the given information, 1/2 teaspoon of baking soda is enough to neutralize one cup of sour milk. The density of baking soda is 2.16 g/cm³. Volume of 1/2 teaspoon of baking soda = 1/2 * 4.93 cm³ = 2.465 cm³ Mass of 1/2 teaspoon of baking soda = Volume * Density = 2.465 cm³ * 2.16 g/cm³ = 5.3246 g Now, convert the mass of baking soda to moles: Molar mass of NaHCO3 = 23(g/mol) + 1(g/mol) + 12(g/mol) + 3 * 16(g/mol) = 84 g/mol Moles of NaHCO3 = 5.3246 g / 84 g/mol ≈ 0.06339 moles Since 1 mole of NaHCO3 neutralizes 1 mole of lactic acid: moles of lactic acid = moles of NaHCO3 = 0.06339 moles Now, calculate the concentration of lactic acid in one cup of sour milk (=236.6 mL): Concentration of lactic acid = moles / volume = 0.06339 moles / 236.6 mL = 0.2679 mol/L
03

Calculate the volume of carbon dioxide gas produced

To find the volume of carbon dioxide gas produced, we will use the ideal gas equation: PV = nRT Where: P = pressure = 101.3 kPa n = moles of CO2 formed = 0.06339 moles (since 1 mole of NaHCO3 produces 1 mole of CO2) R = gas constant = 8.314 J/(mol·K) (converting kPa to J, multiply by 1000: 8.314 * 1000 = 8.314 kPa·L/(mol·K)) T = temperature = 177°C + 273.15 = 450.15 K Rearrange the formula to solve for the volume, V: V = nRT/P V = (0.06339 moles) × (8.314 kPa·L/(mol·K)) × (450.15 K) / (101.3 kPa) V ≈ 0.1052 L The volume of carbon dioxide gas produced is approximately 0.1052 L, at a pressure of 101.3 kPa and a temperature of 177°C.

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Key Concepts

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

Neutralization Reactions
Neutralization reactions take place when an acid and a base interact to form water and a salt. In the baking soda and lactic acid example provided in the exercise, sodium bicarbonate (baking soda) acts as the base and lactic acid is the acid. When they react, they produce sodium lactate, water, and carbon dioxide gas. This carbon dioxide gas is what causes baked goods like cakes to rise.

Here is the balanced chemical equation for this reaction:
  • NaHCO₃ + CH₃CH(OH)COOH → H₂O + CO₂ + CH₃CH(OH)COONa
This equation shows that one mole of sodium bicarbonate reacts with one mole of lactic acid to produce one mole each of water and carbon dioxide, as well as a mole of sodium lactate.

Neutralization reactions are significant in not only culinary applications but also many industrial and laboratory processes. They often help in adjusting pH levels and in preparing salts and other chemicals.
Mole Concept
The mole concept is a core principle in chemistry that helps us understand the amount of a substance. One mole equates to Avogadro's number, about \(6.022 \times 10^{23}\) entities (atoms, molecules, etc.). In this exercise, the mole concept helps calculate how much lactic acid is needed to neutralize the given amount of baking soda.

To find out how many moles of baking soda are in 1/2 teaspoon, you multiply the volume by the substance's density to get the mass, and then divide that mass by the molar mass of \(\text{NaHCO}\_3\) which is 84 g/mol. Calculation reveals that there are approximately 0.06339 moles of baking soda available.

Since the reaction is one-to-one, these moles of baking soda will neutralize an equal number of moles of lactic acid. Understanding this helps calculate the concentration of lactic acid in sour milk by using the formula:
  • Concentration (mol/L) = moles/volume (L)
The known volume of sour milk is 236.6 mL (or 0.2366 L), leading to a concentration of approximately 0.2679 mol/L of lactic acid.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry represented by \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. This law describes the behavior of ideal gases by relating these four variables.

In the context of the baking soda and sour milk reaction, once the carbon dioxide gas is produced, it occupies a volume that can be calculated using the ideal gas law. Given that \(P = 101.3\,\text{kPa}\), \(n = 0.06339\,\text{moles}\), \(R = 8.314\,\text{kPa} \cdot \text{L}/(\text{mol} \cdot \text{K})\), and \(T = 450.15\,\text{K}\), the volume \(V\) can be found.

Solving for volume gives:
  • \(V = \frac{nRT}{P} = \frac{(0.06339) \cdot (8.314) \cdot (450.15)}{101.3} \approx 0.1052\,\text{L}\)
Thus, approximately 0.1052 L of carbon dioxide gas would be produced under these conditions, illustrating how the ideal gas law is instrumental in predicting gas behavior in chemical reactions.

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Most popular questions from this chapter

Assume that \(30.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of a weak base \(\mathrm{B}\) that accepts one proton is titrated with a \(0.10 \mathrm{M}\) solution of the monoprotic strong acid HA. (a) How many moles of HA have been added at the equivalence point? (b) What is the predominant form of B at the equivalence point? (a) Is the \(\mathrm{pH} 7\), less than 7 , or more than 7 at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

A 20.0-mL sample of \(0.150 \mathrm{MKOH}\) is titrated with \(0.125 \mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the pH after the following volumes of acid have been added: \((\mathbf{a}) 20.0 \mathrm{~mL},(\mathbf{b}) 23.0 \mathrm{~mL},\) (c) \(24.0 \mathrm{~mL}\) (d) \(25.0 \mathrm{~mL},\) (e) \(30.0 \mathrm{~mL}\).

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(a) Calculate the percent ionization of \(0.0085 \mathrm{Mbutanoic}\) acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.0085 \mathrm{M}\) butanoic acid in a solution containing \(0.075 M\) sodium butanoate.

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