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Chapter 8: Problem 88

When a solution prepared by dissolving \(4.00 \mathrm{g}\) of an unknown monoprotic acid in \(1.00 \mathrm{L}\) of water is titrated with \(0.600 \mathrm{M} \mathrm{NaOH}, 38.7 \mathrm{mL}\) of the \(\mathrm{NaOH}\) solution is needed to neutralize the acid. Determine the molarity of the acid solution. What is the molar mass of the unknown acid?

Short Answer

Expert verified
The molarity of the acid is 0.0232 M, and the molar mass is 172.2 g/mol.

Step by step solution

01

Identify the Reaction and Write the Balanced Equation

The reaction is a neutralization between a monoprotic acid (\[\mathrm{HA}\]) and \[\mathrm{NaOH}\]. The balanced chemical equation is:\[\mathrm{HA} + \mathrm{NaOH} \rightarrow \mathrm{NaA} + \mathrm{H_2O}\]Since the acid is monoprotic, it reacts in 1:1 molar ratio with the base (\[\mathrm{NaOH}\]). This means one mole of acid reacts with one mole of \[\mathrm{NaOH}\].
02

Calculate the Moles of NaOH Used

Use the formula to find moles:\[\text{Moles of } \mathrm{NaOH} = \text{Volume (in liters)} \times \text{Molarity (M)}\]Given that the volume of \(\mathrm{NaOH}\) used is 38.7 mL (convert to 0.0387 L) and the molarity is 0.600 M:\[\text{Moles of } \mathrm{NaOH} = 0.0387\, \mathrm{L} \times 0.600\, \mathrm{M} = 0.02322\, \text{moles}\]
03

Calculate the Moles of Acid

Since the reaction between the acid and the base is 1:1, the moles of acid (\[\mathrm{HA}\]) in the solution is equal to the moles of \[\mathrm{NaOH}\] used:\[\text{Moles of } \mathrm{HA} = 0.02322\, \text{moles}\]
04

Determine Molarity of the Acid Solution

Molarity (\[\mathrm{M}\]) is calculated using the formula:\[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]Plug in the values found:\[\text{Molarity of } \mathrm{HA} = \frac{0.02322\, \text{moles}}{1.00\, \text{L}} = 0.02322\, \mathrm{M}\]
05

Determine the Molar Mass of the Acid

Molar mass is determined using the formula:\[\text{Molar mass} = \frac{\text{Mass of solute}}{\text{Moles of solute}}\]Given that the mass of the acid is 4.00 g, use the moles obtained:\[\text{Molar mass of } \mathrm{HA} = \frac{4.00\, \text{g}}{0.02322\, \text{moles}} \approx 172.21\, \text{g/mol}\]

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Key Concepts

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

Molarity
Molarity is a way to measure the concentration of a solution. It tells us how many moles of a solute, such as an acid or base, are present in one liter of solution.

To calculate molarity, use the formula:
  • \( \text{Molarity} (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \)
In our example, we're dealing with a monoprotic acid solution that's neutralized by a NaOH solution.

By finding the number of moles of the acid present in the solution and dividing it by the volume of the solution in liters, you determine its molarity.
This is a crucial component in understanding how titration helps determine concentration in chemical solutions.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

The molar mass tells us the weight of individual molecules or atoms within a chemical compound, which is vital for conversions in stoichiometry. In the exercise, knowing the molar mass of an unknown acid helps identify what the acid could be.

To find the molar mass, divide the mass of your sample by the number of moles:
  • \( \text{Molar mass} = \frac{\text{Mass of solute}}{\text{Moles of solute}} \)
Knowing this allows you to translate chemical formulas into precise masses used in experiments. This concept ties directly with calculating moles to interpret lab data correctly, especially in titration.
Neutralization
Neutralization occurs when an acid and a base react to form water and a salt. This reaction is central to many processes in chemistry and everyday life.

In our exercise, the neutralization involves a monoprotic acid and NaOH, resulting in water and a salt.
  • The balanced equation for this process is: \( \mathrm{HA} + \mathrm{NaOH} \rightarrow \mathrm{NaA} + \mathrm{H_2O} \)
The key aspect here is the molar ratio, which in this case is 1:1 due to the monoprotic nature of the acid.

Understanding neutralization helps in accurately calculating the amounts of reactants and products, crucial in laboratory and industrial chemical processes.
Monoprotic Acid
A monoprotic acid is an acid that can donate only one proton (hydrogen ion) per molecule during a chemical reaction.

This is a simple, yet fundamental concept in chemistry.

Common examples include hydrochloric acid (HCl) and acetic acid (CH鈧僀OOH). In the exercise, knowing the acid is monoprotic means it will react in a 1:1 ratio with the NaOH.
  • This simplifies the calculation of moles during the titration process, as each mole of acid reacts with one mole of base.
Recognizing whether an acid is monoprotic or not is essential for predicting its reactivity and for solving titration problems accurately.

This concept helps in understanding the stoichiometry of neutralization reactions, ensuring that chemical equations balance correctly and responses are logical.

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Most popular questions from this chapter

Write equations to show what happens when, to a buffer solution containing equal amounts of HCOOH and \(\mathrm{HCOO}^{-},\) we add: (a) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (b) \(\mathrm{OH}^{-}\)

A railroad tank car derails and spills 26 tons of concentrated sulfuric acid ( 1 ton \(=907.185 \mathrm{kg}\) ). The acid is \(98.0 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) with a density of \(1.836 \mathrm{g} / \mathrm{mL}\). (a) What is the molarity of the acid? (b) Sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), is used to neutralize the acid spill. Determine the kilograms of sodium carbonate required to completely neutralize the acid. (Chapter 4) (c) How many liters of carbon dioxide at \(18^{\circ} \mathrm{C}\) and \(745 \mathrm{mm} \mathrm{Hg}\) are produced by this reaction? (Chapter 5 )

Which has the larger numerical value? (a) The \(\mathrm{p} K_{\mathrm{a}}\) of a strong acid or the \(\mathrm{p} K_{\mathrm{a}}\) of a weak acid (b) The \(K_{\mathrm{a}}\) of a strong acid or the \(K_{\mathrm{a}}\) of a weak acid

Suppose you have an aqueous solution prepared by dissolving \(0.050 \mathrm{mol}\) of \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) in \(1 \mathrm{L}\) of water. This solution is not a buffer, but suppose you want to make it into one. How many moles of solid \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) must you add to this aqueous solution to make it into: (a) A buffer of pH 7.21 (b) A buffer of pH 6.21 (c) A buffer of \(\mathrm{pH} 8.21\)

Show that the pH of a buffer is 1 unit higher than its \(\mathrm{p} K_{\mathrm{a}}\) when the ratio of \(\mathrm{A}^{-}\) to \(\mathrm{HA}\) is 10 to 1.
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