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A 0.3126 g sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) requires 26.21 mL of a particular concentration of \(\mathrm{NaOH}(\mathrm{aq})\) to complete the following reaction. What is the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

Short Answer

Expert verified
The molarity of the \(\mathrm{NaOH(aq)}\) is 0.265 M.

Step by step solution

01

Convert the mass of oxalic acid to moles.

Using the molar mass of oxalic acid, which is 90.03g/mol, the number of moles of oxalic acid can be calculated as follows: Moles of oxalic acid = mass / molar mass = 0.3126g / 90.03g/mol = 0.00347 mol.
02

Apply stoichiometry.

According to the balanced chemical equation, 1 mol of oxalic acid reacts completely with 2 moles of \(NaOH\). Therefore, the moles of \(NaOH\) required for the complete reaction is double the moles of oxalic acid, i.e., = 2 * 0.00347 mol = 0.00694 mol.
03

Calculate molarity

The volume of \(NaOH\) solution used is given as 26.21 mL, which should be converted to liters as molarity is expressed in mol/L: Volume = 26.21 mL = 0.02621 L. From the definition of molarity (moles/volume), we can now calculate the molarity of the \(NaOH\) solution: Molarity of \(NaOH\) = moles/volume = 0.00694 mol / 0.02621 L = 0.265 M.

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Key Concepts

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

Stoichiometry
Stoichiometry is a core concept in chemistry, often involved in the calculation of reactants and products in chemical reactions. It provides a quantitative relationship between the substances involved. In the provided example, stoichiometry is used to determine the amount of sodium hydroxide (\(\text{NaOH}\)) needed to react with a known amount of oxalic acid.When looking at chemical equations, stoichiometry helps you understand:
  • The proportion of reactants consumed and products formed.
  • The mole-to-mole ratio, which is essential for converting between different substances.
In this exercise, the balanced equation shows that \(1\) mole of oxalic acid needs \(2\) moles of \(\text{NaOH}\). This relationship helps calculate how much \(\text{NaOH}\) is required when the amount of oxalic acid is known.
Oxalic Acid
Oxalic acid (\(\text{H}_{2} \text{C}_{2} \text{O}_{4}\)) is an organic compound with a relatively simple structure. It is a dicarboxylic acid, meaning it has two carboxyl groups, which can release protons. This makes it a diprotic acid used in various chemical reactions, including titrations.Key characteristics of oxalic acid include:
  • It is a crystalline solid often found in its dihydrate form.
  • It acts as a reducing agent in some chemical processes.
  • Due to its acidic nature, it can react with bases such as sodium hydroxide to produce salts.
In the given problem, oxalic acid reacts with sodium hydroxide in a neutralization reaction. This type of reaction is common in stoichiometric calculations, helping us find the exact amount of base required to react with the acid.
Sodium Hydroxide
Sodium hydroxide (\(\text{NaOH}\)) is a highly caustic base, widely used in various chemical reactions and industrial processes. Often referred to as lye or caustic soda, it easily dissolves in water to form a strongly alkaline solution.Properties of sodium hydroxide include:
  • It is a strong base, meaning it fully ionizes in water.
  • It is used in processes like soap making, paper production, and water treatment.
  • Handling requires care due to its corrosive nature.
In our exercise, sodium hydroxide reacts with oxalic acid to form sodium oxalate and water. Understanding the properties of sodium hydroxide aids in safely conducting experiments and achieving accurate stoichiometric calculations, such as determining its molarity from used volume and moles reacted.

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Most popular questions from this chapter

A \(1.000 \mathrm{g}\) sample of a mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) is analyzed by burning it completely in \(\mathrm{O}_{2}\), yielding \(2.776 \mathrm{g} \mathrm{CO}_{2} .\) What is the percentage by mass of \(\mathrm{CH}_{4}\) in the mixture? (a) \(93 \% ;\) (b) \(82 \% ;\) (c) \(67 \% ;\) (d) \(36 \%\) (e) less than \(36 \%\)

Lead nitrate and potassium iodide react in aqueous solution to form a yellow precipitate of lead iodide. In one series of experiments, the masses of the two reactants were varied, but the total mass of the two was held constant at \(5.000 \mathrm{g}\). The lead iodide formed was filtered from solution, washed, dried, and weighed. The table gives data for a series of reactions. $$\begin{array}{lll} \hline & \text { Mass of Lead } & \text { Mass of Lead } \\ \text { Experiment } & \text { Nitrate, } g & \text { lodide, } g \\ \hline 1 & 0.500 & 0.692 \\ 2 & 1.000 & 1.388 \\ 3 & 1.500 & 2.093 \\ 4 & 3.000 & 2.778 \\ 5 & 4.000 & 1.391 \\ \hline \end{array}$$ (a) Plot the data in a graph of mass of lead iodide versus mass of lead nitrate, and draw the appropriate curve(s) connecting the data points. What is the maximum mass of precipitate that can be obtained? (b) Explain why the maximum mass of precipitate is obtained when the reactants are in their stoichiometric proportions. What are these stoichiometric proportions expressed as a mass ratio, and as a mole ratio? (c) Show how the stoichiometric proportions determined in part (b) are related to the balanced equation for the reaction.

When water and methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}),\) are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes. (Refer to Exercise 99 for an explanation.) When \(72.061 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(192.25 \mathrm{g}\) \(\mathrm{CH}_{3} \mathrm{OH}\) are mixed at \(25^{\circ} \mathrm{C},\) the resulting solution has a density of \(0.86070 \mathrm{g} / \mathrm{mL} .\) At \(25^{\circ} \mathrm{C},\) the densities of water and methanol are \(0.99705 \mathrm{g} / \mathrm{mL}\) and \(0.78706\) \(\mathrm{g} / \mathrm{mL},\) respectively. (a) Calculate the volumes of the pure liquid samples and the solution, and show that the pure liquid volumes are not additive. [ Hint: Although the volumes are not additive, the masses are.] (b) Calculate the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution.

Consider the reaction below. \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) How many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react completely with \(415 \mathrm{mL}\) of \(0.477 \mathrm{M} \mathrm{HCl} ?\) (b) How many kilograms of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react with 324 L of a HCl solution that is 24.28\% HCl by mass, and has a density of \(1.12 \mathrm{g} / \mathrm{mL} ?\)

The minerals calcite, \(\mathrm{CaCO}_{3},\) magnesite, \(\mathrm{MgCO}_{3}\) and dolomite, \(\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3},\) decompose when strongly heated to form the corresponding metal oxide(s) and carbon dioxide gas. A 1.000 -g sample known to be one of the three minerals was strongly heated and \(0.477 \mathrm{g} \mathrm{CO}_{2}\) was obtained. Which of the three minerals was it?

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