Question

What is the volume of \(1.222 \mathrm{M}\) sodium hydroxide required to react with (a) \(32.5 \mathrm{~mL}\) of \(0.569 \mathrm{M}\) sulfurous acid? (One mole of sulfurous acid reacts with two moles of hydroxide ion.) (b) \(5.00 \mathrm{~g}\) of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} ?\) (One mole of oxalic acid reacts with two moles of hydroxide ion.) (c) \(15.0 \mathrm{~g}\) of concentrated acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2},\) that is \(88 \%\) by mass pure?

Short answer

Question: Calculate the volume of 1.222 M sodium hydroxide required to react completely with each of the following: (a) 32.5 mL of 0.569 M sulfurous acid; (b) 5.00 g of oxalic acid; (c) 15.0 g of concentrated acetic acid that is 88% by mass pure. Answer: Perform the following calculations to find the required volume of sodium hydroxide for each case: (a) Volume of sodium hydroxide solution = 0.0186 L or 18.6 mL (b) Volume of sodium hydroxide solution = 0.0183 L or 18.3 mL (c) Volume of sodium hydroxide solution = 0.0107 L or 10.7 mL

Step-by-step solution

Calculate moles of sulfurous acid

Use the formula moles = molarity × volume (in liters). Moles of sulfurous acid = \(0.569\mathrm{M} \times 32.5\mathrm{mL}\) × \(\frac{1\mathrm{L}}{1000\mathrm{mL}}\)

Calculate moles of sodium hydroxide required

According to the stoichiometry given, 1 mole of sulfurous acid reacts with 2 moles of hydroxide ions. Moles of sodium hydroxide = moles of sulfurous acid × 2

Calculate the volume of sodium hydroxide solution

Use the formula volume = \(\frac{\text{moles}}{\text{molarity}}\). Volume of sodium hydroxide = \(\frac{\text{moles of NaOH}}{1.222M}\) (b) \(5.00\mathrm{g}\) of oxalic acid, \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)

Calculate moles of oxalic acid

Use the formula moles = \(\frac{\text{mass}}{\text{molar mass}}\). The molar mass of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) is 90.03g/mol. Moles of oxalic acid = \(\frac{5.00\mathrm{g}}{90.03\mathrm{g/mol}}\)

Calculate moles of sodium hydroxide required

According to the stoichiometry given, 1 mole of oxalic acid reacts with 2 moles of hydroxide ions. Moles of sodium hydroxide = moles of oxalic acid × 2

Calculate the volume of sodium hydroxide solution

Use the formula volume = \(\frac{\text{moles}}{\text{molarity}}\). Volume of sodium hydroxide = \(\frac{\text{moles of NaOH}}{1.222\mathrm{M}}\) (c) \(15.0\mathrm{g}\) of concentrated acetic acid, \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) that is \(88\%\) by mass pure.

Calculate mass and moles of pure acetic acid

Mass of pure acetic acid = \(15.0\mathrm{g} \times 0.88\). The molar mass of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) is 60.05g/mol. Moles of acetic acid = \(\frac{\text{mass of pure acetic acid}}{60.05\mathrm{g/mol}}\)

Calculate moles of sodium hydroxide required

According to the stoichiometry given, 1 mole of acetic acid reacts with 1 mole of hydroxide ions (since 1 mole of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) releases 1 mole of protons). So, moles of sodium hydroxide = moles of acetic acid.

Calculate the volume of sodium hydroxide solution

Use the formula volume = \(\frac{\text{moles}}{\text{molarity}}\). Volume of sodium hydroxide = \(\frac{\text{moles of NaOH}}{1.222\mathrm{M}}\)

Key concepts

Molarity Calculations

Understanding molarity is at the heart of many chemical calculations. Molarity, symbolized by ‘M’, represents the number of moles of solute dissolved per liter of solution. It tells us how concentrated a solution is. To calculate molarity, we use the formula:
\[ Molarity (M) = \frac{moles\text){of solute}}{liters\text){of solution}} \] Calculating molarity is essential when preparing solutions for experiments or when performing a titration, as it allows for precise control of reactant quantities. Imagine a recipe—without knowing the exact amount of each ingredient, the dish might not turn out as intended. Similarly, without precise molarity calculations, our reactions might not proceed as expected.

In the exercise provided, molarity calculations allowed us to determine the amount of sodium hydroxide needed to react with given quantities of different acids, by first converting the volume of the acid solutions to moles using their molarity.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict how much of each substance is needed or produced during a reaction. In stoichiometry, the coefficients in a balanced chemical equation tell us the ratio in which substances react and are formed.
For example, if a chemical equation states that 1 mole of A reacts with 2 moles of B to produce 3 moles of C, the stoichiometric coefficients are 1:2:3. Using this, we can determine the amount of reactant B needed to react with a known amount of A or predict the quantity of the product, C.

In our exercise, stoichiometry is used to relate the moles of acids to the moles of sodium hydroxide. Knowledge of the ratio in which they react—such as two moles of hydroxide ions for each mole of sulfurous or oxalic acid—is vital for calculating the correct volume of sodium hydroxide solution required.

Acid-Base Titration

An acid-base titration is a laboratory method used to determine the concentration of an acid or a base. The process involves adding a solution of known concentration, called the titrant, to a solution of unknown concentration, or analyte, until the reaction reaches its end point. This point is typically indicated by a color change due to an indicator or by reaching a particular pH value.
The titration process relies on the stoichiometric relationship between the acid and base. One key aspect of an acid-base titration is the neutralization reaction, where an acid and a base react to form water and a salt. By knowing the stoichiometry of this reaction, we can use the volume of the titrant added to calculate the concentration of the analyte.

In the context of the original exercise, calculating the volume of sodium hydroxide required to react with the acids is similar to performing a titration. We calculate how much base (the titrant) is needed to completely neutralize the given amount of acid (the analyte), which is analogous to reaching the end point of an acid-base titration. This is an essential skill in chemistry, as it enables accurate determination of substance concentrations, which is critical in fields ranging from medical diagnostics to environmental science.