Problem 83 - Chemistry: The Central Science

Question

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: $$ \begin{aligned} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\\ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

Step-by-Step Solution

Answer

To neutralize the spilled $27\,\mathrm{mL}$ of $6.0\,\mathrm{M}\, \mathrm{H}_{2}\mathrm{SO}_{4}$, a minimum of 27.22 g of sodium bicarbonate must be added.

Step 1: Calculate the moles of sulfuric acid

First, we need to calculate the moles of sulfuric acid using the given concentration and volume. The formula for this is: Moles of solute = Volume of solution (L) × Concentration of solute (moles/L) However, the volume provided is in milliliters, so we need to convert it to liters: \(27 \, mL × \frac{1 \, L}{1000 \, mL} = 0.027 \, L\) Now, we calculate the moles of sulfuric acid: Moles of \(H_2SO_4 = 0.027 \, L × 6.0 \, M\) Moles of \(H_2SO_4 = 0.162 \, mol\)

Step 2: Use stoichiometry to find the moles of sodium bicarbonate needed

As we can see from the balanced chemical equation, two moles of sodium bicarbonate neutralize one mole of sulfuric acid: \(2 \,\mathrm{NaHCO}_3(s) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) + 2 \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{CO}_{2}(g)\) Now, we need to find the number of moles of sodium bicarbonate needed using the stoichiometric relationship between sodium bicarbonate and sulfuric acid: Moles of \(NaHCO_3 = 0.162 \, mol \, H_2SO_4 × \frac{2 \, mol \, NaHCO_3}{1 \, mol \, H_2SO_4}\) Moles of \(NaHCO_3 = 0.324 \, mol\)

Step 3: Convert moles of sodium bicarbonate to mass

Finally, we need to convert moles of sodium bicarbonate to mass using its molar mass. The molar mass of sodium bicarbonate is: \(M_{NaHCO_3} = 22.99 + 1.01 + 12.01 + (16.00 × 3) = 84.01 \, g/mol\) So, the minimum mass of sodium bicarbonate required is: Mass of \(NaHCO_3 = 0.324 \, mol × 84.01 \, g/mol\) Mass of \(NaHCO_3 = 27.22 \, g\) Hence, to neutralize the acid, a minimum of 27.22 g of sodium bicarbonate must be added to the spill.

Key Concepts

Stoichiometry

Stoichiometry is the section of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It is a crucial concept for anyone dealing with chemical reactions because it allows us to predict how much of each substance is needed or produced.

In the case of neutralizing a sulfuric acid spill with sodium bicarbonate, stoichiometry helps determine the exact amount of sodium bicarbonate required to react with a given volume and concentration of sulfuric acid. The balanced chemical equation provides a stoichiometric 'recipe' showing that two moles of sodium bicarbonate are needed for every mole of sulfuric acid. We can think of it as a baking recipe: just as two eggs might be required to bake a cake, two units of sodium bicarbonate are required to fully neutralize one unit of sulfuric acid.

It's essential to start with a balanced chemical equation as it provides the mole ratio between the reactants and products. From this ratio, students can perform mole-to-mole conversions to find out how much reactant is needed to fully react with a given amount of another reactant. Through this method, the exercise guides the student to find the exact amount of sodium bicarbonate necessary to neutralize the given acid.

Molar Mass Calculation

The molar mass of a substance represents the mass of one mole of that substance and is expressed in grams per mole (g/mol). This is a fundamental value used in chemistry to convert between the mass of a substance and the amount of substance in moles.

When dealing with chemical reactions, such as the neutralization of sulfuric acid by sodium bicarbonate, it is essential to know the molar mass of the reactants and products to convert moles to grams (or vice versa). In this example, we calculated the molar mass of sodium bicarbonate by adding the atomic masses of sodium (Na), hydrogen (H), carbon (C), and three oxygen (O) atoms. The molar mass enables us to convert the moles of sodium bicarbonate needed to neutralize the acid into the corresponding mass required.

Having a precise value for molar mass is critical because any error in this calculation will directly affect the accuracy of the final answer. It's like knowing exactly how much each egg weighs to ensure your cake recipe turns out perfectly every time.

Acid-Base Neutralization

Acid-base neutralization is a chemical reaction where an acid and a base react to form water and a salt. It's a type of double displacement reaction that occurs when the H+ ions from the acid combine with the OH- ions from the base to produce water (H2O). In the case of strong acids and bases, the reaction can be vigorous and exothermic, releasing heat.

In the lab spill scenario, sulfuric acid (H2SO4), a strong acid, reacts with sodium bicarbonate (NaHCO3), a weak base. The resulting products are sodium sulfate (Na2SO4), water (H2O), and carbon dioxide (CO2) gas, indicated by the fizzing observed when NaHCO3 is added to the acid. The fizzing stops when there are no more H+ ions available to react, signaling that the acid has been neutralized.

The concept of neutralization is important in real-world applications beyond the classroom. For example, it is used to treat acid spills, like in the exercise, and in our body to maintain pH balance. Antacids, which are base substances, are ingested to neutralize excess stomach acid and relieve heartburn symptoms. This acid-base reaction is a prime illustration of chemistry in action and its importance in everyday life.