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: $$ \begin{array}{r} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\\ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{array} $$ 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

The minimum mass of sodium bicarbonate (\(NaHCO_3\)) needed to neutralize the spilled 27 mL of 6.0 M sulfuric acid (\(H_2SO_4\)) is 27.22 g.

Step 1: Calculate the moles of sulfuric acid

Given the volume and molarity of the spilled sulfuric acid, we can calculate the moles of sulfuric acid using the formula: Moles of acid = Molarity × Volume The volume should be in liters, so we need to convert 27 mL to liters: \( 27 mL = 27 × 10^{-3} L \) Now, we can calculate the moles of sulfuric acid: Moles of \(H_2 SO_4\) = 6.0 M × 27 × \(10^{-3}\) L = 0.162 mol

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

Looking at the balanced chemical equation, we can see that 2 moles of sodium bicarbonate (\(NaHCO_3\)) reacts with 1 mole of sulfuric acid (\(H_2 SO_4\)). Therefore, we can use stoichiometry to find the amount of sodium bicarbonate needed to react with the given moles of sulfuric acid: Moles of \(NaHCO_3\) = 2 × Moles of \(H_2 SO_4\) Moles of \(NaHCO_3\) = 2 × 0.162 mol = 0.324 mol

Step 3: Convert moles of sodium bicarbonate to mass

Now that we know the moles of sodium bicarbonate needed, we can convert it to mass using its molar mass. The molar mass of sodium bicarbonate (\(NaHCO_3\)) can be calculated by summing the molar masses of its elements: Molar mass of \(NaHCO_3 = Na + H + C + (3 × O) = 22.99 + 1.01 + 12.01 + (3 × 16.00) = 84.01 g/mol\) Using the molar mass, we can convert the moles of sodium bicarbonate to mass: Mass of \(NaHCO_3\) = 0.324 mol × 84.01 g/mol = 27.22 g So, the minimum mass of sodium bicarbonate needed to neutralize the spilled sulfuric acid is 27.22 g.

Key Concepts

Stoichiometry in Acid-Base Reactions

Understanding stoichiometry in acid-base reactions is essential for predicting the outcomes of chemical reactions and for tasks such as neutralizing acid spills. In the context of neutralizing a sulfuric acid spill with sodium bicarbonate, stoichiometry allows us to use the balanced chemical equation to determine the precise amount of sodium bicarbonate needed to react completely with the spilled acid.

Let's delve into the specifics. A balanced chemical equation like the one provided, \[2 \mathrm{NaHCO}_3(s) + \mathrm{H}_2 \mathrm{SO}_4(aq) \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4(aq) + 2 \mathrm{H}_2 \mathrm{O}(l) + 2 \mathrm{CO}_2(g)\], demonstrates the stoichiometric relationship between reactants and products. In this case, two molecules (or moles) of sodium bicarbonate neutralize one molecule (or mole) of sulfuric acid to produce sodium sulfate, water, and carbon dioxide gas.

To apply this relationship, first, calculate the moles of the spilled acid. Then, knowing the stoichiometry (in this case, a 2:1 ratio), you can determine the moles of sodium bicarbonate needed for neutralization. This stoichiometric 'roadmap' guides us to the solution and is a fundamental concept in chemistry.

Molar Mass Calculation

The molar mass calculation is critical for converting between moles and grams in chemical reactions. It's the mass of one mole of a substance and is expressed in grams per mole (g/mol). Calculating molar mass involves summing the atomic masses of all atoms in the molecule, as found on the periodic table.

For example, sodium bicarbonate (\(\mathrm{NaHCO}_3\)) is composed of sodium (Na), hydrogen (H), carbon (C), and three oxygen (O) atoms. The molar mass of each element, often found on the periodic table, is summarized as follows: \[\mathrm{Na} = 22.99\, \mathrm{g/mol},\]\[\mathrm{H} = 1.01\, \mathrm{g/mol},\]\[\mathrm{C} = 12.01\, \mathrm{g/mol},\]\[\mathrm{O} = 16.00\, \mathrm{g/mol}.\]
Adding these values together, considering that there are three oxygen atoms in sodium bicarbonate, provides us with the molar mass of the compound:\[\mathrm{Molar\, mass\, of\, NaHCO}_3 = 22.99 + 1.01 + 12.01 + (3 \times 16.00) = 84.01\, \mathrm{g/mol}.\]
Understanding how to calculate molar mass is invaluable for converting back and forth between the mass of a substance and the amount of substance in moles.

Converting Moles to Mass

Once we have the molar mass and the stoichiometry of the reaction figured out, we can convert moles to mass, completing the neutralization equation for an acid spill. This conversion is a direct application of the formula:\[\mathrm{Mass} = \mathrm{Moles} \times \mathrm{Molar\, Mass}.\]

Let's use our sodium bicarbonate example. After calculating the moles of sodium bicarbonate needed to neutralize the sulfuric acid using stoichiometry, we use the compound's molar mass to determine the mass required for a complete reaction. With a mole-to-mole comparison from the stoichiometry and the molar mass of sodium bicarbonate, the mass can be found as follows:\[\mathrm{Mass\, of\, NaHCO}_3 = 0.324\, \mathrm{mol} \times 84.01\, \mathrm{g/mol} = 27.22\, \mathrm{g}.\]
Therefore, to neutralize the spill, we would need at least 27.22 grams of sodium bicarbonate. Mastery of moles-to-mass conversion is vital for practical applications in laboratory settings and ensures safety and accuracy in scientific endeavors.