Question

How many milliliters of \(0.100 \mathrm{M} \mathrm{NaOH}\) are needed to completely neutralize \(25.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} ?\) The reaction is $$ 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}(a q) \stackrel{\longrightarrow}{\mathrm{Na}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}(a q)+2 \mathrm{H}_{2} \mathrm{O}} $$

Short answer

125 milliliters of \(0.100 \mathrm{M} \mathrm{NaOH}\) are needed.

Step-by-step solution

Write down the balanced equation

The balanced chemical equation is already given: \[ 2 \mathrm{NaOH}(aq) + \mathrm{H}_2\mathrm{C}_4\mathrm{H}_4\mathrm{O}_6(aq) \rightarrow \mathrm{Na}_2\mathrm{C}_4\mathrm{H}_4\mathrm{O}_6(aq) + 2 \mathrm{H}_2\mathrm{O}(l) \] This reaction shows that 2 moles of NaOH are required to neutralize 1 mole of \(\mathrm{H}_2\mathrm{C}_4\mathrm{H}_4\mathrm{O}_6\).

Calculate moles of \(\mathrm{H}_2\mathrm{C}_4\mathrm{H}_4\mathrm{O}_6\)

Use the molarity and volume to calculate the moles of \(\mathrm{H}_2\mathrm{C}_4\mathrm{H}_4\mathrm{O}_6\): \[ \text{Moles} = \text{Molarity} \times \text{Volume} = 0.250 \, \mathrm{M} \times 25.0 \, \mathrm{mL} = 0.250 \times 0.0250 \, \mathrm{mol} = 0.00625 \, \mathrm{mol} \] Note: Convert volume from milliliters to liters by dividing by 1000.

Calculate moles of NaOH needed

According to the balanced chemical equation, 2 moles of NaOH are needed for every 1 mole of \(\mathrm{H}_2\mathrm{C}_4\mathrm{H}_4\mathrm{O}_6\). Therefore, the moles of NaOH needed are: \[ 2 \times \text{Moles of } \mathrm{H}_2\mathrm{C}_4\mathrm{H}_4\mathrm{O}_6 = 2 \times 0.00625 \, \mathrm{mol} = 0.0125 \, \mathrm{mol} \]

Calculate volume of NaOH solution required

Now, use the molarity of NaOH to find the volume that contains 0.0125 moles of NaOH: \[ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.0125 \, \mathrm{mol}}{0.100 \, \mathrm{M}} = 0.125 \, \mathrm{liters} = 125 \, \mathrm{mL} \]

Key concepts

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is based on the conservation of mass and the stoichiometric coefficients in a balanced chemical equation. In a neutralization reaction, it helps us understand how much of each reactant is needed to produce a certain amount of product. For example, in the provided exercise, the balanced equation indicates that two moles of NaOH are necessary to neutralize one mole of \(H_2C_4H_4O_6\). This 2:1 ratio allows us to perform calculations to determine the volume of NaOH solution needed to achieve complete neutralization.

When solving stoichiometry problems, ensure that the chemical equation is balanced, then use the coefficients to set up mole ratio comparisons. For the given reaction, knowing the amount of tartaric acid (\(H_2C_4H_4O_6\)) allows us to calculate the required moles of NaOH using the mole ratio derived from the balanced equation. This step is foundational and guides the subsequent calculations.

Molarity and Volume Relationship

The relationship between molarity and volume is crucial when dealing with solutions in chemistry. Molarity (M) is the measure of the concentration of a solute in a solution and is defined as the number of moles of solute per liter of solution. It can be mathematically represented as:\[ M = \frac{moles}{liters} \]

Understanding this relationship is key when you're required to mix or dilute solutions to achieve a desired concentration, as in the case of the neutralization reaction from the exercise. To find the volume of NaOH needed, we start by calculating the moles of \(H_2C_4H_4O_6\) using its molarity and volume. Then, leveraging the stoichiometry from the balanced chemical equation, we find the moles of NaOH required. Finally, we rearrange the molarity formula to solve for the volume of NaOH:
\[ Volume = \frac{Moles}{Molarity} \]
By converting this volume into liters or milliliters as necessary, we can understand practical measures such as how many milliliters of a solution to use.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is obeyed. This law states that matter cannot be created or destroyed in a chemical reaction. Hence, the number of atoms of each element must be the same on both sides of the equation. In the context of the neutralization reaction:\[ 2 \text{NaOH} (aq) + \text{H}_2\text{C}_4\text{H}_4\text{O}_6 (aq) \to \text{Na}_2\text{C}_4\text{H}_4\text{O}_6 (aq) + 2 \text{H}_2\text{O} (l) \]

The equation is already balanced, showing that two sodium hydroxide molecules react with one molecule of tartaric acid to produce sodium tartrate and water. If an equation is unbalanced, it's essential to adjust the coefficients—the numbers in front of the chemical formulas—to ensure that the number of atoms of each element is equal on both sides. Balancing chemical equations is not only a requirement for proper chemical notation but also forms the basis for stoichiometric calculations.