Chapter 8: Problem 35
A sample of \(27.0 \mathrm{mL}\) of \(0.310 \mathrm{M} \mathrm{NaOH}\) is titrated with \(0.740 M \mathrm{H}_{2} \mathrm{SO}_{4} .\) How many milliliters of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution are required to reach the end point?
Short Answer
Expert verified
5.65 mL of the Hâ‚‚SOâ‚„ solution is required.
Step by step solution
01
Write the Balanced Chemical Equation
First, write the balanced chemical equation for the reaction between NaOH and Hâ‚‚SOâ‚„. The equation is:\[2 \mathrm{NaOH} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O}\] This shows that 2 moles of NaOH react with 1 mole of Hâ‚‚SOâ‚„.
02
Calculate Moles of NaOH
Use the molarity and volume of the NaOH solution to find the number of moles present. The formula is:\[\text{Moles of NaOH} = \text{Molarity} \times \text{Volume in L}\]Here, volume should be converted to liters:\[27.0 \text{ mL} = 0.0270 \text{ L}\]Thus, moles of NaOH:\[0.310 \, \mathrm{M} \times 0.0270 \, \mathrm{L} = 0.00837 \, \text{moles}\]
03
Use Stoichiometry to Find Moles of Hâ‚‚SOâ‚„
From the balanced equation, we know that 2 moles of NaOH react with 1 mole of Hâ‚‚SOâ‚„. Use this stoichiometric relation to calculate the moles of Hâ‚‚SOâ‚„ needed:\[\text{Moles of H}_{2} \mathrm{SO}_{4} = \frac{1}{2} \times \text{Moles of NaOH} = \frac{1}{2} \times 0.00837 \]This gives:\[0.004185 \, \text{moles of H}_{2} \mathrm{SO}_{4}\]
04
Find Volume of Hâ‚‚SOâ‚„ Solution Needed
Now, use the molarity of the Hâ‚‚SOâ‚„ solution to find the volume needed:\[\text{Volume (L)} = \frac{\text{Moles of } \mathrm{H}_{2} \mathrm{SO}_{4}}{\text{Molarity}} = \frac{0.004185}{0.740}\]This equation provides the volume in liters. Perform the calculation:\[0.00565473 \, \text{L}\]Convert it to milliliters by multiplying by 1000:\[0.00565473 \, \text{L} \times 1000 = 5.65 \, \text{mL}\]
05
Conclusion
From the calculations, you need 5.65 mL of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution to reach the end point of the titration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Balanced Chemical Equation
A balanced chemical equation is essential in any chemical reaction because it represents the conservation of mass. Each side of the equation must have the same number of atoms of each element. In the reaction between sodium hydroxide (NaOH) and sulfuric acid (Hâ‚‚SOâ‚„), the balanced equation is \(2 \mathrm{NaOH} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O}\).
This equation indicates that two moles of NaOH react with one mole of Hâ‚‚SOâ‚„ to produce one mole of sodium sulfate (Naâ‚‚SOâ‚„) and two moles of water. Balanced equations are crucial because they allow chemists to understand the proportions of reactants and products involved. Understanding the stoichiometry, which is derived from the balanced equation, helps to predict how much product can form from a given set of reactants. It ensures that no atoms are lost or gained; they are merely rearranged.
For complex equations, it is vital to practice balancing by meticulously counting the numbers of each type of atom on both sides until they match. This practice helps in tackling more advanced problems in chemistry.
This equation indicates that two moles of NaOH react with one mole of Hâ‚‚SOâ‚„ to produce one mole of sodium sulfate (Naâ‚‚SOâ‚„) and two moles of water. Balanced equations are crucial because they allow chemists to understand the proportions of reactants and products involved. Understanding the stoichiometry, which is derived from the balanced equation, helps to predict how much product can form from a given set of reactants. It ensures that no atoms are lost or gained; they are merely rearranged.
For complex equations, it is vital to practice balancing by meticulously counting the numbers of each type of atom on both sides until they match. This practice helps in tackling more advanced problems in chemistry.
Molarity
Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution, and it's denoted as \(\mathrm{M}\), such as \(0.310 \mathrm{M} \mathrm{NaOH}\). Molarity is calculated using the formula:
For example, in the exercise given, the NaOH solution has a molarity of \(0.310\), which means that there are \(0.310\) moles of NaOH in one liter of solution. When dealing with titrations, it's important to keep track of the volume in liters, which sometimes involves converting from milliliters to liters by dividing by \(1000\). This conversion ensures all units are consistent, making calculations straightforward.
- \(\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)
For example, in the exercise given, the NaOH solution has a molarity of \(0.310\), which means that there are \(0.310\) moles of NaOH in one liter of solution. When dealing with titrations, it's important to keep track of the volume in liters, which sometimes involves converting from milliliters to liters by dividing by \(1000\). This conversion ensures all units are consistent, making calculations straightforward.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. Derived from the Greek words "stoicheion" (element) and "metron" (measure), it plays a crucial role in understanding chemical processes. By using stoichiometry, chemists can predict the amounts of substances consumed and produced in a given reaction.
In the titration problem presented, stoichiometry is crucial to determining how much sulfuric acid (Hâ‚‚SOâ‚„) is needed to neutralize the sodium hydroxide (NaOH) solution. The balanced chemical equation provides the stoichiometric ratios: 2 moles of NaOH react with 1 mole of Hâ‚‚SOâ‚„. By using these ratios, we can convert moles of one substance to moles of another:
In the titration problem presented, stoichiometry is crucial to determining how much sulfuric acid (Hâ‚‚SOâ‚„) is needed to neutralize the sodium hydroxide (NaOH) solution. The balanced chemical equation provides the stoichiometric ratios: 2 moles of NaOH react with 1 mole of Hâ‚‚SOâ‚„. By using these ratios, we can convert moles of one substance to moles of another:
- If you have \(0.00837\) moles of NaOH, the moles of Hâ‚‚SOâ‚„ needed is half of that, because the ratio from the balanced equation is 2:1.
- This leads to \(0.004185\) moles of Hâ‚‚SOâ‚„.