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What mass of \(\mathrm{NaOH}\) is required to neutralize \(1.8 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{HCl}\) ?

Short Answer

Expert verified
144 grams of NaOH are required to neutralize 1.8 L of 2.0 M HCl.

Step by step solution

01

Calculate the moles of HCl

The formula for molarity \(M\) is defined as moles of solute (\(n\)) divided by the volume of the solution in litres (\(V\)): \(M = n / V\). We can rearrange this to find the number of moles: \(n = M \cdot V\). Given that the HCl solution has a molarity of 2.0 M and the volume is 1.8 L, we get: \(n_{HCl} = 2.0 M \cdot 1.8 L = 3.6 moles\).
02

Apply stoichiometric relations

NaOH and HCl react in a 1:1 ratio to give H2O and NaCl, according to the reaction: NaOH + HCl -> H2O + NaCl. Therefore, the moles of NaOH required to completely neutralize the HCl are equal to the moles of HCl. So we need \(n_{NaOH} = 3.6 moles\).
03

Calculate the mass of NaOH

Now, we can find the mass of NaOH needed using the formula relating mass (m), moles (n), and molar mass (M): \(m = n \cdot M\). The molar mass of sodium hydroxide, \(M_{NaOH}\), is approximately 40 g/mol. Thus, the mass of NaOH required is: \(m_{NaOH} = n_{NaOH} \cdot M_{NaOH} = 3.6 moles \cdot 40 g/mol = 144 g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Neutralization reaction
A neutralization reaction is a chemical process where an acid and a base react with each other to form water and salt. In our example, hydrochloric acid (HCl), which is a strong acid, reacts with sodium hydroxide (NaOH), a strong base. This type of reaction is particularly straightforward because both reactants interact in a 1:1 mole ratio to completely neutralize each other.
Neutralization reactions are important in various real-world applications, such as in industrial processes or environmental science, where controlling pH is crucial. By forming water and a salt, these reactions help us manage acidity and basicity in different solutions.
  • For instance, the equation is: \[\text{NaOH} + \text{HCl} \rightarrow \text{H}_2\text{O} + \text{NaCl}\]
  • Since the equation is balanced, each mole of NaOH neutralizes one mole of HCl.
This balance helps us calculate exactly how much base is needed to neutralize a given amount of acid.
Molarity
Molarity is a way to express the concentration of a solution. It is the number of moles of solute (the substance dissolved) per liter of solution. When we say a solution has a molarity of 2.0 M, it means there are 2 moles of the solute in every liter of solution.
Understanding molarity is crucial for performing calculations that involve chemical solutions. In our exercise, we use molarity to calculate the moles of hydrochloric acid present in the solution.
  • Molarity (M) is given by the formula:\[M = \frac{n}{V}\]
  • Here, \( n \) is the number of moles of solute, and \( V \) is the volume in liters.
  • We rearrange the formula to find the moles: \[n = M \cdot V\]
This step allows us to understand and calculate how much of a particular substance is present in solution, enabling precise reactions and predictions.
Moles calculation
Moles are a fundamental concept in chemistry. They help us quantify the amount of substance we have. In stoichiometry, moles are used to relate quantities of reactants and products in chemical reactions.
Calculating moles is essential for understanding how much of a substance will react, and in our example, it helps determine how much NaOH is needed to neutralize HCl.
To calculate the mass of a compound from moles, we use the formula:
  • \[m = n \cdot M\]
Here, \(m\) is mass, \(n\) is the number of moles, and \(M\) is molar mass.
In our scenario, we calculated that we need 3.6 moles of NaOH to match the 3.6 moles of HCl. Knowing the molar mass of NaOH is 40 g/mol allows us to find the total mass required, which is 144 grams.
Each step in these calculations helps build a clearer picture of the chemical reaction at play and ensures accuracy in experimental and practical applications.

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Most popular questions from this chapter

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