Question

Suppose that \(250.0 \mathrm{~mL}\) of a basic solution is \(0.100 \mathrm{M}\) in \(\mathrm{KOH}\). What volume of a \(0.200 \mathrm{MHCl}\) solution would be required to completely neutralize the basic solution?

Short answer

125 mL of 0.200 M HCl is required to neutralize 250.0 mL of 0.100 M KOH.

Step-by-step solution

Write the neutralization reaction

First, write the balanced chemical equation for the neutralization reaction between KOH and HCl: KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l).The reaction shows that one mole of KOH reacts with one mole of HCl to produce one mole of KCl and one mole of water.

Calculate moles of KOH

Using the concentration and volume of KOH, calculate the number of moles of KOH present: Moles of KOH = concentration of KOH (M) * volume of KOH (L).Moles of KOH = 0.100 M * 0.250 L = 0.025 moles.

Determine moles of HCl needed

From the balanced equation, we know that 1 mole of KOH reacts with 1 mole of HCl. Therefore, the number of moles of HCl needed is equal to the number of moles of KOH.Moles of HCl needed = moles of KOH = 0.025 moles.

Calculate volume of HCl required

Now, use the concentration of HCl to find the volume needed to provide 0.025 moles of HCl:Volume of HCl (L) = moles of HCl / concentration of HCl (M).Volume of HCl = 0.025 moles / 0.200 M = 0.125 L.

Convert volume from liters to milliliters

Finally, convert the volume of HCl from liters to milliliters by multiplying by 1,000 mL/L (since 1 L = 1,000 mL):Volume of HCl (mL) = 0.125 L * 1,000 mL/L = 125 mL.The volume of 0.200 M HCl solution required to neutralize 250.0 mL of 0.100 M KOH is 125 mL.

Key concepts

Understanding Stoichiometry in Neutralization Reactions

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the context of a neutralization reaction, such as the one between potassium hydroxide (KOH) and hydrochloric acid (HCl), stoichiometry allows us to predict the volume of acid needed to neutralize a given volume of base.

To apply stoichiometry, we first need to write a balanced chemical equation. The neutralization reaction between KOH and HCl can be written as:
\[ \text{KOH(aq)} + \text{HCl(aq)} \rightarrow \text{KCl(aq)} + \text{H}_2\text{O(l)} \]
From this balanced equation, we see that the molar ratio of KOH to HCl is 1:1. This means one mole of KOH reacts with one mole of HCl to completely neutralize each other. Using the provided concentrations and volumes, we calculate the moles of KOH and then use the stoichiometric ratio to find the moles of HCl required for neutralization.

Molarity Calculations for Chemical Reactions

Molarity, also known as molar concentration, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute dissolved in one liter of solution. This concept is crucial for dilution and concentration calculations in chemistry, particularly in reactions like neutralizations.

In our example, the molarity of KOH is given as 0.100 M, which means there are 0.100 moles of KOH in every liter of solution. To find the moles present in 250.0 mL of this solution, we convert the volume to liters (0.250 L) and use the formula:
\[ \text{Moles of KOH} = 0.100 \text{ M} \times 0.250 \text{ L} = 0.025 \text{ moles} \]
The volume of HCl required to neutralize the KOH can then be calculated using the molarity of the HCl solution and the number of moles needed. Molarity calculations help ensure that reactions proceed to completion by providing the right proportions of reactants.

Acid-Base Reactions and Neutralization

Acid-base reactions are a type of chemical reaction that involves the transfer of protons (H+) from the acid to the base. Neutralization is a specific kind of acid-base reaction where an acid reacts with a base to produce a salt and usually water. Understanding these reactions is crucial for many fields, including titrations in analytical chemistry.

The example reaction between KOH and HCl represents a straightforward neutralization reaction:
\[ \text{KOH(aq)} + \text{HCl(aq)} \rightarrow \text{KCl(aq)} + \text{H}_2\text{O(l)} \]
In every neutralization reaction, the acid (HCl, in this case) donates a proton to the base (KOH), forming water and a salt (KCl). Knowing this, students can grasp the one-to-one relationship necessary for stoichiometry and understand why equivalent moles of acid and base are required for neutralization in our example. When the ratios are right, they result in a neutral solution, important for processes in industrial chemistry to environmental studies.