Question

An acid solution is \(0.100 \mathrm{M}\) in \(\mathrm{HCl}\) and \(0.200 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) What volume of a \(0.150 \mathrm{M}\) KOH solution would completely neutralize all the acid in \(500.0 \mathrm{~mL}\) of this solution?

Short answer

1667 mL of a 0.150 M KOH solution is needed to neutralize the acid.

Step-by-step solution

Write the chemical equations

Before calculating the volume of KOH solution needed, we need to write the balanced chemical equations for the neutralization reactions:- For HCl: \( \text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O} \) - For \( \text{H}_2\text{SO}_4 \): \( \text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O} \)

Calculate moles of HCl and \(H_2SO_4\)

The number of moles of each acid in 500 mL of solution can be calculated using their molarities (M = moles/L): - Moles of HCl: \( 0.100 \, \text{M} \times 0.500 \, \text{L} = 0.050 \, \text{moles of HCl} \) - Moles of \(H_2SO_4\): \( 0.200 \, \text{M} \times 0.500 \, \text{L} = 0.100 \, \text{moles of } H_2SO_4 \)

Calculate total moles of KOH needed

Using the balanced equations, we determine the total moles of KOH needed to neutralize the acids:- Moles of KOH needed for HCl: \( 1:1 \) ratio, so \( 0.050 \, \text{moles} \) - Moles of KOH needed for \(H_2SO_4\): \( 2:1 \) ratio, so \( 2 \times 0.100 \, \text{moles} = 0.200 \, \text{moles} \) Total moles of KOH = \( 0.050 + 0.200 = 0.250 \, \text{moles} \)

Calculate the volume of KOH solution

Knowing the molarity of KOH and the total moles needed, we can find the volume: \( \frac{\text{moles}}{\text{Molarity}} = \text{Volume in L} \) Therefore, \( \frac{0.250 \, \text{moles}}{0.150 \, \text{M}} = 1.667 \, \text{L} = 1667 \, \text{mL} \) of KOH solution.

Key concepts

Chemical Equations

Chemical equations are representations of chemical reactions, with the reactants on the left and the products on the right, separated by an arrow indicating the direction of the reaction. For the neutralization reaction between hydrochloric acid (HCl) and potassium hydroxide (KOH), we write:
\[ \text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
Similarly, for sulfuric acid (\( \text{H}_2\text{SO}_4 \)) reacting with KOH, the balanced equation is:
\[ \text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O} \]
Each equation is balanced to comply with the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means the number of atoms of each element must be the same on both sides of the equation.

Molar Concentration

Molar concentration, often referred to as molarity, is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute dissolved in one liter of solution. It is represented by the letter 'M' and is expressed in units of moles per liter (mol/L).
For example, a hydrochloric acid solution with a molarity of \(0.100\, \text{M}\) contains \(0.100\) moles of HCl in every liter of solution.
To find the number of moles of a solute in a given volume of solution, you multiply the molarity by the volume (in liters), like this:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (L)} \]
This simple relation is fundamental in calculating the amounts of reactants and products in a chemical reaction.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using the balanced chemical equations to calculate the masses, volumes, and number of moles of reactants and products.
For a neutralization reaction, stoichiometry allows us to calculate the exact amount of an acid that a given quantity of base can neutralize, and vice versa. We use the coefficients of the balanced equation to determine the mole ratio between reactants and products. For instance, the reaction between HCl and KOH has a \(1:1\) mole ratio. Conversely, the reaction with \( \text{H}_2\text{SO}_4 \) has a \(1:2\) mole ratio with KOH, meaning it takes two moles of KOH to neutralize one mole of \( \text{H}_2\text{SO}_4 \).
By applying stoichiometry, we can calculate the precise volume of KOH needed to neutralize a known volume and molarity of an acid solution.

Acid-Base Reactions

Acid-base reactions, also known as neutralization reactions, involve the reaction of an acid with a base to produce a salt and water. These reactions are exothermic, meaning they release heat.
In our exercise, we deal with two acids: HCl, a strong acid; and \( \text{H}_2\text{SO}_4 \), also a strong acid. KOH, the base, is a strong base. The acids donate protons (\( \text{H}^+ \)), which are accepted by the hydroxide ions (\( \text{OH}^- \)) from the base, forming water. The remaining ions from the acid and base combine to form a salt.
Understanding the nuances of acid-base chemistry is crucial for various applications, including titration experiments in analytical chemistry and the manufacturing of numerous household products.