Question

A $0.4324\mathrm{g}$ sample of a potassium hydroxide-lithium hydroxide mixture requires $28.28\mathrm{mL}$ of $0.3520\mathrm{M}\mathrm{HCl}$ for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?

Short answer

0 %

Step-by-step solution

Calculate moles of HCl

Molarity is defined as the number of moles of solute per litre of solution. So, the amount of moles of HCl can be calculated by multiplying the molarity by the volume (in litres). Moles of $HCl=0.3520M\times 0.02828L=0.00995456mol$.

Determine moles of OH-

In a titration to the equivalence point with HCl, the moles of hydronium ion (H+) from HCl will equal the moles of hydroxide ion (OH-) from the base there was originally in the mixture. Hence, moles of OH- = 0.00995456 mol.

Calculate mass of each hydroxide

Potassium hydroxide (KOH) and lithium hydroxide (LiOH) both provide one mole of OH- per mole. We can now write an equation representing the total OH- from KOH and LiOH as: Mass = moles OH- \times molar mass = 0.00995456 mol \times 56.1 g/mol (for KOH) + 0.00995456 mol \times 23.95 g/mol (for LiOH) = 0.5586 g + 0.23833 g = 0.79693 g. This is the maximum total mass of KOH and LiOH that could be in the sample.

Calculate mass percent of LiOH

Knowing maximum total mass of KOH and LiOH that could be in the sample and the actual mass of the sample, we can now subtract to get the mass of the unknown component. Subtracting from the total mass yields an estimate of the mass of LiOH: Mass of LiOH = total mass of sample - mass of KOH = 0.4324 g - 0.5586 g = -0.1262 g. Since the calculated mass of LiOH is negative, this suggests that the sample doesn't contain LiOH. Therefore, the percent by mass of LiOH in the mixture would be 0.

Key concepts

Understanding Molarity

Molarity is a key concept in chemistry used to describe the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution. Molarity is crucial in titration chemistry as it helps determine the amount of reactant needed to reach the equivalence point.
Here is a simple way to think about it:

• A solute is the substance dissolved in a solution, like sugar in water.
• The number of moles tells us how many molecules or atoms we have, not just how much it weighs.

The formula is usually written as: $$M=\frac{n}{V}$$ Where:

• $M$ is the molarity.
• $n$ is the number of moles of the solute.
• $V$ is the volume of the solution in liters.

This equation allows chemists to easily calculate the amount of chemicals or the concentration needed in a reaction. For example, in the original exercise, molarity helps determine the moles of hydrochloric acid (HCl) used.

Significance of Equivalence Point

The equivalence point in titration is when the amount of titrant added is exactly enough to neutralize the analyte solution. This is vital in determining the endpoint of a titration, where the reactants perfectly balance each other.
In the given exercise, the equivalence point is achieved when the moles of hydrochloric acid ( HCl ) added equal the amount of hydroxide ions ( OH^- ) in the base mixture. This confirms that all the reactive species have reacted completely.
Reaching this point means:

• No excess reactant is left in the solution.
• The measured volume corresponds precisely to the stoichiometry of the reaction.

The concept is used to calculate unknown concentrations or to determine the composition of a mixture, like the lithium and potassium hydroxides in the sample. Equivalence points can be indicated using indicators that change color or through more precise methods like pH meters.

Role of Hydroxide Ions in Neutralization

Hydroxide ions ( OH^- ) are basic anions central to neutralization reactions in titration chemistry. During titration, these ions react with hydrogen ions ( H^+ ) from acids to form water, a neutral compound. The basic principle of titration is based on this neutralization.
Consider these points about hydroxide ions:

• They are derived from bases like potassium hydroxide (KOH) and lithium hydroxide (LiOH).
• Each hydroxide molecule contributes one OH^- ion.

In titration exercises like the one provided, understanding how many hydroxide ions are present allows us to match them with equivalent hydrogen ions. This process tells us how much base or acid was originally in the sample. As OH^- ions react with H^+ ions from HCl , their complete neutralization indicates that all the acid or base has reacted, precisely at the equivalence point. This concept is key in determining the concentrations of solutions and mixtures in chemistry.