Problem 89

Question

A \(0.5895-\mathrm{g}\) sample of impure magnesium hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of \(0.2050 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then needs \(19.85 \mathrm{~mL}\) of \(0.1020 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percent by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

Step-by-Step Solution

Verified
Answer
The percent by mass of magnesium hydroxide in the sample is approximately 91.33%.
1Step 1: Write the balanced equations for the reactions
We need the balanced equations for the reactions: 1. Between magnesium hydroxide and HCl: \(Mg(OH)_2(aq) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)\) 2. Between excess HCl and NaOH: \(HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)\)
2Step 2: Calculate the moles of HCl used in both reactions
Moles of HCl reacted with the impure sample = Volume of HCl solution × Molarity of HCl solution \(= 100.0 \, \text{mL} \times 0.2050 \, \text{M} = 0.2050 \times 10^2 \, \mathrm{mmol} = 20.50 \, \mathrm{mmol}\) Moles of excess HCl = Volume of NaOH needed for titration × Molarity of NaOH \(= 19.85 \, \text{mL} \times 0.1020 \, \text{M} = 0.1020 \times 19.85 \, \mathrm{mmol} = 2.0257 \, \mathrm{mmol}\)
3Step 3: Calculate the moles of HCl that reacted with magnesium hydroxide
Moles of HCl that reacted with magnesium hydroxide = Total moles of HCl - Moles of excess HCl \(= 20.50 \, \mathrm{mmol} - 2.0257 \, \mathrm{mmol} = 18.4743 \, \mathrm{mmol}\)
4Step 4: Calculate the moles of magnesium hydroxide
Using the balanced equation of reaction between magnesium hydroxide and HCl, we know that 1 mole of magnesium hydroxide reacts with 2 moles of HCl. Thus: Moles of magnesium hydroxide = Moles of HCl reacting with magnesium hydroxide / 2 \(= 18.4743 \, \mathrm{mmol} / 2 = 9.23715 \, \mathrm{mmol}\)
5Step 5: Calculate the mass of magnesium hydroxide
Now, calculate the mass of magnesium hydroxide: Mass of magnesium hydroxide = Moles of magnesium hydroxide × Molar mass of magnesium hydroxide \(= 9.23715 \, \mathrm{mmol} \times 58.32 \, \mathrm{g\,mol^{-1}} = 538.383 \, \mathrm{mg}\)
6Step 6: Calculate the percentage of magnesium hydroxide in the sample
Finally, calculate the percent by mass of magnesium hydroxide in the sample: Percentage of magnesium hydroxide = (Mass of magnesium hydroxide / Mass of the sample) × 100% \(= (538.383 \, \mathrm{mg} / 589.5 \, \mathrm{mg}) \times 100\% = 91.33\%\) The percent by mass of magnesium hydroxide in the sample is approximately 91.33%.

Key Concepts

StoichiometryMolarityNeutralization Reaction
Stoichiometry
Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. Understanding stoichiometry allows you to predict how much product will be formed from given reactants or how much of a reagent is required to react completely with another reagent. This involves using balanced chemical equations to determine the mole relationships between substances.
For instance, in the reaction between magnesium hydroxide and hydrochloric acid provided in the exercise, the balanced equation is \(Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O\). This tells us that one mole of magnesium hydroxide reacts with two moles of hydrochloric acid. By understanding this stoichiometric relationship, we can calculate the amount of magnesium hydroxide that reacted by halving the moles of HCl that participated in the reaction.
  • Begin by finding the moles of all reactants based on the volumes and molarities provided.
  • Identify the stoichiometric ratios from the balanced equation.
  • Use these ratios to calculate unknown quantities, such as moles of magnesium hydroxide in this case.
Mastery of stoichiometry is crucial for solving complex chemical reactions and determining the yield of reactions in a laboratory setting.
Molarity
Molarity is a measure of concentration that expresses the number of moles of a solute present in a liter of solution. It is denoted by the symbol "M" and is formulated as \(M = \frac{moles \ of \ solute}{liters \ of \ solution}\). This concept is vital in chemistry as it helps determine how solutions will react when mixed.
In our example, the molarity of the HCl solution was provided as 0.2050 M, which denotes that 0.2050 moles of HCl are present in each liter of solution. Similarly, the NaOH solution used had a molarity of 0.1020 M. Although the volumes used were not in liters, converting milliliters to liters by dividing by 1000 helps in using the molarity directly for calculations.
  • To find the moles of a reactant, multiply the molarity by the volume (in liters) used in the reaction.
  • This allows you to determine how much of the substance is available to react.
  • Molarity is crucial for titration calculations, where precise concentration measurements are required to determine unknown quantities.
Understanding molarity and its application in calculating the amounts of substances in reactions is a staple in chemistry studies.
Neutralization Reaction
A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. This process is central to many analytical methods in chemistry, particularly in titrations where it helps determine the concentration of unknown solutions.
The exercise involves two such reactions. First, the reaction of magnesium hydroxide, a base, with HCl, an acid, is an example of a neutralization reaction. The second occurs when excess HCl is titrated with NaOH, another base, to find how much acid remains unreacted. Both follow the pattern of an acid-base reaction: \(HCl + NaOH \rightarrow NaCl + H_2O\).
  • In neutralization, equal moles of acid and base make water and a salt.
  • The procedure determines the point of neutralization, known as the equivalence point.
  • Titration data, such as the volume of NaOH used, helps calculate the excess or stoichiometric amounts of reactants involved.
Understanding neutralization reactions is key for performing accurate titrations and determining the properties of the substances involved.