Question

What mass of \(\mathrm{NaOH}(s)\) must be added to \(1.0 \mathrm{~L}\) of \(0.050 \mathrm{M} \mathrm{NH}_{3}\) to ensure that the percent ionization of \(\mathrm{NH}_{3}\) is no greater than \(0.0010 \%\) ? Assume no volume change on addition of \(\mathrm{NaOH}\).

Short answer

To ensure that the percent ionization of $\mathrm{NH}_{3}$ is no greater than 0.0010% when mixed with $1.0 \mathrm{~L}$ of $0.050 \mathrm{M} \mathrm{NH}_{3}$, \(2.0 \times 10^{-5}\) grams of $\mathrm{NaOH}$ must be added, based on stoichiometric calculations and the given ionization percentage.

Step-by-step solution

Write the reaction between NH3 and NaOH

Write the balanced chemical equation for the reaction between ammonia (NH3) and sodium hydroxide (NaOH): NH3(aq) + NaOH(aq) → NH4+(aq) + OH-(aq) + Na+(aq)

Calculate the initial moles of NH3

Given the volume of the solution 1.0 L and the concentration of NH3 is 0.050 M, calculate the initial moles of NH3: Moles of NH3 = Concentration × Volume Moles of NH3 = 0.050 mol/L × 1.0 L = 0.050 mol

Calculate the moles of NH3 ionized

Since the percent ionization of NH3 should be no greater than 0.0010%, calculate the moles of NH3 that are ionized: Moles of ionized NH3 = Initial moles of NH3 × (percent ionization / 100) Moles of ionized NH3 = 0.050 mol × (0.0010 / 100) = 5.0 × 10^(-7) mol

Determine the moles of NaOH required

According to the balanced chemical equation, the stoichiometric ratio between NH3 and NaOH is 1:1, so the moles of NaOH required will be equal to that of moles of ionized NH3. Moles of NaOH = Moles of ionized NH3 = 5.0 × 10^(-7) mol

Calculate the mass of NaOH

Using the molar mass of NaOH (22.99 g/mol for Na, 15.999 g/mol for O, and 1.008 g/mol for H), calculate the mass of NaOH required: Mass of NaOH = Moles of NaOH × Molar mass of NaOH Mass of NaOH = (5.0 × 10^(-7) mol) × (22.99 + 15.999 + 1.008) g/mol Mass of NaOH = 5.0 × 10^(-7) mol × 40.0 g/mol = 2.0 × 10^(-5) g Therefore, 2.0 × 10^(-5) g of NaOH must be added to 1.0 L of 0.050 M NH3 to ensure that the percent ionization of NH3 is no greater than 0.0010%.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry focused on calculating the quantities of reactants and products in a chemical reaction. When dealing with stoichiometry, it is important to consider the balanced chemical equation for the reaction. This equation provides the molar ratios of reactants and products which are essential for solving problems.

• For example, in the reaction given, NH3 reacts with NaOH to form NH4+, OH-, and Na+. The stoichiometric coefficients show a 1:1 ratio between NH3 and NaOH.
• This means if you know how much NH3 reacts, you can determine the amount of NaOH needed just by looking at the coefficients.

The step-by-step solution to the problem relies on this stoichiometric relationship to calculate how much NaOH is required to maintain a given ionization level of NH3.

Percent Ionization

Percent ionization is a measure of the extent to which a substance dissociates in solution into ions. It's particularly important in weak acids and bases. In this exercise, NH3 is partially ionized in the solution when contacted with NaOH.
To calculate percent ionization, use the formula:
\[ \text{Percent Ionization} = \frac{\text{moles ionized}}{\text{initial moles}} \times 100 \% \]

• In the given problem, NH3 has to be ionized less than or equal to 0.0010%.
• The initial moles of NH3 are calculated, and then, using the percent ionization, you discover how many moles of NH3 are involved in the reaction with NaOH.

Maintaining such a small percent ionization implies controlling the reactant quantities carefully, which showcases the precision stoichiometry seeks to achieve.

Chemical Equations

Chemical equations represent chemical reactions in terms of symbols and formulas. They are essential for visualizing and calculating the changes in a chemical reaction.

• A properly balanced chemical equation ensures the law of conservation of mass is followed, meaning the same number of each type of atom is present before and after the reaction.
• In this example, NH3 is reacting with NaOH to yield NH4+, OH-, and Na+.

Understanding chemical equations is critical because it forms the basis for stoichiometry and subsequent calculations. They help you determine how much of each reactant is necessary and what products will form, as seen in the problem solved.