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

In order to keep the percent ionization of \(NH_3\) no greater than 0.0010%, we must add a mass of approximately \(0.000020 \mathrm{~g}\) of solid \(NaOH\) to the 1.0 L solution of 0.050 M \(NH_3\).

Step-by-step solution

- Determine the concentration of NH3 after ionization

Given that the percent ionization of NH3 is no greater than 0.0010%, first, we need to find out the concentration of NH3 after ionization. The initial concentration of NH3 is 0.050 M. Percent ionization = (Amount ionized / Initial amount) × 100 Let x be the amount ionized, 0.0010% = (x / 0.050) × 100 By solving this equation, we will get the value of x.

- Calculate the amount ionized

To calculate x, first divide by 100 on both sides of the equation: 0.0010%/100 = (x / 0.050) After that, multiply both sides by 0.050 to isolate x: 0.000010 × 0.050 = x Now, calculate x by multiplying: x = 0.0000005 mol So, 0.0000005 mol of NH3 has been ionized in 1.0 L of the solution.

- Calculate the moles of NaOH needed via stoichiometry

The balanced chemical equation for the reaction between NH3 and NaOH is: NH3(aq) + NaOH(aq) → Na+(aq)+ NH4OH(aq) From the balanced equation, we can see that one mole of NH3 reacts with one mole of NaOH. Therefore, the moles of NaOH required for the reaction will be equal to the moles of NH3 ionized. Moles of NaOH required = Moles of NH3 ionized = 0.0000005 mol

- Calculate the mass of NaOH needed

Now that we have the moles of NaOH needed for the reaction, we can calculate the mass of NaOH to be added. The molar mass of NaOH is 22.99 g/mol (for Na) + 15.999 g/mol (for O) + 1.008 g/mol (for H) = 39.997 g/mol. Mass of NaOH = Moles of NaOH × Molar mass of NaOH = 0.0000005 mol × 39.997 g/mol Now, calculate the mass: Mass of NaOH = 0.0000199985 g

- Round the answer to an appropriate number of significant figures

The given values in the problem have 2 significant figures each (1.0 L and 0.050 M), so our final answer should have 2 significant figures as well. Final answer: Mass of NaOH to be added = 0.000020 g (rounded to 2 significant figures)

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry that involves using balanced chemical equations to calculate the relative quantities of reactants and products involved in a chemical reaction. To master stoichiometry, one must first understand the concept of the mole, which is the fundamental unit of measurement for the amount of substance in chemistry.

Considering the exercise where we aim to limit the percent ionization of ammonia (NH3) upon the addition of sodium hydroxide (NaOH), stoichiometry plays a crucial role. In this case, the balanced chemical equation shows a straightforward 1:1 molar ratio between NH3 and NaOH. This means that for every mole of NH3 that ionizes, one mole of NaOH is required to ensure the percent ionization does not exceed the specified limit. By understanding the stoichiometric relationship in the chemical equation, we can confidently calculate the amount of NaOH needed to be added.

When approaching stoichiometry problems, start by writing down the balanced chemical equation. Next, use the mole ratio from the equation to determine how many moles of one substance react with moles of another. The stepwise solution in the exercise perfectly demonstrates this approach by matching the moles of ionized NH3 directly to the moles of NaOH needed to maintain the desired percent ionization.

Molar Mass Calculation

The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's a crucial concept for stoichiometry, as it allows for the conversion between moles and grams. To calculate molar mass, sum the atomic masses of all the atoms in a molecule. Each element's atomic mass can be found on the periodic table.

For instance, the molar mass of sodium hydroxide (NaOH) used in our example is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H). Breaking it down: Na has a molar mass of 22.99 g/mol, O is 15.999 g/mol, and H is 1.008 g/mol. Adding these together yields NaOH's molar mass of 39.997 g/mol. The calculation of molar mass enables us to determine the exact mass of NaOH required to add to the ammonia solution, which is essential to controlling the percent ionization.

By utilizing this concept, students can solve the exercise by converting moles of NaOH needed into grams. It's a pivotal step in the procedure linking the theoretical reaction stoichiometry to practical, measurable quantities in the laboratory or industrial setting.

Weak Base Ammonia

Ammonia (NH3) is a common example of a weak base in chemistry. A weak base does not fully dissociate into its ions in an aqueous solution, which means its ionization is incomplete in water. The percent ionization refers to the fraction of the initial amount of the base that has ionized, presented as a percentage. This property is crucial when studying the behavior of weak bases like ammonia in solution.

To control the percent ionization of ammonia in our exercise, the addition of a strong base, sodium hydroxide (NaOH), is necessary. NaOH is a strong base that completely ionizes in solution, thereby shifting the equilibrium of the NH3 ionization reaction. By careful calculation of the amount of NaOH needed, we can keep the ionization level of NH3 within the desired range.

This method offers a practical application of the Le Chatelier's principle, which states that if a dynamic equilibrium is disturbed, the system will adjust to counteract the disturbance and reestablish equilibrium. By providing just enough NaOH to react with the ionized NH3, we effectively use chemistry's governing principles to manipulate the extent of a reaction in a predictable way.