Question

Calculate the mass of \(\mathrm{HONH}_{2}\) required to dissolve in enough water to make 250.0 \(\mathrm{mL}\) of solution having a pH of 10.00\(\left(K_{\mathrm{b}}\right.\) \(=1.1 \times 10^{-8} )\)

Short answer

The mass of HONH₂ required to make a 250.0 mL solution with a pH of 10.00 is approximately 0.0009 g (rounded to four decimal places).

Step-by-step solution

Calculate pOH from given pH

The relationship between pH and pOH is: pH + pOH = 14 We are given pH = 10, so: pOH = 14 - 10 = 4

Calculate the concentration of OH⁻ ions

We can determine the concentration of OH⁻ ions from the pOH using the following formula: \[ OH^⁻ = 10^{-pOH} \] \[ OH^⁻ = 10^{-4} = 1 \times 10^{-4} \, M\]

Calculate the concentration of HONH₂ using the Kb expression

Let's represent the dissociation of HONH₂ as: HONH₂ (aq) + H₂O (l) ⇌ HONH⁻ (aq) + OH⁻ (aq) The Kb expression can be written as: \[ K_b = \frac{[HONH⁻][OH-]}{[HONH_2]} \] We know the Kb value and the concentration of OH⁻. Assuming x mol/L is the concentration of HONH₂ in equilibrium, we have: \[ 1.1 \times 10^{-8} = \frac{x(1 \times 10^{-4})}{x - 1 \times 10^{-4}} \] Now we can solve for x, which represents the equilibrium concentration of HONH₂. The concentration of HONH₂ is very small compared to the initial concentration; hence x - 1 × 10^(-4) ≈ x \[ 1.1 \times 10^{-8} = \frac{x \times 1 \times 10^{-4}}{x} \] \[ x = 1.1 \times 10^{-4} \, M\]

Determine the mass of HONH₂

Now that we have the concentration of HONH₂, we can calculate the mass required. We know that the volume of the solution is 250.0 mL. First, let's convert the volume into liters: Volume = 250.0 mL × (1 L / 1000 mL) = 0.250 L Next, we can calculate the moles of HONH₂ using the concentration: Moles of HONH₂ = concentration × volume Moles of HONH₂ = (1.1 × 10⁻⁴ M) × (0.250 L) = 2.75 × 10⁻⁵ mol Finally, we can determine the mass by multiplying the moles by the molecular weight of HONH₂ (Molar mass of HONH₂ = 1(H) + 15(N) + 16(O) + 1(H) + 1(N) + 2(H) = 33 g/mol): Mass of HONH₂ = moles × molecular weight Mass of HONH₂ = (2.75 × 10⁻⁵ mol) × (33 g/mol) ≈ 0.0009075 g Therefore, the mass of HONH₂ required to make a 250.0 mL solution with a pH of 10.00 is approximately 0.0009 g (rounded to four decimal places).

Key concepts

pH and pOH Calculation

Understanding the relationship between pH and pOH is crucial when dealing with acid-base chemistry. pH measures the acidity of a solution, while pOH indicates its basicity. These two are connected through a simple equation:

• pH + pOH = 14

This equation lets us determine the pOH if the pH is known, and vice versa. In this exercise, the pH is given as 10. Therefore, the pOH is calculated as:

• pOH = 14 - 10 = 4

It means the solution is more basic because the pH is greater than 7.

Why is pOH Important?

The pOH provides insight into the concentration of hydroxide ions often denoted as OH⁻. A low pOH indicates a high concentration of OH⁻ ions, characterizing the solution's basic nature. Understanding both pH and pOH helps predict and control chemical reactions, especially in equilibrium scenarios.

Base Dissociation Constant (Kb)

The Base Dissociation Constant, represented as Kb, is vital for understanding the strength of a base. It indicates how well a base dissociates in water, forming hydroxide ions. In this scenario, we deal with the base dissociation of hydroxylamine, HONH₂.

• HONH₂ (aq) + H₂O (l) ⇌ HONH⁻ (aq) + OH⁻ (aq)

The equilibrium expression for this reaction is:

• \( K_b = \frac{[HONH⁻][OH⁻]}{[HONH₂]} \)

Given the value of Kb (1.1 × 10⁻⁸), we can estimate the concentration of the dissociating base in equilibrium conditions.

Using Kb in Calculations

The Kb helps calculate unknown concentrations if the concentrations of other species are known. Here, it helped determine the concentration of HONH₂ necessary to achieve a desired concentration of OH⁻ ions in solution. A small Kb value indicates a weak base, which only partially dissociates in water.

Molarity and Concentration

Molarity is an expression of concentration, calculated as moles of solute per liter of solution (mol/L). In equilibrium problems, molarity helps describe how concentrated a solution is. For instance, in our exercise:

• The concentration of OH⁻ ions was calculated as 1 × 10⁻⁴ M using the pOH.

This value represents the molarity of OH⁻ ions in the solution.

Calculating Molarity and Concentration

Molarity allows easy conversion between volume and moles, crucial when preparing solutions of a specific concentration. To find the concentration of HONH₂, we assumed that at equilibrium, the concentration slightly above the dissociated OH⁻ can give us

• the concentration of HONH₂ needed to reach the equilibrium concentration.

Eventually, understanding molarity helps understand many chemical processes and reactions in a quantitative manner.

Stoichiometry Calculation

Stoichiometry involves quantifying relationships between reactants and products in a chemical reaction. In equilibrium calculations like this one, stoichiometry is used to calculate the mass of a compound required to produce a particular molarity.

• Once the concentration of HONH₂ was found, it was used to calculate the moles of HONH₂ needed for our solution volume.

The steps involved were:

• Moles of HONH₂ = concentration × volume
• From the moles, determining the mass required using the molar mass of HONH₂

The Importance of Stoichiometry

By knowing the moles and using the molar mass to find the mass, stoichiometry turns abstract concentration values into tangible quantities. By understanding this, you can calculate the exact amount of substance required, ensuring precise experimental results.