Question

Calculate the number of grams of \(\mathrm{NH}_{4} \mathrm{Cl}\) that must be added to \(2 \ell\) of \(0.1 \mathrm{M} \mathrm{NH}_{3}\) to prepare a solution of \(\mathrm{pH}=11.3\) at \(25^{\circ} \mathrm{C}\). The \(\mathrm{K}_{\mathrm{b}}\) for \(\mathrm{NH}_{3}=1.8 \times 10^{-5}\). The equilibrium equation is: \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}\)

Short answer

Approximately 0.192 grams of NH4Cl must be added to the 2L of 0.1 M NH3 solution to prepare a solution with a pH of 11.3 at 25°C.

Step-by-step solution

Convert pH to pOH and find [OH-] concentration

To find the concentration of OH- ions, we need to convert the given pH to pOH. The relationship between pH and pOH is given by: pOH = 14 - pH Substitute the given pH value. pOH = 14 - 11.3 pOH = 2.7 Now we have the pOH, we can find the [OH-] concentration using the following equation: [OH-] = 10^(-pOH) Substitute the pOH value. [OH-] = 10^(-2.7) [OH-] ≈ 0.0020 M

Use Kb to find [NH4+] concentration

We are given the Kb value of NH3 and the equilibrium equation: NH3 + H2O ⇌ NH4+ + OH-. We can write the equilibrium expression: Kb = {([NH4+][OH-]) / ([NH3])} We know the [OH-] concentration from step 1, and the initial NH3 concentration is given as 0.1 M. The [NH4+] concentration formed can be represented as x. Then we can write the equation: Kb = {[x * 0.0020] / (0.1 - x)} Substitute the given Kb value: \(1.8 \times 10^{-5}\) = {[x * 0.0020] / (0.1 - x)} Now we can solve for x to find the [NH4+] concentration. x ≈ 0.0018 M

Calculate moles and grams of NH4Cl needed

We know the concentration of NH4+ needed, as well as the volume of the solution (2L). We can now find the moles of NH4Cl that need to be added. moles = [NH4+] × volume Substitute the values: moles = 0.0018 M × 2 L moles ≈ 0.0036 mol Now we can find the number of grams of NH4Cl needed by using the molar mass of NH4Cl (53.49 g/mol): grams = moles × molar mass Substitute the values: grams = 0.0036 mol × 53.49 g/mol grams ≈ 0.192 g Therefore, approximately 0.192 grams of NH4Cl must be added to the 2L of 0.1 M NH3 solution to prepare a solution with a pH of 11.3 at 25°C.

Key concepts

Acid-Base Equilibrium

Understanding the acid-base equilibrium is crucial for calculating the pH of solutions. Acid-base equilibrium refers to the state of balance between the concentration of acid and base species in a solution. In a simple case represented by the reaction \[\mathrm{HA} \rightleftarrows \mathrm{H}^+ + \mathrm{A}^-\] where HA is the acid and A- is its conjugate base, equilibrium will be established according to the acid dissociation constant, Ka, defined by the concentration of products over reactants. However, with bases like NH3, we consider the base dissociation constant, Kb, which corresponds to the equilibrium \[\mathrm{NH}_3 + \mathrm{H}_2O \rightleftarrows \mathrm{NH}_4^+ + \mathrm{OH}^-\]The Kb value is essential for calculating how much NH4+ will be formed when NH3 and H2O come to equilibrium. In the above exercise, we use the Kb of NH3 to determine the concentration of NH4+ needed to achieve a specific pH. Higher Kb values indicate stronger bases, which are more likely to accept protons and form the conjugate acid.

pH to pOH Conversion

In chemistry, pH and pOH are measures of the acidity and basicity of a solution respectively. It's important to note that pH refers to the negative logarithm of the hydrogen ion concentration, while pOH refers to the negative logarithm of the hydroxide ion concentration. Their relationship is governed by the basic formula:\[\text{pH} + \text{pOH} = \text{pK}_w\]where pKw is the ionic product of water (\[\text{pK}_w = 14\]at 25°C). To convert from pH to pOH, or vice versa, use the equation above. In the given exercise, pOH was calculated by subtracting the given pH from 14, and from the pOH, the concentration of OH- ions was found using \(10^{-\text{pOH}}\). This step is fundamental in acidic solutions where the pH is given, but the solution's basicity needs to be known for further calculations.

Molar Concentration

Molar concentration, often referred to as molarity, is the number of moles of a solute per liter of solution. It is a critical concept in chemistry because it allows for the quantification of substances in a precise manner, making stoichiometric calculations possible. The unit for molar concentration is moles per liter (M). In the exercise, the molar concentration of NH4+ needed to be calculated in order to determine how many grams of NH4Cl must be added to achieve the desired equilibrium state. By knowing the volume of the solution and the required concentration of NH4+, we simply multiply these two values to find the moles of NH4Cl needed. Finally, by multiplying the moles by the molar mass of NH4Cl, we convert moles to grams, bridging the gap between the abstract concept of moles and the practical measurement in grams.