Problem 46 - General Chemistry: Principles and Modern Applications

Question

Determine the following characteristics of the titration curve for \(20.0 \mathrm{mL}\) of \(0.275 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq})\) titrated with \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) (a) the initial \(\mathrm{pH}\) (b) the volume of \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) at the equivalence point (c) the \(\mathrm{pH}\) at the half-neutralization point (d) the \(\mathrm{pH}\) at the equivalence point

Step-by-Step Solution

Answer

The initial pH of the solution is around 11.23. The volume of 0.325 M HI(aq) required at the equivalence point is approximately 16.92 mL. The pH at the half-neutralization point is around 8.75 and the pH at the equivalence point is approximately 5.27.

Step 1: Calculate Initial pH

The initial pH is calculated using the concentration of the weak base; NH3 is a weak base and thus will form a basic solution when dissolved in water. The expression for the ionization of NH3 in water is given as: NH3 +H2O ↔ NH4+ + OH-. From this, the Kb expression can be formed as \([NH4^+][OH^-]/[NH3]\). Replace known values into Kb expression and find [OH-]. Next, Use the OH- concentration to calculate pOH using the formula -log[OH-] and then find pH by subtracting pOH from 14.

Step 2: Calculate the volume of the titrant at equivalence point

The volume of the strong acid needed to completely neutralize weak base is calculated by applying the equivalence of moles at the equivalence point. According to the titration stoichiometry, 1 mole of HI neutralizes 1 mole of NH3. From this, n(NH3) = V(NH3)*C(NH3) and n(HI) = V(HI)*C(HI). Because at the equivalence point n(HI)=n(NH3), we can express V(HI) as n(NH3)/C(HI). Plug in the values and calculate V(HI).

Step 3: Calculate pH at the half neutralization point

At the half-neutralization point, the pOH is equal to the pKb of the weak base. Hence, we can calculate the pKb of NH3 using the given Kb value in textbooks or chemical literature, then calculate pH, pH = 14 - pKb.

Step 4: Calculate pH at the equivalence point

At the equivalence point, all the weak base has been reacted with the strong acid to produce its conjugate acid (NH4+). This situation represents a solution of a weak acid, so we have to determine pH of this solution. The expression for the ionization of NH4+ in water is given as: NH4+ ↔ NH3 + H3O+. From this, the Ka expression can be formed as [NH3][H3O+]/[NH4+]. Replace the known values into Ka expression and find [H3O+]. Convert H3O+ concentration to pH by using -log[H3O+].

Key Concepts

Initial pH

To understand the initial pH of a solution, consider the nature of the solute. In our case, we have ammonia, NH3, which is a weak base. When NH3 is dissolved in water, it partially ionizes to form hydroxide ions, OH⁻, and ammonium ions, NH4⁺. The chemical equation for this reaction is:NH3 + H2O ↔ NH4⁺ + OH⁻
Using this equation, we can find the equilibrium expression known as the Kb expression: \[K_b = \frac{[NH4^+][OH^-]}{[NH_3]}\]By replacing the known concentrations into this expression, you can solve for the hydroxide ion concentration, [OH⁻]. Once you have this value, you can calculate the pOH using the formula:- \[ \text{pOH} = -\log[OH^-] \]Finally, to get the initial pH, subtract the pOH from 14:- \[ \text{pH} = 14 - \text{pOH} \]This process will give you the initial pH of the ammonia solution before any titration occurs.

Equivalence Point

The equivalence point in a titration is reached when the amount of titrant added is stoichiometrically equivalent to the amount of substance in the solution being titrated. For our ammonia titration with hydroiodic acid, HI, this means the point where all NH3 has reacted with HI to form NH4+.Key points to consider at the equivalence point:

• The moles of acid equal the moles of base.
• The chemical reaction follows the equation: HI + NH3 → NH4I.

To calculate the volume of HI needed to reach the equivalence point, use the mole relationship:\[n(NH_3) = V(NH_3) \times C(NH_3) \]At the equivalence point, the number of moles of HI added equals the moles of NH3 present, and we get:\[V(HI) = \frac{n(NH_3)}{C(HI)}\]By plugging in the known concentrations and solving for the volume, you obtain the exact amount of titrant needed to reach the equivalence point.

Half-neutralization Point

The half-neutralization point occurs halfway to the equivalence point. This is where half of the base, NH3, has been converted to its conjugate acid, NH4+. It's an important concept in titration as at this point, the concentration of the base equals the concentration of its conjugate acid.An essential feature of this point in the titration of a weak base with a strong acid is that:- The pOH equals the pKb of the weak base.Therefore, to find the pH at this half-neutralization point, you utilize the relation:\[\text{pH} = 14 - \text{pKb}\]This provides an easy way to determine the pH at this specific stage without performing extensive calculations.

pKb

The strength of a weak base like ammonia is expressed by its base dissociation constant, Kb. The pKb is simply the negative logarithm of this constant:\[\text{pKb} = -\log(K_b)\]For ammonia, you'll find the Kb value from a Chemistry textbook or reliable chemical data source. Ammonia's Kb is typically around 1.8 x 10⁻⁵.Understanding the pKb helps predict how a base behaves in solution:

• Lower pKb values indicate stronger bases.
• Higher pKb values suggest weaker bases.

In the context of titration, knowing the pKb assists in calculating the pH at various stages, especially at the half-neutralization point where pOH is equal to pKb.

Ka Expression

Once at the equivalence point, the originally present base (NH3) has been completely converted to its conjugate acid (NH4+). Calculating the pH of the solution at this point involves using the Ka of the conjugate acid.The ionization of NH4+ in water is represented by:NH4⁺ ↔ NH3 + H3O⁺From this, the expression for the acid dissociation constant, Ka, is:\[K_a = \frac{[NH3][H3O^+]}{[NH4^+]}\]Since NH4+ is the product at the equivalence point, you can use its concentration to find [H3O⁺]. Substituting into the Ka expression allows the calculation of [H3O⁺], which you then convert to pH using:\[\text{pH} = -\log[H3O^+]\]This approach is crucial for understanding how the formation of NH4+ alters the solution's acidity, making it vital for accurate pH determination at the equivalence point.