Question

Repeat the procedure in Exercise $67,$ but for the titration of 25.0 $\mathrm{mL}$ of 0.100$M{\mathrm{NH}}_3(K_{\mathrm{b}}=1.8\times {10}^{-5})$ with 0.100 $\mathrm{M}$ $\mathrm{HCl}.$

Short answer

In the titration of 25.0 mL of 0.100 M NH3 with 0.100 M HCl, we need to calculate the pH of the solution at different stages of the titration. First, write the chemical reaction between NH3 and HCl. Next, calculate the moles of NH3 and HCl at each stage (before adding any HCl, halfway point, equivalence point, and past the equivalence point). Finally, use the moles of NH3, HCl, and NH4+ to calculate the pH at each stage, using $K_b$, Henderson-Hasselbalch equation, and dissociation constants as needed.

Step-by-step solution

Write the chemical reaction and initial conditions

The chemical reaction between NH3 and HCl can be written as: $N{H}_3(aq)+HCl(aq)\to N{H}_4^{+}(aq)+C{l}^{-}(aq)$ Initial conditions are as follows: Volume of $N{H}_3$ solution: 25.0 mL Molarity of $N{H}_3$ solution: 0.100 M $K_b(N{H}_3)=1.8\times {10}^{-5}$ Volume of $HCl$ solution: To be determined Molarity of $HCl$ solution: 0.100 M

Calculate the moles of NH3 and HCl

Calculate the initial moles of NH3: moles of $N{H}_3$ = volume × molarity moles of $N{H}_3$ = (25.0 mL) × (0.100 mol/L) moles of $N{H}_3$ = 0.00250 mol moles of $HCl$ = 0 (Initially)

Consider reaction progress and find moles at equilibrium

As the titration progresses, HCl is added to the solution and reacts with the NH3. At each stage of the titration, we will calculate the moles of NH3 and HCl in different scenarios. Scenario 1: Before adding any HCl moles of $N{H}_3$ = 0.00250 mol moles of $HCl$ = 0 mol Scenario 2: Halfway point (12.5 mL of HCl added) moles of $HCl$ added = (12.5 mL) × (0.100 mol/L)= 0.00125 mol moles of $N{H}_3$ remaining = 0.00250 mol - 0.00125 mol = 0.00125 mol Scenario 3: Equivalence point (25.0 mL of HCl added) moles of $HCl$ added = (25.0 mL) × (0.100 mol/L) = 0.00250 mol moles of $N{H}_3$ remaining = 0.00250 mol - 0.00250 mol = 0 mol moles of $N{H}_4^{+}$ formed = 0.00250 mol Scenario 4: Past the equivalence point (30.0 mL of HCl added) moles of $HCl$ added = (30.0 mL) × (0.100 mol/L) = 0.00300 mol moles of $N{H}_3$ remaining = 0 moles of $N{H}_4^{+}$ formed = 0.00250 mol excess moles of $HCl$ = 0.00300 mol - 0.00250 mol = 0.00050 mol

Calculate the pH at each scenario

In each scenario, we will use the moles of NH3, HCl, and NH4+ to find the pH of the solution. Scenario 1: Before adding any HCl Since HCl is not added yet, the pH of the solution depends on the basicity of NH3. Use $K_b$ to calculate the concentration of $O{H}^{-}$ formed, and then find the pH using the $pOH$. Scenario 2: Halfway point (12.5 mL of HCl added) Use the Henderson-Hasselbalch equation to calculate the pH: $pH=p{K}_a+log\frac{[base]}{[acid]}$ Here, $p{K}_a$ is the negative logarithm of the acid dissociation constant for $N{H}_4^{+}$, and the ratio will be the ratio of moles of NH3 to NH4+. Scenario 3: Equivalence point (25.0 mL of HCl added) Since all NH3 has reacted, we have only NH4+ in the solution. Calculate the concentration of $N{H}_4^{+}$, use its dissociation constant to find the $H^{+}$ concentration, and finally, calculate the pH. Scenario 4: Past the equivalence point (30.0 mL of HCl added) Calculate the excess moles of HCl, and find the $H^{+}$ concentration in the solution. Then, calculate the pH. By following the steps above, you should be able to find the pH at each stage of the titration of NH3 with HCl.

Key concepts

Acid-Base Chemistry

Acid-base chemistry involves the study of how acids and bases interact with each other. In this context, an acid is a substance that can donate a proton (H^+), while a base is a substance that can accept a proton. When acids and bases react, they form water and a salt, a process called neutralization. In the titration of ${\text{NH}}_3$, a base, with $\text{HCl}$, an acid, we observe this neutralization principle in action. This reaction results in the formation of ${\text{NH}}_4^{+}$ ions and chloride ions (${\text{Cl}}^{-}$). Each step in titration involves the transfer of protons, causing changes in the pH of the solution.

• Acids release ${\text{H}}^{+}$ ions.
• Bases accept ${\text{H}}^{+}$ ions.
• Neutralization leads to the formation of water and salts.

Equivalence Point

The equivalence point in a titration is the stage where the amount of acid equals the amount of base in a solution. Practically, it indicates that all of the original substance, such as ${\text{NH}}_3$, has reacted with the titrant, like $\text{HCl}$ in this case. At this point, we observe a significant change in the pH of the solution. For ${\text{NH}}_3$ titration with $\text{HCl}$, the equivalence point is reached when 25.0 mL of $\text{HCl}$ is added. This is because the initial moles of ${\text{NH}}_3$ are completely converted to ${\text{NH}}_4^{+}$ ions, marking a crucial turning point in the titration process. Understanding this point is important for determining the necessary amount of titrant needed to fully react with the analyte.

pH Calculation

Calculating the pH throughout a titration allows us to understand the progress of the reaction. The pH is a measure of the acidity or basicity of a solution. During titration, the pH changes as more $\text{HCl}$ is added to the ${\text{NH}}_3$ solution. Various methods are used to determine the pH at different stages:

• Prior to adding any $\text{HCl}$: The pH is based on the basic nature of ${\text{NH}}_3$.Using its $K_b$, you can find the ${\text{OH}}^{-}$ concentration and thus the pH.
• Halfway to equivalence: Here, the concentration of base and acid components causes pH to be calculated via the Henderson-Hasselbalch equation: $pH=p{K}_a+\log \left(\frac{\text{base}}{\text{acid}}\right)$
• At equivalence point: Consisting purely of ${\text{NH}}_4^{+}$, pH is calculated using the ice table and $K_a$ values derived from $K_w/K_b$.
• Beyond equivalence: When excess $\text{HCl}$ remains, the pH is determined directly from the excess ${\text{H}}^{+}$ concentration.

Ammonia Titration

Ammonia titration involves the careful addition of an acid to a solution of ammonia to understand its neutralizing capabilities. In this titration exercise, ammonia (${\text{NH}}_3$) is the weak base being titrated with the strong acid $\text{HCl}$. Ammonia titration is exemplary of how weak bases can be analyzed using strong acids. ${\text{NH}}_3$, with a known $K_b$, is neutralized by $\text{HCl}$, forming ${\text{NH}}_4^{+}$. Key points to consider:

• Weak bases like ${\text{NH}}_3$ are neutralized in stages as they encounter strong acids.
• The reaction with $\text{HCl}$ is complete at the equivalence point, forming a salt.
• Partial progress in titration involves balance between remaining base and generated conjugate acid.

Understanding ammonia titration helps illustrate the concept of weak base titration, providing insights into behavior under acidic conditions.