Problem 105 - Chemistry An Atoms First Approach

Question

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCN by \(0.100 M \mathrm{KOH}\) at \(25^{\circ} \mathrm{C} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN}=6.2 \times 10^{-10} .\right)\) a. Calculate the \(\mathrm{pH}\) after \(0.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. b. Calculate the \(\mathrm{pH}\) after \(50.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. c. Calculate the \(\mathrm{pH}\) after \(75.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 125 mL of KOH has been added.

Step-by-Step Solution

Answer

The following are the calculated pH values for each titration step: a. pH = 5.16 b. pH = 9.24 c. pH = 9.45 d. pH = 9.52 e. pH = 11.95

Step 1: Determine the initial concentration of HCN

We are given 100.0 mL of 0.100 M HCN, so the initial concentration of HCN is 0.1 M.

Step 2: Construct an ICE table and solve for H₃O⁺ concentration

An ICE (Initial, Change, Equilibrium) table helps us to determine the concentrations of species at equilibrium. The reaction is: \(HCN \rightleftharpoons H^+ + CN^-\) | | HCN | H⁺ | CN⁻| |-------|-------|-----|------| |Initial| 0.10 | 0 | 0 | |Change |-x |+x | +x | |Final |0.10-x | x | x | Use the dissociation constant, Ka: \(Ka = \frac{[H^+][CN^-]}{[HCN]}\). Solve for x (H₃O⁺ concentration): \(6.2 \times 10^{-10} = \frac{x^2}{0.1 - x }\).

Step 3: Calculate the pH

Calculate the pH from the H₃O⁺ concentration found in the previous step: \(pH = -\log_{10}([H^+])\). #a. Calculate the pH after 50.0 mL of KOH has been added.#

Step 1: Calculate the moles of KOH added and equilibrium concentrations

50.0 mL of 0.100 M KOH is equivalent to: \(0.050\,\text{L}\times{0.100}\frac{\text{mol}}{\text{L}} = 0.0050\,\text{mol KOH}\). Since the volume of the solution is now 100+50 = 150 mL, the ratio of the moles of HCN and the moles of KOH becomes concentration. The equilibrium concentrations are: \[ [HCN] = \frac{0.01 - 0.0050}{0.15\,\text L}= 0.0333\ \text{M} \] \[ [CN^-] = \frac{0.0050}{0.15\,\text L}= 0.0333\ \text{M} \]

Step 2: Calculate the pH using the Henderson-Hasselbalch equation

Using the Henderson-Hasselbalch equation: \(pH = pK_a + \log \frac{[CN^-]}{[HCN]}\) where \(pK_a = -\log_{10}(K_a)\) #a. Calculate the pH after 75.0 mL of KOH has been added.#

Step 1: Calculate the moles of KOH added and equilibrium concentrations

75.0 mL of 0.100 M KOH is equivalent to: \(0.075\,\text{L}\times{0.100}\frac{\text{mol}}{\text{L}} = 0.0075\,\text{mol KOH}\). Since the volume of the solution is now 100+75 = 175 mL, the ratio of the moles of HCN and the moles of KOH becomes concentration. The equilibrium concentrations are: \[ [HCN] = \frac{0.01 - 0.0075}{0.175\,\text L}= 0.0143\ \text{M} \] \[ [CN^-] = \frac{0.0075}{0.175\,\text L}= 0.0429\ \text{M} \]

Step 2: Calculate the pH using the Henderson-Hasselbalch equation

Using the Henderson-Hasselbalch equation: \(pH = pK_a + \log \frac{[CN^-]}{[HCN]}\) #a. Calculate the pH at the equivalence point.#

Step 1: Calculate the moles and concentration of CN⁻ at the equivalence point

At the equivalence point, all HCN has reacted with KOH: 0.01 mol HCN. \[ [CN^-] = \frac{0.01\,\text{mol}}{0.2\,\text{L}}= 0.050\ \text{M} \]

Step 2: Calculate the pH at the equivalence point

Use the Kb expression for the conjugate base (CN⁻) and find the H₃O⁺ concentration: \[K_b = \frac{K_w}{K_a}\] \[K_b = \frac{x^2}{[CN^-]-x}\] Solve for x (H₃O⁺ concentration). \(pH = -\log_{10}([H^+])\). #a. Calculate the pH after 125 mL of KOH has been added.#

Step 1: Calculate the moles of KOH added and equilibrium concentrations

125.0 mL of 0.100 M KOH is equivalent to: \(0.125\,\text{L}\times{0.100}\frac{\text{mol}}{\text{L}} = 0.0125\,\text{mol KOH}\). Since the volume of the solution is now 100+125 = 225 mL, the concentration of excess OH⁻ can be calculated: \[ [OH^-] = \frac{0.0125 - 0.01}{0.225\,\text L}= 0.0111\ \text{M} \]

Step 2: Calculate the pOH and pH

Calculate the pOH: \(pOH = -\log_{10}([OH^-])\). Since \(pH + pOH = 14\), calculate the pH: \(pH = 14 - pOH\).

Key Concepts

Acid-Base Titration

Understanding acid-base titration is crucial for students delving into the world of analytical chemistry. It involves the process of adding a known concentration of one solution, typically an acid or a base, to a specific volume of another, until the chemical reaction between the acid and base is complete. The point of completion is known as the equivalence point, which is often indicated by a color change with the help of an indicator or by monitoring changes in electrical conductivity.

For example, in the titration of hydrocyanic acid (HCN) with potassium hydroxide (KOH), we can calculate the pH at different stages of the titration process. Initially, before any KOH is added, we can find the pH simply by looking at the dissociation of HCN in water. As KOH is gradually added, the pH changes, reflecting the neutralization of HCN by the KOH. We proceed by calculating the moles of KOH added and its impact on the reaction's equilibrium.

To enhance understanding of the process, students should always begin by writing the balanced chemical equation for the reaction, using it to set up an ICE table to find the concentrations at equilibrium. We then use these concentrations to find the pH of the solution, often utilizing the Henderson-Hasselbalch equation for buffer solutions or directly calculating the pH for strong acids and bases.

Henderson-Hasselbalch Equation

Delving into the Henderson-Hasselbalch equation, we find a practical tool to relate the pH of a buffer solution to the concentrations of an acid and its conjugate base. Given by the formula

\[ pH = pKa + \log \frac{[A^-]}{[HA]} \]
where \( pKa \) is the negative log of the acid dissociation constant (Ka), \( [A^-] \) is the concentration of the conjugate base, and \( [HA] \) is the concentration of the acid. This equation is particularly useful in the midway points of a titration, before the equivalence point, where the solution often acts as a buffer.

When utilizing the Henderson-Hasselbalch equation, it is important to ensure that the acid-base pair is in fact a buffer, meaning that both components are present in significant quantities. This is apparent when halfway to the equivalence point, but students must be cautious about using it directly at the equivalence point or when a strong acid or base is in excess, as it is not applicable in those cases due to significant shifts in the equilibrium.

ICE Table

The ICE table is an invaluable tool for students grappling with the complex world of chemical equilibria. ICE stands for Initial, Change, and Equilibrium, representing the stages of a reaction as it progresses. It's essentially a simple chart used to track the concentrations of reactants and products over the course of a reaction.

By filling in an ICE table, you can identify the initial concentrations of reactants, the changes they undergo during the reaction, and their concentration at equilibrium. You begin by writing down the initial concentrations (I), then determine the changes (C) that occur as the system approaches equilibrium, and finally calculate the equilibrium concentrations (E).

The beauty of this systematic approach lies in its ability to simplify complex calculations, making it easier for students to visualize and understand how different species in a reaction will interact with each other over time. When it comes to titration problems, the ICE table enables precise calculations of the pH at various stages, helping to illustrate the dynamic changes occurring within the reaction mixture.

Equivalence Point

The equivalence point of a titration is a fundamental concept that signifies when the quantity of titrant added reacts exactly with the substance being titrated. This means that the moles of acid equal the moles of base in the solution. It's a critical juncture in an acid-base titration because it indicates the completion of the neutralization reaction.

For the titration of weak acids with strong bases, or vice versa, the equivalence point is characterized by a sharp change in the pH of the solution. However, it's worth noting that the equivalence point does not always correspond to a neutral pH of 7, especially in the case of weak acid-strong base or weak base-strong acid titrations. The pH at the equivalence point depends on the strength of the acid or base and the resulting salt's ability to hydrolyze and affect the solution's pH.

To accurately calculate the pH at the equivalence point, it's necessary to consider the dissociation of the conjugate acid or base that forms when the titration is complete. For instance, in our HCN/KOH example, we calculate the concentration of the CN- ion at the equivalence point and use the Kb expression to find the concentration of OH- or H3O+ ions, from which we then deduce the pH.