Question

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

Short answer

The short answers for the given problem are as follows: a. pH after 0.0 mL of KOH has been added: \( \approx 5.10 \) b. pH after 50.0 mL of KOH has been added: \( \approx 9.21 \) c. pH after 75.0 mL of KOH has been added: \( \approx 5.38 \) d. pH at the equivalence point: \( \approx 8.40 \) e. pH after 125 mL of KOH has been added: \( \approx 12.05 \)

Step-by-step solution

Determine the initial concentrations of HCN and OH-

Since no KOH has been added at this point, the solution consists only of HCN. The initial concentration of HCN is given as 0.100 M.

Set up an equilibrium expression to calculate the concentration of CN-

The dissociation of HCN into H+ and CN- can be represented as follows: HCN ⇌ H+ + CN- The equilibrium constant, Ka, for this reaction is given as 6.2 × 10^-10. Set up an equilibrium expression relating HCN, H+, and CN-: Ka = [H+][CN-] / [HCN] Since H0 is assumed to be 0 initially, we can rewrite the expression as: Ka = x^2 / ([HCN] - x) Where x represents the change in concentration of H+ (and CN-).

Solve for the concentration of H+ and calculate the pH

Since Ka is very small, we can assume that x is much less than [HCN], so the equation becomes: Ka ≈ x^2 / [HCN] Solve for x: x = √(Ka × [HCN]) = √(6.2 × 10^-10 × 0.100) ≈ 7.87 × 10^-6 M Now, we can calculate the pH using the formula: pH = -log[H+] pH = -log(7.87 × 10^-6) ≈ 5.10 So, the pH after 0.0 mL of KOH has been added is approximately 5.10. #b. Calculate the pH after 50.0 mL of KOH has been added.#

Determine the stoichiometry of the reaction and the amounts of HCN and OH-

At this point, 50.0 mL (0.050 L) of 0.100 M KOH has reacted with the HCN. The moles of OH- added are: moles OH- = volume × concentration = 0.050 L × 0.100 M = 0.005 mol Since 100.0 mL (0.100 L) of 0.100 M HCN was initially present, there are 0.010 mol of HCN initially. The stoichiometry of the reaction between HCN and OH- is 1:1, so after the addition of 0.005 mol of OH-, the solution consists of 0.005 mol HCN and 0.005 mol CN-.

Use the Henderson-Hasselbalch equation

Since we have a buffer solution consisting of a weak acid (HCN) and its conjugate base (CN-), we can use the Henderson-Hasselbalch equation to determine the pH: pH = pKa + log([CN-] / [HCN]) First, let's find pKa: pKa = -log(Ka) = -log(6.2 × 10^-10) ≈ 9.21 Now, solve for pH: pH = 9.21 + log(0.005 / 0.005) = 9.21 So, the pH after 50.0 mL of KOH has been added is approximately 9.21. #c. Calculate the pH after 75.0 mL of KOH has been added.# Since the steps are similar to the ones performed in the previous sections, we can present the final results: x ≈ 4.13 × 10^-6 pH = -log(4.13 × 10^-6) ≈ 5.38 So, the pH after 75.0 mL of KOH has been added is approximately 5.38. #d. Calculate the pH at the equivalence point.# The equivalence point is reached when 100.0 mL of KOH has been added (as it is the same concentration as HCN). At the equivalence point, all the HCN has reacted with the OH- to form CN-.

Determine the concentration of CN- and set up an ICE table

At this point, 0.010 mol of CN- has been formed, so [CN-] = 0.010 mol / 0.200 L = 0.050 M. A reaction table can be set up: HCN + OH- → CN- I: - - 0.050 M C: +x +x -x E: x x 0.050 - x However, since there is no more HCN left, the reverse reaction (CN- + H+ → HCN) is now dominant and responsible for the pH of the solution.

Set up an equilibrium expression to calculate the concentration of H+

Using the Kb expression for CN- (the conjugate base of HCN) and the relation between Ka and Kb: Kb = Kw / Ka Where Kw = 1 × 10^-14 is the water dissociation constant. Calculate Kb: Kb = (1 × 10^-14) / (6.2 × 10^-10) ≈ 1.61 × 10^-5 Now set up the equilibrium expression: Kb = [HCN][OH-] / [CN-] Kb = x^2 / (0.050 - x) We can assume that x is much less than [CN-] and simplify the expression: Kb ≈ x^2 / 0.050 Solve for x: x = √(Kb × 0.050) = √(1.61 × 10^-5 × 0.050) ≈ 2.53 × 10^-6 Since x represents the concentration of [OH-] in this case, we can calculate the pOH and then the pH: pOH = -log[x] = -log(2.53 × 10^-6) ≈ 5.60 pH = 14 - pOH = 14 - 5.60 ≈ 8.40 So, the pH at the equivalence point is approximately 8.40. #e. Calculate the pH after 125 mL of KOH has been added.#

Calculate the moles and the concentration of excess OH-

Since 125.0 mL of KOH has been added, the moles of OH- are: moles OH- = 0.125 L × 0.100 M = 0.0125 mol Since there were 0.010 mol of HCN to start with, there is an excess of 0.0025 mol of OH- in the 225.0 mL of the total solution. Excess OH- concentration = 0.0025 mol / 0.225 L ≈ 0.0111 M

Calculate the pOH and then the pH

Now, we can calculate the pOH: pOH = -log[OH-] = -log(0.0111) ≈ 1.95 Then the pH: pH = 14 - pOH = 14 - 1.95 ≈ 12.05 So, the pH after 125 mL of KOH has been added is approximately 12.05.

Key concepts

Buffer Solution

A buffer solution is a mixture that can resist changes in pH upon the addition of small amounts of acids or bases. In the context of titration, a buffer is typically made up of a weak acid and its conjugate base. For example, when HCN is partially neutralized by KOH, a buffer consisting of HCN and CN⁻ (its conjugate base) is created.
The presence of both components allows the solution to neutralize added H⁺ or OH⁻ ions, thus maintaining a relatively constant pH.
Buffer solutions are important in titrations because they help stabilize the pH during the addition of the titrant. This is especially helpful in calculating the pH after various additions of titrant, without requiring a complete reaction.

Equivalence Point

In a titration, the equivalence point is reached when the amount of titrant added is just enough to completely neutralize the analyte solution. At this point, the number of moles of acid equals the number of moles of base. This is significant because it signifies that the reaction is complete.
In the specific case of titrating HCN with KOH, reaching the equivalence point means all the HCN has been converted to CN⁻. This conversion leads to changes in pH properties because the CN⁻ would then act as a base in water, affecting the solution's pH.
The equivalence point should not be confused with the endpoint, which is the point where the indicator used in the titration changes color.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula used to estimate the pH of a buffer solution. It relates the pH of the solution to the pKa (the negative logarithm of the acid dissociation constant) and the ratio of the concentrations of the conjugate base to the weak acid. The equation is:\[ pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right) \]In this equation:

• \([A^-]\) is the concentration of the conjugate base
• \([HA]\) is the concentration of the weak acid

This equation is particularly useful in titration scenarios when a buffer solution is formed, allowing you to easily calculate the pH based on known concentrations without solving complex equilibrium expressions.

pH Calculation

Calculating pH is a critical component in understanding the chemical nature of a solution. The pH is a measure of the acidity or basicity of a solution, on a scale from 0 to 14. In a titration context, calculating the pH at various points helps to track the reaction's progress.
For example, before any KOH is added to the HCN solution, you'd use the acid dissociation constant \(K_a\) to find the concentration of hydrogen ions and compute the pH. During the titration, if a buffer solution is formed, the Henderson-Hasselbalch equation becomes handy. After the equivalence point, you focus on excess base calculations to find the pOH, and thus the pH.
Accurate pH calculations enable precise chemical analysis during titration processes, ensuring you can determine the endpoint and equivalence point of the reaction accurately.