Question

Consider the titration of a \(35.0-\mathrm{mL}\) sample of \(0.175 \mathrm{M} \mathrm{HBr}\) with \(0.200 \mathrm{M}\) KOH. Determine each quantity. a. the initial \(\mathrm{pH}\) b. the volume of added base required to reach the equivalence point c. the \(\mathrm{pH}\) at \(10.0 \mathrm{~mL}\) of added base d. the \(\mathrm{pH}\) at the equivalence point e. the \(\mathrm{pH}\) after adding \(5.0 \mathrm{~mL}\) of base beyond the equivalence point

Short answer

a. Initial pH = 0.756, b. Volume of KOH = 30.625 mL, c. pH at 10.0 mL of added base = 1.28, d. pH at the equivalence point = 7, e. pH after adding 5.0 mL of base beyond the equivalence point = 12.26

Step-by-step solution

- Calculate the initial pH

Hydrobromic acid (HBr) is a strong acid and completely dissociates in water to H+ and Br-. To find the initial pH, calculate the concentration of H+ ions, which is the same as the concentration of HBr. Use the pH formula \( pH = -\log[H+] \) to find the initial pH.

- Calculate the volume of added base to reach the equivalence point

At the equivalence point, moles of HBr equals moles of KOH. Use the formula \( n_{HBr} = n_{KOH} \) and the given concentrations and volume of HBr to find the volume of KOH added.

- Calculate the pH at 10.0 mL of added base

Calculate the moles of HBr initially and the moles of KOH added, then subtract the moles of KOH from the moles of HBr to find the remaining moles of HBr. The concentration of H+ will be equal to the remaining moles of HBr divided by the new total volume (sample volume + added KOH volume). Finally, use the pH formula to calculate the pH.

- Calculate the pH at the equivalence point

At the equivalence point, all the HBr has been neutralized. The pH is determined by the autoionization of water and the properties of the resulting salt, which in this case is KBr, a neutral salt. Therefore, the pH will be 7.

- Calculate the pH after adding 5.0 mL of base beyond the equivalence point

After equivalence point, the solution is basic due to the presence of excess KOH. Calculate the moles of KOH in excess, then find the concentration of OH- ions by dividing the moles of excess KOH by the new total volume. Use the pOH formula \( pOH = -\log[OH-] \) and then \( pH = 14 - pOH \) to calculate the final pH.

Key concepts

Chemical Equilibrium in Acid-Base Titrations

In acid-base titrations, chemical equilibrium is a fundamental concept that occurs when a reaction reaches a state in which the rate of the forward reaction equals the rate of the reverse reaction. Although strong acids like hydrobromic acid (HBr) fully dissociate in solution, causing no real equilibrium to form, the point of equilibrium interest in a titration occurs at the equivalence point.

At this juncture during the titration of HBr with KOH, an equal number of moles of acid and base have reacted. Until this point, the added base reacts completely with the acid present, but once past the equivalence point, any additional base will remain in the solution and impact the resulting pH. Understanding the behavior of the reaction as it approaches and passes this point is crucial for analyzing the titration curve and predicting the outcome of the titration process.

pH Calculation Throughout a Titration

Calculating pH is an essential skill in titrations to determine the acidity or basicity of the solution at different stages. Initially, the pH is determined by the strong acid, HBr, present in solution. HBr, being a strong acid, dissociates completely, so the initial pH directly corresponds to its concentration. This is calculated with the familiar formula: \( pH = -\text{log}[H^+] \).

As KOH is added during the titration, the pH calculation requires a consideration of the stoichiometry of the neutralization reaction. Before the equivalence point, the pH depends on the concentration of remaining HBr. To calculate the pH at various volumes of KOH added, you must find the remaining moles of HBr, account for the dilution effect by adding the volume of KOH, and then use the pH formula. Beyond the equivalence point, the excess KOH dictates the pH, and this is where the calculation involves finding the pOH first and then using \( pH = 14 - pOH \) to find the pH of the basic solution.

Understanding the relationship between the added volume of a titrant, the reaction stoichiometry, and the resulting solution pH gives students a clear approach to predicting the outcome of a titration reaction at any point along the curve.

The Equivalence Point in Acid-Base Titrations

The equivalence point is a critical moment in a titration where the amount of titrant added is stoichiometrically equal to the quantity of substance present in the sample being titrated. In our exercise, it's the precise moment where all the HBr has been neutralized by KOH. For strong acid-strong base titrations, as the one described by the exercise, the equivalence point pH is neutral, typically around 7, because the salt produced and water do not affect the pH of the solution.

This is an idealized scenario; in reality, slight deviations can occur depending on the strength of acids and bases and the temperature, but understanding this foundational concept facilitates comprehension of more complex titration systems involving weak acids or bases where the equivalence point can shift and result in a different pH. Ultimately, grasping the concept of the equivalence point provides a framework for understanding reaction completion and the behavior of species in solution post-titration.