Question

Consider the titration of $40.0\mathrm{mL}$ of $0.200\mathrm{M}{\mathrm{HClO}}_4$ by $0.100$ $M$ KOH. Calculate the $\mathrm{pH}$ of the resulting solution after the following volumes of KOH have been added. a. $0.0\mathrm{mL}$ d. $80.0\mathrm{mL}$ b. $10.0\mathrm{mL}$ e. $100.0\mathrm{mL}$ c. $40.0\mathrm{mL}$

Short answer

The pH values for the given volumes of KOH are as follows: a. $0.0\mathrm{mL}$: pH = 0.70 b. $10.0\mathrm{mL}$: pH = 1.30 c. $40.0\mathrm{mL}$: pH = 7.00 d. $80.0\mathrm{mL}$: pH = 13.07 e. $100.0\mathrm{mL}$: pH = 13.52

Step-by-step solution

Calculate the initial moles of HClO4

Firstly, we need to determine the initial number of moles of HClO4 in the solution. The moles of a substance can be calculated using the formula: moles = concentration × volume. Here, the concentration of HClO4 is 0.200M and the volume is 40.0 mL (which we'll convert to L by dividing by 1000). Initial moles of HClO4 = (0.200 mol/L) × (40.0 mL ÷ 1000) = 0.008 mol

Calculate the moles of KOH added

For each volume of KOH, we need to calculate the moles of KOH added to the solution. The moles of KOH can be calculated using the formula: moles = concentration × volume. Here, the concentration of KOH is 0.100M. We will do this step for each of the given volumes: a. 0.0 mL b. 10.0 mL c. 40.0 mL d. 80.0 mL e. 100.0 mL

Balance the moles of acid and base

For each case, we need to find the difference between the moles of HClO4 and the moles of KOH added, to determine the remaining moles of acid or base in the solution. For each volume, calculate the remaining moles of HClO4 and KOH: Remaining moles of HClO4 = Initial moles of HClO4 - Moles of KOH added

Calculate the new concentration of acid or base

Now we need to calculate the new concentration of HClO4 or OH- ions in the solution after the addition of the respective volumes of KOH. New concentration = (Remaining moles of HClO4 or OH-) ÷ (Initial volume of HClO4 + Added volume of KOH) Calculate the new concentration for each case.

Calculate the pH of the solution

Finally, we will use the new concentrations of HClO4 or OH- ions to calculate the pH of the solution in each case. If HClO4 is present in excess, we can calculate the pH directly using: pH = -log10([HClO4]) If OH- ions are in excess, we can calculate the pOH first: pOH = -log10([OH-]) And then, convert it to pH using the formula: pH = 14 - pOH Find the pH for each case. Please note that the steps to calculate pH may require additional steps in some cases. OnTriggerEnterRestBoundary_Error, but this approach should suffice for this exercise.

Key concepts

pH calculation

Understanding pH calculation is essential when dealing with acid-base titrations. The pH scale is a measure of how acidic or basic a solution is, ranging from 0 (very acidic) to 14 (very basic). In this exercise, we calculate the pH of solutions at different stages of titration by examining the concentration of hydrogen ions $[H^{+}]$ or hydroxide ions $[O{H}^{-}]$. For strong acids like $ext{HClO}_4$, the pH is found by the formula $extpH=-ext{log}_{10}[H^{+}]$. Conversely, when the solution contains excess OH⁻ ions, we use $extpOH=-ext{log}_{10}[O{H}^{-}]$ and then convert pOH to pH: $extpH=14-extpOH$. This recap helps us to systematically calculate the pH values at each stage by considering the prevailing concentrations post-reaction.

moles calculation

The moles calculation is a critical step in titrations, as it establishes the amount of reactants involved in the reaction. The number of moles can be determined using the formula:

• $extmoles=extconcentration\times extvolume$

Both the concentration and volume must be in appropriate units, specifically volume in liters. In this exercise, for example, we start with $0.008$ mol of $ext{HClO}_4$, calculated from its initial conditions ($40.0\text{ mL}=0.040\text{ L}$). Similarly, for KOH, moles must be recalculated for each volume point provided ($0.0\text{ mL},10.0\text{ mL},40.0\text{ mL},80.0\text{ mL},100.0\text{ mL}$). This provides insight into the stoichiometry of the titration and guides the subsequent steps to determine whether an excess of acid or base remains.

neutralization reaction

In a neutralization reaction, an acid reacts with a base to form water and a salt. For this titration, the reaction can be written as: $ext{HClO}_4+extKOH\to ext{H}_{2}extO+ext{KClO}_4$. Here, each mole of $ext{HClO}_4$ neutralizes one mole of $extKOH$, illustrating a 1:1 reaction ratio.

• This means that the number of moles of acid and base involved in the reaction are directly comparable.
• When these moles are equal, complete neutralization occurs, typically producing neutral water and a salt.

Understanding this concept is crucial because it determines the point at which pH starts altering significantly, either from acidic to neutral or from neutral to basic. By knowing how much acid and base are present, we can predict whether the resulting solution will require pH adjustment calculations for excess ions.

HClO4 and KOH

HClO₄ and KOH are strong acids and bases, respectively. Strong acids like $ext{HClO}_4$ completely dissociate in water to yield $ext{H}^{+}$ ions, while strong bases like KOH dissociate to give $ext{OH}^{-}$ ions. This complete dissociation is crucial in titration, as it ensures that each mole of acid or base contributes fully to the reaction.

• The uniform behavior of strong acids and bases allows for precise predictions in pH changes during the titration process.
• In this exercise, the initial strong acid, $ext{HClO}_4$, reacts with the strong base, KOH, to form water and the salt $ext{KClO}_4$.

The knowledge that these substances are strong and fully dissociate supports reliable calculations during each step of the titration and ensures the predictability of neutralization points and pH changes.