Calculate the pH values and draw the titration curve for the titration of 500 mL of 0.010 M acetic acid (pKa 4.76) with 0.010 MKOH.
Calculate the pH of the solution after 490 mL of the titrant have been added.
Calculate the pH of the solution after 500 mL of the titrant have been added.
Calculate the pH of the solution after 510 mL of the titrant have been added.
Calculate the pH of the solution after 750 mL of the titrant have been added.
If you are able to help with the first question, showing it step-by-step, I might be able to do the rest.
Find the mole first
To calculate the pH values and draw the titration curve for the given titration, we need to determine the species present at each stage and calculate their concentrations using the principles of acid-base equilibria. Here's how to approach each part of the question:
1. Before any titrant is added:
The acetic acid is the only species present. Since acetic acid is a weak acid, we can assume it only partially dissociates into acetate ions (CH3COO-) and hydrogen ions (H+). To calculate the pH, we need to determine the concentration of acetate ions and hydrogen ions.
The initial concentration of acetic acid is 0.010 M, but since it only partially ionizes, the concentration of acetate ions and hydrogen ions will be less than this value. We can use the Henderson-Hasselbalch equation to calculate the concentrations:
pH = pKa + log([A-]/[HA])
Where pKa is the negative logarithm of the acid dissociation constant (4.76 in this case), [A-] is the concentration of acetate ions, and [HA] is the concentration of acetic acid.
At the start, there is no acetate ion present, so [A-] = 0. The concentration of acetic acid [HA] is equal to the initial concentration, so [HA] = 0.010 M.
Plugging in these values into the Henderson-Hasselbalch equation, we can calculate the initial pH.
2. After 490 mL of titrant have been added:
At this point, the reaction has gone through almost complete titration, and we have added enough KOH to completely neutralize most of the acetic acid.
To calculate the pH, we need to determine the number of moles of remaining acetic acid and acetate ions formed from its dissociation.
The number of moles of acetic acid before titration can be calculated using the initial concentration and volume (0.010 M x 0.500 L).
Since we have added 490 mL of titrant, we need to calculate the amount of moles of KOH that has reacted with the acetic acid. From the balanced chemical equation (CH3COOH + KOH -> CH3COOK + H2O), we know that one mole of KOH reacts with one mole of acetic acid.
Subtracting the moles of KOH reacted from the initial moles of acetic acid will give us the moles of remaining acetic acid. The volume of the solution is still 500 mL, so we can use this along with the remaining moles to calculate the new concentration of acetic acid. We can then use the Henderson-Hasselbalch equation again to find the pH.
3. After 500 mL of titrant have been added:
At this point, we have just reached the equivalence point, where all the acetic acid has been neutralized by the KOH. The reaction has formed an equal number of moles of acetate ions and water molecules.
After reaching the equivalence point, the solution will become a buffer solution as a result of the presence of the conjugate base (acetate ion) and the weak acid (acetic acid). Buffer solutions resist large changes in pH when small amounts of acid or base are added.
To calculate the pH, we can again use the Henderson-Hasselbalch equation, this time with the concentration of the acetate ions (equal to the concentration of acetic acid at the equivalence point) and the concentration of acetic acid (which would be close to zero).
4. After 510 mL of titrant have been added:
Beyond the equivalence point, excess KOH has been added, resulting in the hydrolysis of the acetate ions. The acetate ions will react with water to form acetic acid and hydroxide ions.
To calculate the pH, we need to determine the concentration of hydroxide ions formed during hydrolysis. Since KOH is a strong base, we can calculate the moles of KOH added and use it to find the new concentration of hydroxide ions.
Then, we can use the equation Kw = [H+][OH-] to calculate the concentration of hydrogen ions ([H+]) and determine the pH.
5. After 750 mL of titrant have been added:
The solution will be a fully basic solution at this point since a large excess of KOH has been added. This means that all the acetic acid has been neutralized, and the solution will be dominated by OH- ions.
To calculate the pH, we can use the equation pOH = -log[OH-] and then convert it to pH using the relation pH + pOH = 14.
Note: The provided directions assume ideal conditions without any deviations or issues such as incomplete reactions or different levels of dissociation. In real-life scenarios, small variations might occur.
To draw the titration curve, plot the pH values against the volume of added titrant (KOH in this case).
Keep in mind that accurate calculations and precise plotting require more detailed analysis and intermediate points. The provided methodology gives a general approach to calculate pH at specific volumes of the titrant.