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.
To calculate the pH values and draw the titration curve for the titration of acetic acid with KOH, we need to consider the acid-base reaction between acetic acid (CH3COOH) and hydroxide ions (OH-).
Step 1: Determine the initial moles of acetic acid
Given:
Volume of acetic acid solution (V1) = 500 mL
Concentration of acetic acid (C1) = 0.010 M
Using the formula:
moles = concentration * volume
moles of acetic acid = 0.010 M * 0.500 L
moles of acetic acid = 0.005 mol
Step 2: Calculate the number of moles of hydroxide ions needed for complete neutralization.
The balanced chemical equation for the reaction between acetic acid and KOH is:
CH3COOH + KOH -> CH3COOK + H2O
From the balanced equation, we know that 1 mole of acetic acid reacts with 1 mole of hydroxide ions.
moles of KOH needed = moles of acetic acid = 0.005 mol
Step 3: Calculate the volume of KOH solution needed for complete neutralization.
Given:
Concentration of KOH (C2) = 0.010 M
Using the formula:
volume = moles / concentration
volume of KOH solution needed = 0.005 mol / 0.010 M
volume of KOH solution needed = 0.500 L = 500 mL
Step 4: Drawing the titration curve
The titration curve plots the pH of the solution (y-axis) against the volume of KOH solution added (x-axis). It shows the gradual change of pH as the neutralization reaction progresses.
To draw the titration curve, we need to know the pKa value of acetic acid, which is given as 4.76. The pKa value represents the acidity of acetic acid and can be used to determine the pH at certain points during the titration.
At the start of the titration (before any KOH is added), the pH can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
where [A-] is the concentration of the conjugate base (CH3COO-) and [HA] is the concentration of the acid (CH3COOH).
Since no KOH has been added yet, the concentrations of the acid and conjugate base are equal, and the equation becomes:
pH = pKa + log (1) = 4.76 + 0 = 4.76
This is the pH of the solution initially.
As KOH is added, it reacts with acetic acid, forming the conjugate base (CH3COO-) and water. The pH will change as the reaction progresses, until the stoichiometric point is reached. At the stoichiometric point, the moles of KOH added are equal to the moles of acetic acid initially present.
Beyond the stoichiometric point, the excess OH- ions will start to contribute to the pH of the solution.
Step 5: Calculate the pH of the solution after adding 490 mL of KOH
Based on the given information, after adding 490 mL of the titrant, the total volume of KOH solution added is 500 mL (initial volume) + 490 mL = 990 mL = 0.990 L.
To calculate the moles of OH- ions added:
moles = concentration * volume
moles of OH- ions = 0.010 M * 0.990 L = 0.0099 mol
Since we know that 1 mole of OH- reacts with 1 mole of acetic acid, the moles of acetic acid remaining will be 0.005 mol (initial moles) - 0.0099 mol (moles of OH- added) = -0.0049 mol.
Since the number of moles cannot be negative, it means that all the acetic acid has been neutralized by the KOH, and we are in the excess OH- region of the titration curve.
At this point, the pH will depend on the excess OH- ions present in solution. To calculate the pH, we need to use the concentration of OH- ions and the dissociation constant of water (Kw = 1.0 x 10^-14).
Known:
Concentration of OH- ions = 0.010 M
Using the equation for the autoionization of water:
[H+][OH-] = Kw
[H+] = Kw / [OH-]
[H+] = (1.0 x 10^-14) / (0.010)
[H+] = 1.0 x 10^-12
pH = -log [H+]
pH = -log (1.0 x 10^-12)
pH ≈ 12.0
Therefore, the pH of the solution after adding 490 mL of KOH is approximately 12.0.
Note: The negative value for the moles of acetic acid remaining indicates that the total number of moles in solution is based on the excess OH- ions after complete neutralization of the acetic acid.