In the following questions, assume that no excess distilled water was added from a wash bottle during the titration of 10.00 mL of vinegar with your standardized solution of NaOH.
A. calculate the pH of vinegar.
B. what was the pH of the solution after 4.00 mL of titrant was added?
C. What was the pH at the equivalence point?
D. Why is the solution basic at the equivalence point?
E. What was the pH of the solution after 35.00 mL of titrant was added?
F. Draw a titration curve for this titration?
I suspect there is a lot of data missing.
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A. To calculate the pH of vinegar, we need to consider its acidic components, namely acetic acid (CH3COOH). Acetic acid is a weak acid, so we can calculate its pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
The pKa of acetic acid is 4.76, and since vinegar is a dilute solution, we can assume that the concentration of acetic acid ([HA]) is equal to the concentration of acetate ion ([A-]). The equation simplifies to:
pH = pKa + log (1)
Since log(1) is equal to 0, we can conclude that the pH of vinegar is equal to its pKa, which is 4.76.
B. To determine the pH of the solution after adding 4.00 mL of titrant, we need to consider the reaction between NaOH and acetic acid. The balanced equation is:
CH3COOH + NaOH → CH3COONa + H2O
Since NaOH is a strong base and acetic acid is a weak acid, the reaction goes to completion. We are left with sodium acetate (CH3COONa) and water. Sodium acetate is a salt of a weak acid and the conjugate base of acetic acid, so it dissociates partially in water, resulting in a slightly basic solution.
To calculate the pH, we need to know the concentration of sodium acetate and the dissociation constant (Ka) of acetic acid. Assuming the solution is dilute, we can use the approximation that the concentration of sodium acetate (CH3COONa) is equal to the concentration of acetate ion ([A-]). We also know that the pKa of acetic acid is 4.76.
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Since NaOH has reacted with all the acetic acid, the concentration of acetic acid ([HA]) becomes zero. We are left with the concentration of acetate ion ([A-]). Thus, the equation simplifies to:
pH = pKa + log ([A-]/0)
Any value divided by zero is undefined, so we cannot calculate the pH precisely. However, since we added a small amount of titrant (4.00 mL) and vinegar is a weak acid, the pH should be slightly basic, but still close to the original pH of vinegar (around 4.76).
C. At the equivalence point, the moles of acid and base are stoichiometrically balanced. In this case, it means that the moles of acetic acid are exactly neutralized by the moles of NaOH. At this point, we have completed the reaction and are left with only the sodium acetate salt and water.
Sodium acetate is the conjugate base of acetic acid and can hydrolyze in water, resulting in the generation of hydroxide ions (OH-) and the regeneration of acetic acid. Hydroxide ions increase the concentration of hydroxide in the solution, making it more basic.
Therefore, at the equivalence point, the pH will be basic, typically around 7-9 depending on the specific acid and base used.
D. The solution is basic at the equivalence point because the presence of the sodium acetate salt and the hydrolysis of its ions (conjugate base of acetic acid) generate hydroxide ions (OH-) when dissolved in water. These hydroxide ions increase the concentration of hydroxide in the solution, making it basic.
E. To calculate the pH of the solution after adding 35.00 mL of titrant, you need to follow similar steps as in question B. Determine the moles of NaOH added, calculate the moles of excess NaOH, and use the Henderson-Hasselbalch equation to calculate the pH of the remaining solution.
F. Unfortunately, as an AI text-based bot, I am unable to draw images or create visuals. However, you can search online for "titration curve for acetic acid and sodium hydroxide" to find graphical representations of similar titrations. The titration curve generally shows the change in pH of the solution as the volume of titrant (NaOH) increases.