A certain acetic acid solution has pH = 2.14. Calculate the volume of 0.0900 M KOH required to reach the equivalence point in the titration of 35.0 mL of the acetic acid solution.
Let acetic acid (CH3COOH) = HAc
......HAc + KOH ==> KAc + H2O
pH = 2.14 so (H^+) = approx 7.2E-3,
I think you need the (HAc)= ?
To calculate the volume of 0.0900 M KOH required to reach the equivalence point in the titration, you will need to know the molar ratio between acetic acid (CH3COOH) and potassium hydroxide (KOH) in the balanced chemical equation for the reaction. Since acetic acid is a weak acid and KOH is a strong base, the balanced chemical equation for the titration reaction is:
CH3COOH + KOH -> CH3COOK + H2O
From the equation, you can see that one mole of acetic acid reacts with one mole of KOH to produce one mole of CH3COOK (potassium acetate) and one mole of water (H2O).
Given that the volume of the acetic acid solution is 35.0 mL and its concentration is not given, we need to first calculate the number of moles of acetic acid in the solution using the formula:
Moles = Volume (in liters) x Concentration (in M)
Converting the volume of the acetic acid solution to liters:
Volume = 35.0 mL = 35.0 / 1000 = 0.035 L
Since the concentration of the acetic acid is not given, we will use the pH value provided to calculate its concentration using the fact that:
pH = -log[H+]
First, convert the given pH value to the concentration of hydrogen ions ([H+]):
[H+] = 10^(-pH)
[H+] = 10^(-2.14)
[H+] ≈ 0.00600 M
Now that we know the concentration of acetic acid ([CH3COOH]) is approximately 0.00600 M and the volume of the solution is 0.035 L, we can calculate the number of moles of acetic acid:
Moles of acetic acid = Concentration x Volume
Moles of acetic acid = 0.00600 M x 0.035 L
Moles of acetic acid ≈ 0.00021 moles
Based on the balanced equation, since the molar ratio between acetic acid and KOH is 1:1, we know that 0.00021 moles of acetic acid will react with 0.00021 moles of KOH at the equivalence point.
Now we can calculate the volume of 0.0900 M KOH required to reach the equivalence point. To do this, we use the formula:
Volume of KOH = Moles of KOH / Concentration of KOH
Volume of KOH = 0.00021 moles / 0.0900 M
Volume of KOH ≈ 0.0023 L
Finally, convert the result to milliliters:
Volume of KOH ≈ 0.0023 L * 1000 mL/L
Volume of KOH ≈ 2.3 mL
Therefore, approximately 2.3 mL of 0.0900 M KOH is required to reach the equivalence point in the titration of 35.0 mL of the acetic acid solution.