In this lab the student will determine the equilibrium constant, Ka, of a weak m
ID: 980705 • Letter: I
Question
In this lab the student will determine the equilibrium constant, Ka, of a weak monoprotic acid. The reaction between acetic acid and 0.1M sodium hydroxide to produce sodium acetate and water will be monitored using a pH meter.
CH3COOH + NaOH ?CH3COONa + H2O
DOES 1-3 LOOK CORRECT? I NEED HELP WITH #4 & #5 PLEASE.
1. From your graph, determine the volume of NaOH needed to reach the equivalence point in the titration.
From my graph, the volume of NaOH needed to reach the equivalence point in the titration is around 3.8mL.
2. Calculate the volume needed to reach the half-equivalence point in the titration.
The volume needed to reach the half-equivalence point in the titration is a fraction 3.8/2 = 1.9mL.
3. Find this half-equivalence point on the graph and determine its corresponding pH for each titration.
The corresponding pH to the half-equivalence point on the graph is 4.25.
4. Use the half-point pH value to find the experimental pKa.
To solve, recall that pH = pKa + log [base] / [acid]
5. Use the pH and molarity to determine the Ka of acetic acid.
Recall that Ka = ([H+] [A]) /[HA] and pH = -log [H+]
Explanation / Answer
1. From your graph, determine the volume of NaOH needed to reach the equivalence point in the titration.
From my graph, the volume of NaOH needed to reach the equivalence point in the titration is around 3.8mL.
You are correct, since the pH change is drastical here, menaing th eOH concnetration changes suddenly, that is, no more acid present
2. Calculate the volume needed to reach the half-equivalence point in the titration.
The volume needed to reach the half-equivalence point in the titration is a fraction 3.8/2 = 1.9mL.
True, by definition, the half equivalence point is the half of the total volume of neutralization
3. Find this half-equivalence point on the graph and determine its corresponding pH for each titration.
The corresponding pH to the half-equivalence point on the graph is 4.25.
pH is around 4 and 5, this is not accurate bur anything around 4.50 will do
4. Use the half-point pH value to find the experimental pKa.
To solve, recall that pH = pKa + log [base] / [acid]
since [base] / [acid] = 1 in th ehalf equvialence point
then
log(1) = 0
so pH = pKa
pKa = 4.25
5. Use the pH and molarity to determine the Ka of acetic acid.
pKa = -log(Ka)
4.25 = -log(Ka)
Ka = 10^-4.25 = 0.00005623413252 = 5.6*10^-5
which is pretty near to that of 1.8*10^-5 real value
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