In this lab the student will determine the equilibrium constant, Ka, of a weak m
ID: 980199 • 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
1. Find the half-equivalence point on the graph and determine its corresponding pH for each titration.
2. Use the half-point pH value to find the experimental pKa. To solve, recall that pH = pKa + log [base] / [acid]
3. Use the pH and molarity to determine the Ka of acetic acid. Recall that Ka = ([H+] [A]) /[HA] and pH = -log [H+]
4. Look up the actual value for the equilibrium constant (Ka; acid ionization constant) for acetic acid. Compare your answer, Do you notice any variation?
Syringe Reading
pH After Each 0.5ml Increment
15mL
3.5
14.5mL
3.8
14.0mL
4.2
13.5mL
4.4
13.0mL
4.5
12.5mL
4.8
12.0mL
5.3
11.5mL
6.0
11.0mL
9.8
10.5mL
10.0
Syringe Reading
pH After Each 0.5ml Increment
15mL
3.5
14.5mL
3.8
14.0mL
4.2
13.5mL
4.4
13.0mL
4.5
12.5mL
4.8
12.0mL
5.3
11.5mL
6.0
11.0mL
9.8
10.5mL
10.0
Explanation / Answer
1. The equivalence point on graph is that point where pH suddenly increases
pH at 3.5 mL NaOH = 6.0
pH at 4.0 mL NaOH = 9.8 (sudden increase from pH = 6.0)
The equivalence point lies at volume between 3.5 mL and 4.0 mL or equivalence point = 3.75 mL
Half equivalence point lies at half volume of equivalence point = 3.75mL / 2 = 1.875 mL
Hence half equivalence point = 1.875mL volume NaOH and corresponding pH is approximately = 4.5
2. pH = pKa + log [base] / [acid]
But at half equivalence point, moles of base = moles of acid or [base] = [acid]
Hence, pH = pKa
pKa of acetic acid = pH at half equivalence point = 4.5
3. Now we know that pH = -log [H+]
Similarly pKa = -log [Ka]
4.5 = -log [Ka]
log [Ka] = -4.5
Ka = 10(-4.5)
Ka = 3.16 x 10-5 M
4. Actual value of Ka for acetic acid is 1.74 x 10-5 M
% error in Ka = [ (Actual value of Ka - experimental value of Ka) / Actual value of Ka ] x 100
% error in Ka = [ (1.74 x 10-5 M - 3.16 x 10-5 M) / 1.74 x 10-5 M ] x 100
% error in Ka = 81.6 %
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