Part B: Determining the pKa of an Unknown Acid pH of Unknown Acid: 302 Unknown #
ID: 560068 • Letter: P
Question
Part B: Determining the pKa of an Unknown Acid pH of Unknown Acid: 302 Unknown # | 4 3 lution PH (0.25 pt) Solution pH (0.25 pt) After 2.5 mL NaOH 355 After 5.0 mL NaOH After 17.5 mL NaOH 12.12 After 20.0 mL NaOH 2.4 342 After 7.5 mL NaOH 4 2After 22.5 mL NaOH 12.58 After 10.0 mL NaOH 4.53 After 12.5 mL NaOH 4 After 15.0 mL NaOH6 Observations (0.5 pt): The pH bo a big irceosed After 25.0 mL NaOH 12.6 After 27.5 mL NaOH | 12-ju After 30.0 mL NaOH/ 12.78 after i5mL ott, and the Solution 3. (1 pt) Determine the pKa of the unknown acid using your results from part B (Explain). Answer 4. (1 pt) Using Table 5.1 and your data from Part B, determine the identity of your unknown acid. Bubble in your answer at the right. A. KHP B. lactic acid C. acetic acid 5. (2 pts) Calculate the concentration of your unknown acid solution (show your work)?Explanation / Answer
3) As per the given data, the equivalence point is reached when 15 ml of NaOH is added.
Now, pH = pka at half equivalance point which is at V(NaOH) = 1/2 V(NaOH at eq point) = 15/2 = 7.5 ml
The pH at V = 7.5 ml = 4.22. Therefore, the pKa of unknown acid = 4.22
4) pKa = -logKa
therefore, Ka = 10-pKa = 10-4.22 = 6.03 * 10-5 . This is close to the ka of acteic acid. Hence, the weak acid is acetic acid.
5) The given acid is CH3COOH of pH = 3.02
therefore [H+] = 10-pH = 10-3.02 = 9.55 * 10-4 M
CH3COOH ------ CH3COO- + H+
Ka = [CH3COO-]eq[H+]eq/[CH3COOH]eq
therefore [CH3COOH]eq = (9.55 * 10-4)2/6.03*10-5 = 1.51 * 10-2 M
[CH3COO]initial = 9.55 * 10-4 + 1.51 *10-2 = 0.0161M