Question

Let's say you did this part of the experiment and you obtained the following information: - You used 0.192 g of unknown acid for your titration. - It took you 26.42 mL of 0.05 M NaOH to reach the equivalence point. - The pH at half-equivalence point was 3.18. Questions: 1. How many moles of NaOH were required to reach equivalence point? See your lab manual for help (page 97). (2 points) 2. How many moles (or equivalents) of unknown acid were at the equivalence point? Hint: your unknown acid is monoprotic. (1 point) 3. What is pka of your unknown acid? (1 point) 4. What is the equivalent mass of your unknown acid? See your lab manual for help (page 97). (2 points) 5. What is the identity of your unknown acid? Use Table 1 (page 93) in your lab manual. Hint: you should use both equivalent mass and pka to figure this out. Also: the numbers that you got may not match theoretical ones exactly. (2 points) 6. Assuming that you correctly identified your unknown acid, calculate the percentage error for each of the following: a. Equivalent mass of the unknown acid (2 points) b. pk, of the unknown acid (2 points) 7. A student performed this experiment and recorded the NaOH concentration as 0.060 M instead of the correct value of 0.050 M. Assuming that there were no other errors, would each of the following experimental results be (HIGHER/LOWER/UNAFFECTED) by the error? Explain your answer in 1-2 sentences. (1 pt.) a. Equivalent mass of the unknown acid b. PK, of the unknown acid

Answer 1

IML = 0.001L 2642 ml = 0.02642 L Molaity y Nacht: 0.Osm mol of Naol used to reach equivalence point = Molwi ty x volumelu = 0.05 moj x 0.036424 mol y No 0.00132 mol 0.00132 Mop if acid is monopronic HC HA HA + Naot NaA +H2O - Imoly Naou rept ImoluA 0.00132 mol og Naon reput 0:00/32001 HA molg acid = 0.00132 At halt equivalence point so [HA]= CA] pu=pka ph = 3:18 man J unknown acid = 0.1928 molgaced = molman j acid = mass = = 145.5 9 anę mo 0.1929 0.001324 mol moi