Acid-Base Equilibria - Lab Report Example

Acid-base Titration I Maximum score: 90 points Partner: Lab Section (Day & Time):Attach your plots from the three titrations and answer all of the following questions.1. Attach your annotated plots and compare the titration curves for HCl, KHP, and one of the unknown acid, ►Near the start of each titration, ►Near the half-equivalence point, ►At the equivalence point, and ►For the region beyond the equivalence point.(40 points)AcidTrialpH at start of titrationpH at half-equivalence pointpH at equivalence pointpH beyond equivalence pointHCl2.742.896.4312.28KHP5.085.788.0112.01Unknown(averaged)3.063.218.2112.51HClKPHUnknown The titration curves for HCL and the Unknown are almost the same.

This means that the PH of HCL is closer to that of the Unknown sample.2. Calculate the molarity and % difference for the molarities of the two known acids solutions and explain why your values may differ from the true value. Show all calculations. (10 points).The number of moles of base needed to reach the endpoint: 0.1800 mmoles/mL*40 mL = 7.2 mmol OH- Because HCL and NaOH react in a 1 to 1 ratio, it means that the number of moles of vinegar that reacted with the base. 7.2 mmol OH- x 1 H3O+ / 1 OH- = 7.

2 mmol H3O+ The concentration of HCL 7.2 mmol H3O+ / 10.0 mL = 0.72 M 3. Calculate the molarity of the unknown acid stock solution. This requires accounting for the 1:10 dilution. Show all calculations. (10 points).The number of moles of base needed to reach the endpoint: 0.1800 mmoles/mL*40 mL = 7.2 mmol OH- Because Unknown and NaOH react in a 1 to 1 ratio, it means that the number of moles of vinegar that reacted with the base. 7.2 mmol OH- x 1 H3O+ / 1 OH- = 7.2 mmol H3O+ The concentration of Uknown 7.

2 mmol H3O+ / 10.0 mL = 0.72 M 4. Suggest one source of error in an acid-base titration that could result in a molarity of acid that is too low and one source of error that could result in a molarity of acid that is too high, assuming that the titrant is a base. (10 points)One of the sources of error is the addition of too much base or acid. This will raise the milliliters or the volume of the titrant added, decreasing the molarity or making it too low. That is the reason why one want it a very light color and not a dark rich color.

The other source of error is the use of impure titrant. For instance, in case one has to be using 1M NaOH to titrate HCL and the 1M NaOH has something like a little vinegar because the beaker was not cleaned well during the last lab, molarity will be much smaller since there are impurities in the titrant. The calculated molarity will be too low. 5. How does your pKa and Ka value for KHP compare with the accepted values? Explain using % difference. Show all calculations. (10 points).6. Report the pKa and Ka value that you determined for the unknown acid.

List the three acids in order of increasing acid strength. (10 points)