Raw Data (and assumed quantities) Name of Acid (or Acid Salt):ENOIC ACAD Volume

Question

Raw Data (and assumed quantities) Name of Acid (or Acid Salt):ENOIC ACAD Volume of water used to dissolve acid: 40.0 mL pH and 'Volume of NaOH(aq) added" values: Attach table of raw data and plot (titration curve) NaoH(Assumed quantity, this is not raw data because it wasn't measured by you) Data Analysis and Calculations 1.(a) Use your plot, your raw data table, and definitions from class to determine the following as precisely as your data allow. pH at equivalence point1 pH at half-equivalence point h57 Vop (Vat equivalence point)- 58 Vis p (Vat halif-equivalence point) of acid (from experiment): pKa of acid (from experiment): Show how you calculated K (b) Find a published pk, value (or find a K value and calculate the pks yourselfh for your acid (check your book's ). Comment on the agreement between your appendix, the CRC handbook, or another reference provided value and the published one. Suggest ideas why the values might not be identical. pK of acid (published value):8 Comments Sourcer lahs ohemnesb.edu 2. (a) initial moles of acid = (DO NOT use mass and molar mass here! Use the data obtained by titration!) Show calculation: Initial concentration of acid, [acid]s- Show calculation:

Explanation / Answer

1. a) At the half-equivalence point, 2*no. of moles of acid = no. of moles of base

i.e. no. of moles of salt (nsalt) = no. of moles of unreacted acid (nacid)

i.e. nsalt/nacid = 1

According to Henderson-Hasselbulch equation: pH = pKa + Log(nsalt/nacid)

Here, pH = 4.57

i.e. 4.57 = pKa + Log(1)

i.e. 4.57 = pKa + 0

i.e. pKa = 4.57

Formula: pKa = -Log(Ka)

i.e. 4.57 = -Log(Ka)

i.e. Ka = 10-4.57

i.e. Ka = 2.69*10-5

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