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7.811.761.36E-117.911.831.17E-11811.978.57E-12
01000 2000 3000 4000 5000 6000 7000 8000 9000000000f(x) = - 0x + 0Vb*10^(-pH)Vb (µL)Vb*pH1.Equivalence point (Point D): 7.4mLa.Point A: ¼(7.4) = 1.85b.Point B: ½ (7.4) = 3.7c.Point C: ¾ (7.4) = 5.52.ΔpH = 5.75(±0.002) – 4.95(±0.002) = 0.8(±0.002) Monoprotic3.[HA] = [NaOH] = 0.1M4.Moles NaOH = Moles HA = 0.1M * 0.0074L = 7.4*10-4 mol HA5.Molecular weight = Mass of Acidusedtitrationmoles of acid= 0.1595g7.4E-4mol=215.54g/mola.Mass (acid) = 1.595g250mL25mL=0.1595g6.pH (equivalence point) = 7Ka = 10-pH pKa = -log[10-pH] pH = pKa + log [7.4E-47.4E-4]pH = pKa = - log [10-7] = 7Based on the molecular weight and pKa value, the unknown sample is KHP.Discussion: 1.Could visual acid base indicator be used in this experiment?
2.In general, what techniques are used for the determination of pKa values?
Signed and Dated by Analyst: References: Wei, S. Laboratory Manual for CHEM 3401, Lamar University: Beaumont, Texas, 2017.

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