Question

Molarity of NaOH: 0.1388 M

Unknown Acid Buret:

	Trial 1	Trial 2
Initial Buret Reading	27.11 ml	17.91 ml
Final Buret Reading	47.31 ml	42.01 ml
Volume of acid added	20.10 ml	24.10 ml

Results:

	Trial 1	Trial 2
Volume of NaOH at equivalence point	4.75 mL	14.03 mL
Volume of NaOH at one-half the equivalence point	2.37 mL	7.015 mL
pH at half equivalence point	4.88	4.21
pKa of unknown acid
Average pKa
Average Ka
Moles of unknown acid	Moles	Moles
M of unknown acid	M	M
Average M of unknown acid

can you show the steps as to how to find the answers for the missing boxes, work included as I would like to learn? and a few sentences explaining the results.

Answer 1

From the given table values,

pH at half equivalence point = 4.88 = pKa of the acid.

since from the formula pH = pKa + log[salt]/[Acid]

at half equivalence point [Salt] = [Acid] so {salt]/[Acid] = 1 then,

pH = pKa

simiraly in second case pH = 4.21 = pKa

Average pKa = (4.88+4.21)/2

pKa = 4.545

-log(Ka) = 4.545

Ka = 10^-4.545

Ka = 2.851*10^-5

at equivalence point number of moles of acid is equal to number of moles of base.

number of moles of NaOH = 4.75mL * 0.1388 mol/L = 0.6593 mmol

number of moles of Acid = 0.6593 mmol

Molarity of unknown acid = 0.6593 mmol/20.10mL = 0.0328 mol/L

similarly in second case,

number of moles of NaOH = 14.03*0.1388 = 1.9474mmol

number of moles of the acid = 1.9474mmol/24.1mL = 0.081 mol/L

Avrage concentration = (0.0328+0.081)/2 = 0.0569 mol/L