#1-6 Titration of 10.0mL of an unknown diprotic acid with 1.00M NaOH 13 pH- 9.27

Question

#1-6 Titration of 10.0mL of an unknown diprotic acid with 1.00M NaOH 13 pH- 9.27 - 4.01 20 25 30 10 15 Using the titration curve shown above I. Calculate the concentration of the acid using the data from equivalence point #1. 2. Calculate the concentration of the sulfurous acid using the data from equivalence point #2. 3. Calculate the volumes for midpoint #1 and midpoint #2. 4, Calculate the value of pKal from midpoint #1 and pKa2 from midpoint #2. 5. Calculate the value of pKal using the data from 3.0mL. 6. Calculate the value of pKa2 from the data at 17.0mL

Explanation / Answer

1) from the graph total 20 mL of 1 M NaOH required to reach endpoint

total mole of NaOH reacted = 20*10^-3 * 1 = 20*10^-3 moles

since it is diprotic acid so 1 mole is reacting with 2 mole of NaOH

so moles of acid reacted = 20*10^-3 / 2 = 10*10^-3 moles

concentration = moles / volume = 10*10^-3 / 10*10^-3 =   1 M   ( 10 mL of acid is used )

2)

M1V1 = M2V2 ( M2 = molarity of NaOH = 1 , V2 = volume of NaOH = 10mL , V1 = volume of acid ,

M1 = molarity of acid )

here the HSO4- is the monoprotic acid

concentration of sulfurous acid ( HSO4- ) = 10 * 1 / 20    = 0.5 M

after first deprotonation the volume of solution becomes ( 10 ml initial acid + 10 ml added NaOH )

                               = 20 ml

3) volume for midpoint 1 = 5 mL

volume at midpoint 2 = 15 mL

4)

pK1 = 2                        pK2 = 6

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