Acid-base titration Hi, In this lab, I was using pH electrode to find the pH dur

Question

Acid-base titration

Hi,

In this lab, I was using pH electrode to find the pH during titration. I was given an unknown base. The mass of the unknown base is 0.610 g. I then dissolved with water in 250-mL volumetric flask. After that I pipet 100 mL of that unknown solution and titrated with standard 0.1 M HCl.

After the titration, I graphed my volume added and pH in Excel. I found that the equivalence point is at 19.17 mL, and the pH of it is 5.12. From this how can I find the mass of unknown.

I know the equation is:

HCl + B- ---> HB +Cl-

Kb= HB/H+ + B-

Thanks

Thanks

Explanation / Answer

At equivalence point

moles of acid = moles of base

so

moles of HCl = moles of base

moles of HCl = molarity x volume / 1000

moles of HCl = 0.1 x 19.17 / 1000

moles of HCl = 1.917 x 10-3

so

moles of Base = 1.917 x 10-3

molarity of Base = moles x 1000 / volume

molarity of base = 1.917 x 10-3 x 1000 / 100

molarity of base = 1.917 x 10-2 M

now the apply the same for initial base

molarity = moles x 1000 / volume

1.917 x 10-2 = moles x 1000 / 250

moles = 4.7925 x 10-3

then

molar mass = mass / moles

molar mass = 0.61 / 4.7925 x 10-3

molar mass = 127.28 g

so the molar mass of unknown base is 127.28 g

2)

HCl + B- ---> HB + C-

form the above reaction

moles of HB formed = moles of HCl reacted = 1.917 x 10-3

now this HB undergoes salt hydrolysis

HB ---> H+ + B-

Ka = [H+] [B-] / [HB]

at equilibrium

Ka = x*x / [ 1.917 *10-3 - x )

Ka = x2 / ( 1.917 * 10-3 - x )

as HB is a weak acid , x is so small

so

Ka = x2 / 1.917 * 10-3

x = sqrt ( Ka * 1.917 * 10-3 )

but given

pH = 5.12

-log [H+] = 5.12

[H+] = x = 7.58 * 10-6

so

7.58 * 10-6 = sqrt ( Ka * 1.917 * 10-3 )

solving we get

Ka = 3 x 10-8

Kb = Kw / Ka

Kb = 10-14 / 3 x 10-8

Kb = 3.33 x 10-7

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