Question

A 25.0 mL sample of a 0.2800 M solution of aqueous trimethylamine is titrated with a 0.3500 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25°C.

pH after 20.0 mL of acid have been added =

Answer 1

pKb = 4.19

millimoles of base = B = 25 x 0.2800 = 7.00

1 ) 10 mL acid added

millimoles of acid = 10 x 0.3500 = 3.50

This is half - equivalence point :

pOH = pKb

pOH = 4.19

pH = 9.81

2 ) 20. mL acid added

millimoles of acid = 20 x 0.35 = 7

This is equivalence point :

here salt only remains.

salt concentration = 7 / 20 + 25 = 0.156 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (4.19 + log 0.156)

pH = 5.31

3 ) 30 mL acid added

millimoles of acid = 30 x 0.35 = 10.5

strong acid is present .so [H+] = 0.0636 M

pH = -log [H+] = -log (0.0636)

pH = 1.20