A 25.0 ml sample of a 0.1700 m solution of aqueous trimethylamine is titrated with a 0.2125 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.

Volume required for neutralization V will be:

V * 0.2125 M HCl = 25 mL * 0.17 M

V = 20 ml

First part:

When 10 mL is added we can apply Henderson equation to get the result, so:

The pH will be of basic buffer

pOH = pKb + log (salt/base)

or pOH = 4.19 + log (0.2125*10 / 25*0.17 - 10*0.2125)

pOH = 4.19 and pH = 14 - 4.19 = 9.81

Second part:

When 20 ml is added, there is only salt formed

The pH will be salt of strong acid and weak base

So pH = 7 - 0.5 pKb - 0.5 log C

where C is the concentration of the salt formed so:

pH = 7 - (0.5*4.19) - (0.5 log (25*0.17) / (25+20))

= 5.42

Third part:

When 30 ml of the acid has been added,

The pH will be of the remaining strong acid

pH = - log (0.2125*10 / 25 + 30)

= 1.326