The next problems deal with the titration of 75 mL of 1.55 M diethylamine, (CH3C

Question

The next problems deal with the titration of 75 mL of 1.55 M diethylamine, (CH3CH2)2NH (Kb = 1.3 x 10-3) with 0.25 M HCl. Water will be a major species throughout the titration. The chemical species, in addition to water, that can be found in this reaction mixture during the titration are:

I. H+ II. OH- III. Cl- IV. (CH3CH2)2NH V. (CH3CH2)2NH2+

a)What is the pH of the solution at the equivalence point?

b) How many mL of HCl will need to be added to the diethylamine, (CH3CH2)2NH solution to reach a pH of 11.11?

Explanation / Answer

The reaction of diethyl amine with HCl will be

(C2H5)2NH(aq) + HCl(aq) --> [(C2H5)2NH2]Cl(aq)

a) pH = pKa + log [salt] / [acid]

pOH = pKb + log [salt] / [Base]

At equivalent point

[(C2H5)2NH2]+ --> [(C2H5)2NH] + H+

So

Ka = Kw / Kb = [(C2H5)2NH] [ H+] / [(C2H5)2NH2]+

The equivalence point is reached when moles of acid = moles of base

Moles of base = 75 X 1.55 = 0.116 moles

Moles of acid = 0.116 = Molarity x volume

Volume = 0.116 / 0.25 = 0.464 L

Concentration of diethylamine ion= moles/ Total volume =0.116 / 0.464 + 0.075 = 0.215

Let the decomposition = x

so [H+] = [C2H5NH] = x

[(C2H5)2NH2]+ = 0.215 - x

Ka = 7.69 X 10^-12 = x^2 / 0.215-x

x<<1

so 7.69 X 10^-12 = x^2 / 0.215

1.653 X 10^-12 = x^2

x = 1.285 X 10^-6 = [H+]

So pH = -log [h+] = 5.89

b) so

pH = pKa + log [salt] / [acid]

11.11 = 11.11 + log[salt] / [acid]

so concentration of salt = concentration of acid

the mL of acid added should be such that it may half neutralize the base

Moles = 0.116 /2 = 0.058 moles

Moles of acid = Molarity X volume = 0.058 = 0.25 X Volume

volume = 0.232 L = 232 mL

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