Question

Consider the titration of 65 mL of 1.75 M diethylamine, (CH3CH2)2NH (Kb = 1.3×10-3) with 0.25 M HCl. How many mL of HCl will need to be added to the diethylamine solution to reach a pH of 11.11?

Answer 1

Consider the titration of 65 mL of 1.75 M diethylamine, (CH3CH2)2NH (Kb = 1.3×10-3) with 0.25 M HCl. How many mL of HCl will need to be added to the diethylamine solution to reach a pH of 11.11?

Given

Volume of diethylamine = 65 mL = 0.065 L and molarity of it is = 1.75 M

Concentration of HCl = 0.52 M

Calculation of moles of diethyl amine

= volume in L * molarity

= 0.065 L * 1.75 M

= 0.11375 mol

pH = 11.11

When HCl is mixed in weak base (diethyl amine ) then there is a formation of buffer.

(CH3CH2)2NH (aq) + HCl (aq) ------- > (CH3CH2)2NH2Cl(aq)

We use Henderson equation equation to get concentrations of salt (CH3CH2)2NH2Cl(aq)

pOH = pKb + log [Acid]/ [base ]

pkb = -log kb

pkb = - log ( 1.3 E-3 )

= 2.89

We know pOH = 14 –pH

= 14 – 11.11

= 2.89

2.89 = 2.89 + log (mol acid / mol base )                ( volume is same )

0 = log (mol acid / 0.11375 mol )

By taking antilog of both side

1 = mol acid / 0.11375 mol

Mol acid = 1 * 0.11375

Now we got moles of acid

We get volume by using its molarity

As molarity = mol / volume in L

Volume in L = mol / molarity

= 0.11375 mol / 0.25 M

= 0.455 L

= 455 mL of HCl will be required

Volume of HCl = 455 mL