Question

10a) A 29.7 mL sample of 0.275 M methylamine, CH3NH2, is titrated with 0.267 M nitric acid. At the equivalence point, the pH is _______.

10B)A 20.4 mL sample of 0.357 M trimethylamine, (CH₃)₃N, is titrated with 0.395 M hydrochloric acid.

After adding 28.0 mL of hydrochloric acid, the pH is____.

Answer 1

10 a Ans.

Volume of methylamine, V₁ = 29.7mL

Concentration of methylamine, C₁ = 0.275M

Concentration of nitric acid, C₂ = 0.267M

At equivalence point, the no. of mol of the acid = no. of mol of the base.

Let V₂ be the volume of the acid

No. of mol of CH₂NH₃⁺ is equal to the no. of mol of the acid or the base.

Therefore, no. of mol of CH₂NH₃⁺ is

Total volume = 29.7mL + 30.6mL = 60.3mL = 60.3 x 10^-3L

Concentration of the conjugate acid is

The conjugate acid will undergo hydrolysis

The ICE table is

	CH2NH3+(aq)	H2O(l)	CH2NH2(aq)	H3O+(aq)
I	0.135M	-	0	0
C	-x	-	+x	+x
E	(0.135-x)M	-	x	x

The K_b of CH₂NH₂ is 4.4 x 10^-4

x in the denominator is neglected since the CH₂NH₂ is a weak base

Therefore, pH at the equivalence point is 5.757

10 b Ans.

Volume of (CH₃)₃N = 20.4mL

Concentration of (CH₃)₃N = 0.357M = 0.357 mol/L

Therefore, no. of mol of (CH₃)₃N is

Volume of HCl = 28.0mL

Concentration of HCl = 0.395M = 0.395 mol/L

No. of mol of HCl is

Excess no. of mol of HCl is

Total volume = 20.4mL + 28.0mL = 48.4mL = 48.4 x 10^-3L

Therefore, concentration of the excess HCl is

Therefore, pH when 28.0 mL of HCl is added = 1.108