Answer to Question #122721 in General Chemistry for Amy

Question

A 25.0 mL sample of a 0.1900 M solution of aqueous trimethylamine is titrated with a 0.2375 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)(3)N = 4.19 at 25 Degrees C

Answer 1

Answer.

(a) (first part)

Concentration of HCl after addition to 25 ml 0.190 M trimethylamine solution = S_Hcl (M) (Let)

Total volume of solution after addition of 10 ml 0.2375 (M) HCl to 25 Ml Buffer solution = (25+10) = 35 ml

According to the formula,

V₁ x S₁ = V₂ x S₂

35 ml x S_Hcl = 10 ml x 0.2375 (M)

S_HCl = 0.067 (M)

Concentration of trimethylamine considering the dilution effect = S_Base (M) (Let)

According to the formula,

V₁ x S₁ = V₂ x S₂

35 ml x S_Base = 25 ml x 0.190 (M)

S_Base= 0.135 (M)

Concentration of salt after addition of HCl to the trimethylamine = 0.067 (M)

Concentration of unreacted trimethylamine = (0.135-0.067) (M) = 0.068(M)

According to Henderson’ s equation for basic buffer,

pOH = pKb + log ([salt]/[base])

pKb = 4.19

pOH = 4.18

pH = (14-4.18) = 9.82 (Ans.)

(b) (second part)

Concentration of HCl after addition to 25 ml 0.190 M trimethylamine solution = S_Hcl (M) (Let)

Total volume of solution after addition of 20 ml 0.2375 (M) HCl to 25 Ml Buffer solution = (25+20) = 45 ml

According to the formula,

V₁ x S₁ = V₂ x S₂

45 ml x S_Hcl = 20 ml x 0.2375 (M)

S_HCl = 0.105 (M)

Concentration of trimethylamine considering the dilution effect = S_Base (M) (Let)

According to the formula,

V₁ x S₁ = V₂ x S₂

45 ml x S_Base = 25 ml x 0.190 (M)

S_Base= 0.105 (M)

Concentration of salt after addition of HCl to the trimethylamine = 0.105 (M)

So, equivalence point has reached

According to formula,

pH = 7- (0.5 x 4.19) – [0.5x log(0.105)] = 5.39 (Ans.)

(c) (third part)

Concentration of HCl after addition to 25 ml 0.190 M trimethylamine solution = S_Hcl (M) (Let)

Total volume of solution after addition of 30 ml 0.2375 (M) HCl to 25 Ml Buffer solution = (25+30) = 55 ml

According to the formula,

V₁ x S₁ = V₂ x S₂

55 ml x S_Hcl = 30 ml x 0.2375 (M)

S_HCl = 0.129 (M)

Concentration of trimethylamine considering the dilution effect = S_Base (M) (Let)

According to the formula,

V₁ x S₁ = V₂ x S₂

55 ml x S_Base = 25 ml x 0.190 (M)

S_Base= 0.086 (M)

Concentration of salt after addition of HCl to the trimethylamine = 0.086 (M)

Concentration of excess strong acid HCl = (0.129 -0.086) (M) = 0.043 (M)

pH= -log(0.043) = 1.37 (Ans.)