Benzylamine C_6H_5CH_2NH_2 is a molecular weak base with a pK_b of 4.67. Show yo

Question

Benzylamine C_6H_5CH_2NH_2 is a molecular weak base with a pK_b of 4.67. Show your work! What is the formula of the related conjugate acid, benzylammonium? What is the pH and % dissociation of a 0.300 M solution of benzylamine? First write a balanced net ionic equation of the equilibrium involved for this base in water, using actual formulas. pH = %dissoc. = What is the pH and %dissociation of a 0.150 M solution of benzylammonium chloride, C_6H_5CH_2NH_3CI? This is the hydrochloric acid salt of benzylamine (the salt that would form if the base is reacted with HCl). (Note, salts like this are often written in an old style formula C_6H_5CH_2NH_2 HCl and named "benzylamine hydrochloride" rather than named "benzylammonium chloride" as chemists would do for simpler salts.) First write a balanced net ionic equation, eliminating spectator ions, for the equilibrium involved for this salt in water. pH = %dissoc.= What is the pH of a buffer solution containing both 0.285 M C_6H_5CH_2NH_2 and 0.196 M C_6H_5CH_2NH_3Cl?

Explanation / Answer

A. the conjugate acid of benzylamine is C6H5CH2NH3+.

C6H5CH2NH2 + H2O ----------> C6H5CH2NH3+ + OH-

B. Benzylamine is a weak base, so

C6H5CH2NH2 + H2O ----------> C6H5CH2NH3+ + OH-

______________________[C6H5CH2NH2]_________________[C6H5CH2NH3+]___________________[OH-]

Initial_________________0.300____________________________0____________________________0

Change_______________-x_______________________________+x____________________________+x

Equilibrium___________0.300-x____________________________x_____________________________x

pKb = 4.67

pKb = -log(Kb)

4.67 = -log(Kb)

Kb = 2.14*10-5

Kb = [C6H5CH2NH3+][OH-]/[C6H5CH2NH2]

2.14*10-5 = x.x/(0.300-x)

Suppose 0.300-x = 0.300

2.14*10-5 = x2/0.300

x2 = 2.14*10-5 *0.300 = 6.41*10-6

x = 0.0025

[OH-] = x = 0.0025

pOH = -log[OH-] = -log(0.0025) = 2.60

pH = 14-pOH = 14-2.60 = 11.40

% dissociation of base = ([OH-]/[C6H5CH2NH2])*100 = (0.0025/0.300)*100 = 0.83 %

C. C6H5CH2NH2.HCl ---------> C6H5CH2NH3+ + Cl-

C6H5CH2NH3+ + H2O --------> C6H5CH2NH2 + H3O+

______________________[C6H5CH2NH3+]_________________[C6H5CH2NH2]___________________[H3O+]

Initial_________________0.150____________________________0____________________________0

Change_______________-x_______________________________+x____________________________+x

Equilibrium___________0.150-x____________________________x_____________________________x

pKb = 4.67

pKb = -log(Kb)

4.67 = -log(Kb)

Kb = 2.14*10-5

Ka = 10-14/2.14*10-5 = 0.467*10-9

Ka = [C6H5CH2NH2][H3O+]/[C6H5CH2NH3+]

0.467*10-9 = x.x/(0.150-x)

Suppose 0.150-x = 0.150

0.467*10-9 = x2/0.150

x2 = 0.070*10-9

x = 8.37*10-6

[H+] = x = 8.37*10-6

pH = -log[H+] = -log(8.37*10-6) = 5.08

% dissociation of acid = ([H+]/[C6H5CH2NH3+])*100 = (8.37*10-6/0.150)*100 = 55.8*10-4 %

D. According to Handerson Hasselbach equation,

pOH = pKb + log [C6H5CH2NH3Cl]/[C6H5CH2NH2]

= 4.67+ log(0.196/0.285)

= 4.67+ (-0.163)

= 4.507 = 4.51

pOH = 4.51

pH = 14-pOH = 14-4.51 = 9.49

pH = 9.49

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