Calculate the pH of a 0.0333 M aqueous solution of dimethylamine ((CH3)2NH, Kb = 5.9×10-4) and...

Calculate the pH of a 0.0333 M aqueous solution of dimethylamine ((CH₃)₂NH, K_b = 5.9×10^-4) and the equilibrium concentrations of the weak base and its conjugate acid.

pH	=
[(CH₃)₂NH]_equilibrium	=	M
[(CH₃)₂NH₂⁺ ]_equilibrium	=	M

Expert Solution

((CH₃)₂NH + H2O--------> [(CH₃)₂NH₂⁺ + OH^-

I 0.0333 0 0

C -x +x +x

E 0.0333-x +x +x

Kb = [(CH₃)₂NH₂+] [OH^-]/[(CH₃)₂NH]

5.9*10^-4 = x*x/0.0333-x

5.9*10^-4 *(0.0333-x) = x^2

x= 0.004147

[OH-] = x = 0.004147M

[(CH₃)₂NH] = 0.0333-x = 0.0333-0.004147 = 0.0292M

[(CH₃)₂NH₂⁺ ] = x= 0.004147M

POH = -log[OH-]

= -log0.004147 = 2.3823

PH = 14-POH

= 14-2.3823 = 11.6177

queen_honey_blossom answered 1 year ago

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