Calculate the pH and concentrations of CH_3NH_2 and CH_3NH_3^+ in a 0.0299 M met

Question

Calculate the pH and concentrations of CH_3NH_2 and CH_3NH_3^+ in a 0.0299 M methylamine (CH_3NH_2) solution. The K_b of CH_3NH_2 = 4.47 Times 10^-4. pH = [CH_3NH_2] = [CH_3NH_3^- = Incorrect. You have calculated the pH at equilibrium correctly. Recall that since the initial concentration of methylamine (0.0299 M) is not sufficiently dilute, [OH^-] = [CH_3NH_3^+] at equilibrium. The [CH_3NH_2] at equilibrium is equal to the initial, formal concentration of methylamine minus the amount that is hydrolized to form CH_3NH_3^+.

Explanation / Answer

CH3NH2 + H2O <====> CH3NH3+ + OH-

[OH-] = [CH3NH3+]

[OH-] = ( 0.0299*4.47*10-4)1/2

[OH-] = 3.66*10-3

pOH = 2.44

pH = 11.56

[CH3NH3+] =  3.66*10-3 M

[CH3NH2] = 0.0299 - 0.00366 M

[CH3NH2] = 2.62*10-2 M

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