Calculate the pH and concentrations of CH3NH2 and CH3NH3\' in a 0.0365 M methyla

Question

Calculate the pH and concentrations of CH3NH2 and CH3NH3' in a 0.0365 M methylamine (CH3NH2) solution. The Kb of CH3NH2 = 4.47x 104. Number pH- 11.61 Incorrect. You have calculated the pH at equilibrium correctly. Recall that since the initial concentration of methylamine (0.0365 M) Number [CH,NH].||4x10 |M| , M is not sufficiently dilute,(OH1-(CH3NHs1 at equilibrium. The [CH3NH2l at equilibrium is equal to the initial, formal concentration of methylamine minus the amount that is hydrolyzed to form CH3NH3 Number CH,NH; 1.0325

Explanation / Answer

   CH3NH2 + H2O ---------> CH3NH3^+ (aq) + OH^-

I         0.0365                                0                           0

C         -x                                      +x                           +x

E      0.0365-x                              +x                           +x

             kb   = [CH3NH3^+][OH^-]/[CH3Nh2]

            4.47*10^-4 = x*x/0.0365-x

          4.47*10^-4*(0.0365-x) = x^2

            x   = 0.00382

      [CH3NH2] = 0.0365-x = 0.0365-0.00382   = 0.03268M

      [CH3NH3^+] = x             = 0.00382M

      [OH^-]           = x             = 0.00382M

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