pling Calculate the pH and concentrations of CH3NH2 and CH3NH3 n a 0.0485 M meth

Question

pling Calculate the pH and concentrations of CH3NH2 and CH3NH3 n a 0.0485 M methylamine (CH3NH2) solution. The Kb of CH3NH 4.47 x 1 Number pH 11.66 Incorrect Number You have calculated the pH at equilibrium correctly. Recall that since the initial concentration of methylamine (0.0485 M) CH, NH C M is not sufficiently dilute, CoH [CH3NH3 at equilibrium. The [CH3NH2] at equilibrium is equal to the initial, formal concentration of methylamine minus the amount that is hydrolyzed to form CH3NH3 Number CH, NH M. Previous Give Up & View Solution Try Again Next Ext

Explanation / Answer

CH3NH2 + H2O ------------------> CH3NH3+ + OH-

0.0485 0 0 -------------> initial

0.0485-x x x --------------> equilibrium

Kb = [CH3NH3+][OH-]/[CH3NH2]

4.47 x 10^-4 = x^2 / 0.0485 -x

x^2 + 4.47 x 10^-4 x - 2.17 x 10^-5 = 0

x = 4.44 x 10^-3

[OH-] = 4.44 x 10^-3 M

pOH = -log [4.44 x 10^-3 ]

pOH = 2.35

pH + pOH = 14

pH = 11.65

[CH3NH3+] = x = 4.44 x 10^-3 M

[CH3NH2] = 0.0485 -x = 0.0441 M

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