Calculate the pH and concentrations of CH_3NH_2 and CH_3NH_3^+ in a 0.0427 M met

Question

Calculate the pH and concentrations of CH_3NH_2 and CH_3NH_3^+ in a 0.0427 M methylamine (CH_3NH_2) solution. The K_b of CH_3NH_2 = 4.47 TIMES 10^-4. Incorrect. You made the assumption that x is small and could be ignored in the denominator when solving the K_b expression. This approximation is only valid if x is less than 1 % of the formal concentration of methylamine (0.0427 M). Here, x (0.00437) is 10.2% of 0.0427, so the assumption that x is small is not justified, and the quadratic formula is needed to solve for x.

Explanation / Answer

pH calculation for CH3NH2

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

let x amount has hydrolyzed

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

4.47 x 10^-4 = x^2/(0.0427 - x)

x^2 + 4.47 x 10^-4x - 1.91 x 10^-5 = 0

x = [OH-] = 4.15 x 10^-3 M

pOH = -log[OH-] = 2.382

pH = 14 - pOH = 11,62

concentration of [CH3NH2] = 0.0427 - 4.15 x 10^-3 = 0.03855 M

concentration of [CH3NH3+] = 4.15 x 10^-3 M

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