Question

Calculate the pH at the equivalence point for the titration of 0.240 M methylamine (CH3NH2) with 0.240 M HCl. The Kb of methylamine is 5.0× 10–4.

Answer #1

Since the concentrations of CH₃NH₂ and HCl are equal (both 0.240M), you need equal volumes of acid and base to react at the equivalence point. I'll assume 1 L for both of them, so the total volume would be 2 L.

Therefore the final concentrations would be 0.240 * (1/2) = 0.120 M for both reactants.

Now at the equivalence point, almost all the base has reacted with the HCl, and we have the following products:

CH₃NH₃⁺ and Cl⁻ and concentrations are already calculated above. so, we need to know the Ka:

Ka = 1x10^{-14} / 5x10^{-4} =
2x10^{-11}

Now the overall reaction and a ICE chart:

r: CH₃NH₃⁺ <--------> CH₃NH_{2} +
H^{+}

i: 0.120 0 0

e: 0.120-x x x

2x10^{-11} = x^{2} / 0.120-x Kb is small so
0.120-x can be neglected and stay as 0.120

2x10^{-11} * 0.120 = x^{2}

x = 1.5491x10^{-6} M = [H^{+}]

pH = -log(1.5491x10^{-6})

pH = 5.81

Hope this helps

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