a) Calculate the pH of a 0.116 M solution of ethylenediamine (H2NCH2CH2NH2). The

Question

a) Calculate the pH of a 0.116 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form of ethylenediamine ( H3NCH2CH2NH3 ) are 6.848 (pKa1) and 9.928 (pKa2).

b) Calculate the concentration of each form of ethylenediamine in this solution at equilibrium.

Calculate the pH of a 0.116 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form of ethylenediamine (H3NCH2CH2NH3 are 6.848 (pKa 1) and 9.928 (pKa2). Number pH Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. Number M H NCH, CH,NH Number H NCH, CH,NH M Number HT NCH,CH, NH M

Explanation / Answer

f the pKa1 value is 6.8, what is the pH of a 10^-3 M solution of this acid?

My solution:
pKa = -log(Ka) = 6.8
log(Ka) = -6.8
10^-6.4 = Ka



Ka = (x)(x)/(10^-3 - x) = 10^-6.8

Ignoring the x in the denominator since it is small relative to 10^-3, we have:

x^2 = (10^-3)(10^-6.8)= 10^-9.4.
x = (10^-9.4)^(1/2) = 10^-4.7

Since x = [H+], pH = -log(x) = -log(10^-4.7) = 4.7

And pH = 4.7 is the correct answer given by Kaplan.

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