So I've been working on this online homework problem and got it wrong 8 times. I need help. Here's the question:
1,2-diaminoethane, H2N-CH2-CH2-NH2, is used extensively in the synthesis of compounds containing transition-metal complexes in water. If pK_b1 = 3.29 and pK_b2=6.44, what is the pH of a 2.58x10^-4 M solution of diaminoethane?
I first solved for Kb1 and Kb2 using the formula K_b1= 1x10^(pK_b1).
After that I made an ICE-table and plugged my values from that table into
K_b1= ( [1,2-diaminoethaneH+] * [OH-])/ [1,2-diaminoethane]
I solved for x using the quadratic formula and got 3.638x10^-4
I then plugged x into the formula pOH=-log(3.638x10^-4) = 3.44
Then finally solved for pH
pH = 14 - pOH
pH = 14 - 3.44 = 10.56
What is it that I'm doing wrong?
1 answer