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?
Based on the information you provided, it seems like you made an error in your calculations. Let's go through the steps one by one to find the mistake.
1. You correctly calculated Kb1 using the formula Kb1 = 1x10^(pKb1). Since pKb1 = 3.29, Kb1 should be equal to 1x10^(3.29) = 1.99x10^(-4).
2. Next, you set up an ICE table and wrote the expression for Kb1 as Kb1 = ([1,2-diaminoethaneH+][OH-])/[1,2-diaminoethane].
3. Then, you solved for x using the quadratic formula. However, you didn't provide the quadratic equation or your calculations, so it's difficult to determine where the mistake might have occurred. Please double-check your calculations for finding x.
4. After solving for x, you correctly calculated pOH as -log(x) and obtained pOH = 3.44.
5. Finally, to find the pH, you used the equation pH = 14 - pOH and got pH = 14 - 3.44 = 10.56, which is correct.
To find the mistake, please review your calculations while solving for x using the quadratic formula. Double-check your equation setup and the values you substituted into the equation. If you're still having trouble, I recommend consulting your textbook, class notes, or seeking additional assistance from your instructor or classmates.