Achieve #7 Week 2

[Kaitlin Joya 1I]

I do not understand where I am going wrong with this problem.

1. I found Kb to be 2.5 x 10^-2.
2. I constructed my ICE table and found x = 0.0089.
3. The equilibrium concentration of [ClO-] = 0.0031. [HClO] = [OH-] = 0.0089.
4. pOH = -log(0.0031) = 2.5.
5. pH = 14 - 2.5 = 11.5.

I also tried:

4. pKa = 7.4
5. pH = 7.4 + log (0.0031/0.0089) = 6.9

[Kaitlin Joya 1I]

I realized I found the wrong Kb, but after correcting it I am still not getting the right answer.

[Lauren Brotman 2F]

I had different numbers but it looks like you calculated pOH using [ClO-] instead of [OH-]. pOH is -log[OH-], so maybe try doing -log(0.0089) instead and going from there.

[Jada Dawson]

Hi,
I got that the Kb=2.5 x 10^-11. Using that Kb, I plugged it back into the equilibrium expression and got that x=5.48 x 10^-7. Since [x]=[OH] you can find the pOH using the equation pOH=-log[OH] and then subtract 14-pOH to find the pH. It might be a calculator issue on your part but it seems you know how to do the problem. Hope this helps!