[hallie3]

Determine the pH of 2.00×10-2 M sodium hypochlorite with pKb= 6.47
 
My attempt:
Kb=antilog(-6.47)
Kb=(3.39E-7)
After completing appropriate ICE chart...
3.39E-7=(x^2/2.00E-2 - x)
x^2 - (6.78E-9)x + 3.39E-7
After completing the quadratic formula (disregarding extraneous negative x value)... 
x= 0.0005822336751302544
pH=-log(0.0005822336751302544)
pH= 7.49
However, I am being told this is incorrect. Where am I going wrong?
Thank you so much in advance!

[bjams12]

So Kb is the constant for the rxn
B- + H2O <-> BH + OH-
With B- being your base, the hypochlorite anion, dissociated from the Na. The ice table you used is great! You found x! But look at the the x corresponds to - OH-. So taking -log(x) like you did led you to find -log(OH-). So you found pOH. So take 14-pOH to find your pH!

[hallie3]

bjams12: Thank you for your response... What you said makes sense, yet when I subtracted my pOH value from 14 it still registers as incorrect... am I overlooking something in this question?

[bjams12]

Hm, I ended up doing the math myself and I think there may be an error in the calculation. I used the quadratic equation (wolfram) and found x = 8.21₄₇₂*10^-6. Using that value for [OH-] gives me a reasonable answer.

[Borek]

10^-6 or 10^-5?

[bjams12]

You're correct, 10^-5. I made a typo, thanks for catching it!