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April 19, 2015

Posted by **Kati** on Friday, February 6, 2009 at 4:25pm.

How does one approach this question?

Kati

- Chemistry, help. -
**DrBob222**, Friday, February 6, 2009 at 4:43pmNaOCl = 5.25% which means 5.25 g NaOCl + 94.75 g of water but since the density of the solution is 1.00 g/mL then that's 94.75 mL water.

moles NaOCl = 5.25 g NaOCl/molar mass NaOCl = 5.25/74.44 = 0.0705 moles NaOCl.

That is in 94.75 mL H2O or 0.09475 L so molarity is 0.0705/09475 = 0.744 M. check my arithmetic.

OCl^- + HOH ==> HOCl + OH^-

Kb = Kw/Ka = (HOCl)(OH^-)/(OCl^-)

Look up Ka for HOCl. Kw = 1 x 10^-14. In the equation you know (HOCl)=(OH^-) so solve for (OH^-), convert that to pOH, and convert that to pH.

Post your work if you get stuck.

- Chemistry, help. -
**Kati**, Friday, February 6, 2009 at 5:06pmOk, so I found the Ka and then found the Kb. I got 2.857x10^-7. I then subbed that into the equation: [x^2]/[0.744-x]...you can use the assumption rule to get rid of the minus x in the denominator. When I solved I got x=0.0000461043, and a pH of 3.34. What went wrong? The answer is supposed to be 10.7 which makes sense, since the products are basic.

Thanks

Kati

- I FIGURED IT OUT! :) -
**Kati**, Friday, February 6, 2009 at 5:29pmI understand what I did! The 3.34 is the pOH thus, 14-pOH is the pH which is 10.66 or 10.7 :)

Thanks very much!!

-Kati

- Chemistry, help. -
**DrBob222**, Friday, February 6, 2009 at 5:30pmNothing went wrong except that you forgot to think about your answer. IF you look at the equation, what did you let X stand for? The answer is that you let it stand for (OH^-). So you found the pOH (when you took the negative log) thinking that was the negative log of H^+. Just subtract that from 14 to get pH. 14 -3.3 = 10.7. Voila! Ain't chemistry interesting? Besides that, it keeps one on one's toes.

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