I have 200mL of HClO 0,64M.
I mix it with 0,1mol of NaClO.
How should I calculate the pH?

My question is about this: should I keep attention about changes made to the solution volume after adding NaClO?

So far this is what I've done.


$\ce{NaClO + H2O -> Na+ + ClO-}$

With n(NaClO) = n(ClO-) = 0.1mol,

I calculated the molarity of the conjugate base:

[ClO-] = 0.1mol/0.2L = 0.5M.

Then I applied the Henderson-Hesselbalch equation:

pH = pKa + log([ClO-]/[HClO]) = 7.53 + log(0.781M) = 7.422

In this case I didn't consider the variation to the solution volume due to the addition of NaClO. So, is this correct?


closed as off-topic by airhuff, Mithoron, Todd Minehardt, Jan, A.K. Nov 26 '17 at 4:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$ – airhuff Nov 24 '17 at 19:48
  • $\begingroup$ I'm a college student, this is not a homework question. I've already solved it but I'm not sure about the result. I would like to compare my result with someone who know exactly how to solve it. $\endgroup$ – Scanfi98 Nov 24 '17 at 19:55
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    $\begingroup$ Please see the homework link in my above comment to learn what qualifies as a homework type of question and how to ask one. You can get help with this here, you just need to follow the guidelines. Best of luck. $\endgroup$ – airhuff Nov 24 '17 at 19:59
  • $\begingroup$ In addition to the problem that this would be considered a homework question, it also qualifies as an amirite question (seeks to consider confirmation or clarification on an already derived answer) which we have determined to be a bad fit for the site. $\endgroup$ – Jan Nov 25 '17 at 13:10

You're close. Everything is correct, except that when you take the ratio of concentrations in the H-H equation that ratio is not in moles. It's just a number, because you divide moles by moles . Rule of thumb: logarithms and exponential should never involve anything with units. So don't include the molar unit under the logarithm and you're good.


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