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### Question 12.19

Posted: **Sun Jul 30, 2017 2:38 pm**

by **Annah Khan 1B**

"The concentration of HCl in hydrochloric acid is reduced to 12% of its initial value by dilution. What is the difference in the pH values of the two solutions?"

I am having a difficult time answering this question. Where do I start? Can someone please explain this to me? Thank you!

### Re: Question 12.19

Posted: **Mon Jul 31, 2017 12:03 am**

by **Ben Rolnik 1D**

yeah this is a tricky log based question!

Essentially, once you understand that the question is basically saying "X" initial HCL and ".12X" change in HCL

then you SECOND molarity will be log(.12x) .... and the principle of logs is: log(A * B) = Log(a) + log(b).

therefore, log(.12x) = log(.12) + log(x).... if you take log(.12) you see it's "- .92" ... so that's the difference... whether or not you add it or subtract it kind of depends on the question

ph = -log(.12x) -log(.12) - log(x)

therefore I think what will end up happening is that you are ADDING two pH values together +.92 and + (-Log(X)) which will increase the pH.

I would double check the answer on this but do the logs make sense at least?

### Re: Question 12.19

Posted: **Sat Dec 02, 2017 5:58 pm**

by **dstemp**

In the solution for this question is says : pH= -log(.12[HCl]/[HCl])=0.92 ..... I understand why you would want to use logs here, but why can't you just do -log(0.12[HCl])? Why is it necessary to divide by the concentration of HCl too?

### Re: Question 12.19

Posted: **Sat Dec 02, 2017 6:42 pm**

by **Maeve Gallagher 1J**

You are solving for the change in pH, so you need to do final-initial. pH final is .12[HCl] and the initial is [HCl], so your equation for the change in pH would be -log(.12[HCl]) - -log([HCl]). Because of the log rules, where subtracting two log equations is the same as the log of the quotient between the two, you get that the change in pH = -log (.12[HCl] / [HCl]). This then allows you to cancel out the [HCl], making it possible to just take the log of .12 to get the change in pH.