## Reduction of Acid 12.19

Kawsar Nasir 3k
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

### Reduction of Acid 12.19

12.19 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 understand how to calculate Kb values, but do I have to subtract by .88? So when I create an ice box, would I write 1-.88 and x-.88 for the other chemicals in the reaction? thanks!

Umair Khan 2G
Posts: 32
Joined: Fri Sep 25, 2015 3:00 am

### Re: Reduction of Acid 12.19

Remember that HCl is a strong acid so it completely dissociates. This mean thats no there is no need for an ice box because whatever the original concentration of HCl is, the value will go to zero.
For example if you have 0.1M of HCl you will get .1M of H30+
If you reduce the concentration by 12%, you will get 0.012M of HCl and H30+.
Doing final-intial of the pH: -log(0.012)-(-log(.1)) which will equal 0.92
Another way of doing this is just taking the negative log of the percent change:
-log(.12)= 0.92

Chem_Mod
Posts: 18880
Joined: Thu Aug 04, 2011 1:53 pm
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### Re: Reduction of Acid 12.19

Hi Kawsar, Umair is correct in that an ICE table is unnecessary since your acid HCl will dissociate completely. To solve this problem, you have to consider the property of logs. Lets say that our initial [HCl]=x so our reduced acid will be [HCl]reduced=0.88x. When we take the logs to calculate pH, they will be -log(x) and -log(0.88x). One property of logs is that $log(xy)=log(x)+log(y)$. So we can expand the equation for our reduced acid to be $-log(0.88)-log(x)$, therefore -log(0.88) would be the pH difference.

Kai_Chiu 1F
Posts: 19
Joined: Fri Sep 29, 2017 7:03 am

### Re: Reduction of Acid 12.19

Chem_Mod wrote:Hi Kawsar, Umair is correct in that an ICE table is unnecessary since your acid HCl will dissociate completely. To solve this problem, you have to consider the property of logs. Lets say that our initial [HCl]=x so our reduced acid will be [HCl]reduced=0.88x. When we take the logs to calculate pH, they will be -log(x) and -log(0.88x). One property of logs is that $log(xy)=log(x)+log(y)$. So we can expand the equation for our reduced acid to be $-log(0.88)-log(x)$, therefore -log(0.88) would be the pH difference.

I don't really understand what you mean when you say that the difference is -log(0.88); how do you conclude that -log(x) = 0? Does x = 1 because the 0.88 is from subtracting 12%?