### page 161 of the course reader

Posted:

**Thu Nov 24, 2016 8:32 am**Why on page 161 when we got x=9.2 x 10^-6

did we take the ph= -log[H3O+]=5.04 and not take 10^-6?

did we take the ph= -log[H3O+]=5.04 and not take 10^-6?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=56&t=17132

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Posted: **Thu Nov 24, 2016 8:32 am**

Why on page 161 when we got x=9.2 x 10^-6

did we take the ph= -log[H3O+]=5.04 and not take 10^-6?

did we take the ph= -log[H3O+]=5.04 and not take 10^-6?

Posted: **Thu Nov 24, 2016 9:56 am**

The question is asking for the pH of NH4Cl (an acidic salt) and because it dissociates into NH3 and H3O+, we need to find the concentration of H3O+ by doing an ICE chart. At equilibrium we find that the concentration of of H3O+ is x, and by solving for x we get 9.2e-6, so by using the equation pH=-log[H30+] we get 5.04. If you're asking why we didn't simply take the exponent on 10 (to get pH=6) then I assume that because 9.2 is very close to 10, just taking the exponent isn't accurate enough. Also, since we're allowed to use a calculator on the midterm, it's safer to plug in the value for H3O+, as it is more accurate.