Destroyer Gchem#81?

[mcdds75]

can someone help me to understand the pH question, I got the correct answer by luck I believe b/c my math isn't the same as the explanations. I stopped at 9x10^-7=x^2 then took the square root of 9 to get 3 and 14-3=11
I have know idea what they did, but my math sucks!

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[I-Baby]

can someone help me to understand the pH question, I got the correct answer by luck I believe b/c my math isn't the same as the explanations. I stopped at 9x10^-7=x^2 then took the square root of 9 to get 3 and 14-3=11
I have know idea what they did, but my math sucks!

At the point where you stopped, you would need to get the square root for both sides of the equation to get x. In this problem x= 9.5*10^-4. This is your concentration of OH-. Now that you have the concentration of OH-, you can find the pOH, and 14-pOH will give you your pH. Hope that helped clarify a little as to how they did it, but your way seemed to work somehow...

 

[mcdds75]

yeah, I got how to get the pH, it is the square root of 9x10^-7 that I don't understand? The next step was 90x10^-8, then took the square root to get -log 9.5x10^-8...why do you go from 9 to 90?

Thanks!!!

 

[mistero]

It is much easier to get the square roots of an even log, so you just change it to 90 from 9 and make x10^-8 instead of 7. Hope that helps.

 

[mcdds75]

Yes, that was the only reasoning that I could come up with, but I needed re-assurance!!
I feel better now, thanks.

 

[msavaya]

Hey.. another reason could have been that they wanted 10^-7 to be an even number, so they made it 10^-8, so that when they did take the square root of it, they can easily divide it by 2 to get 10^-4. Get it?

 

[I-Baby]

Also, what I decided to do was memorize the logs of 1 to 10. They are real simple and follow a trend. To solve for pH or pOH from x:

If x=9.5*10^-4, then pH or pOH (depending on whether you're working with Ka/Kb) will be 4-log 9.5. The log of 9.5 is roughly .9, so 4-0.9 = 3.1.