1. Vinegar has a pH of 2.8. Tomato juice has a pH of 4.0
a) Determine the pH of something that is twice as acidic as tomato juice
b) Determine the pH of something that is one-half as acidic as vinegar
2. In 1995, the world population was 5.8 billion. According to United Nations' forecasts, the population will grow to 8.0 billion by 2025. Suppose the prediction were true. What is the average annual growth rate?
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I don't understand the first one at all. Our class never went over the pH system very well. Only decibels and energy. So I don't get it.
And for the second question, I tried doing this, but I think I'm wrong:
8000000 = 5800000 x N^30
8000000/5800000 = N^30
log(8000000/5800000) = 30 x logN
Can anyone help? I have a math test tomorrow and I need to do good on the rest of them this semester. :'(
I have no idea what the hell to do.
Quote from: Cosmo on November 24, 2008, 05:13:11 PM
I have no idea what the hell to do.
Well, it is grade 12. So it'll probably be too advance for you. But thanks for taking time to try to help me out anyway. giggle;
okay so apparently pH is the negative of the base 10 logarithm of some stuff that determines acidity
so we have:
a) [tex]2.8 = - \log_{10}(blah)[/tex] so uh i guess since smaller numbers equals MORE ACIDY we get [tex]-\log_{10}(2 \cdot blah)=-\log_{10}(blah)-\log_{10}(2)[/tex] so uh then we get [tex]2.8-\log_{10}(2)[/tex] which is approximately 2.499 and uh since i have never dealt with any of this before and am figuring out everything from a sentence i read on wikipedia i'll let you do the rest and i provide no guarantee that this is even close to being right akudood;
Quote from: Taking a coffee break... on November 24, 2008, 05:14:18 PM
Well, it is grade 12. So it'll probably be too advance for you. But thanks for taking time to try to help me out anyway. giggle;
Grade 12 chemistry? Maths?
Okay, let's see if I could do this. :]
So, like pH=-log[H+] or [tex]10^{-pH}=[H+][/tex]; where [H+] denotes the hydrogen ion concentration.
1b. Determine the pH of something that is one-half as acidic as vinegar
Let's determine the [H+] concentration for vinegar:
[tex]10^{-2.8}=1.59*10^{-3}[/tex]
Divide that result by two.
Take the negative log again to get the final result:
3.1
Quote from: Taking a coffee break... on November 24, 2008, 04:56:09 PM
And for the second question, I tried doing this, but I think I'm wrong:
8000000 = 5800000 x N^30
8000000/5800000 = N^30
log(8000000/5800000) = 30 x logN
How you wrote that calculation it appears that you didn't need to use logarithms to solve the problem. You should try again.
What model are they asking you to use?