Can you check if I am right with this pH question!?

This is pretty easy (I think) but I'm trying to get perfect on this assignment, so could you check my reasoning?



The range of blood pH that is considered normal in humans is 7.35 to 7.45. If a person's blood pH changes from 7.35 to 7.45, what is the change in hydronium ion concentration ([H30+])?



a) 1.3 * 10^-8 M

b) 9.2 * 10^-9 M

c) -9.2 * 10^-9 M

d) -1.3 * 10^-8 M

e) -2.3 * 10^-8 M



I get (c) and I found this by actually calculating the hydronium ion concentration at each end of the acceptable pH range and finding the difference. This was done by taking ten to the power of the negative value of the pH. Since an increase in pH results in a decrease in hydronium ion concentration, the answer should be negative. Can you check if I am right!?



ALSO, I tried it this way and it didn't work, and I'm curious why: how come it doesnt work out if you just subtract the pH values and then take the ten to the power of the negative difference, to find the H ion concentration? (do you get understand that? haha)Can you check if I am right with this pH question!?
[H+] at pH 7.35 = 10^-7.35 = 4.46*10^-8

[H+] at pH 7.45 = 10^-7.45 = 3.54*10^-8



Difference = 9.2*10^-9 answer is b)



The difference is always quoted as positive. Think about this:

what is the difference between 2 an 4? answer =2

What is the difference between 4 and 2 answer : = 2

Nowhere do you say negative,



Also Answer: Remember that pH is a logarithmic scale: You will know from your mathematics classes that if you subtract logarithms, you are not finding the difference, but you are dividing. That is why you get an incorrect answer.



If you divide 4.46*10^-8 by 3.54*10^-8 you get 1.259



If you subtract 7.45-7.35 you get 0.1. = 10^0.1 = 1.259



This latter method is not the way to do this type of problem because you are not finding the difference.Can you check if I am right with this pH question!?
Yes. pH of 7.35 is [H+] = 4.47*10^-8;

pH of 7.45 is [H+] = 3.55*10^-8.

Difference is c.



You cannot use your second technique because subtracting logs does not do the same thing as subtracting the exponentials; it is the result of dividing.



10^2 - 10^1 =(100-10) = 90.



Take logs and use your technique gives you 2-1 = 1 which raised to the tenth power is 10 which is not the same thing at all but which is 100/10.
yes, that's correct.

i thought i had to divide -9.2 * 10^-9 by 3 considering there's 3 H in H3O+, but i assumed that there's only 1 H+ and the rest is water there?

..

and ur 2nd question, of course it doesn't work.

is 10^(-7.45 - (-7.35)) same with 10^(-7.45) - 10^(-7.35)?

lol.

xD
/////////////// definitions



pH = -LOG([H(+)])



';pH equals the negative LOG of the Hydrogen ion concentration, expressed in terms of moles per liter';



///////////// calculations



7.35 = -LOG([H(+)])

-7.35 = LOG([H(+)])

10^(-7.35) = [H(+)]

[H(+)] = 4.46E-8



7.45 = -LOG([H(+)])

-7.45 = LOG([H(+)])

10^(-7.45) = [H(+)]

[H(+)] = 3.54E-8



DELTA([H(+)]) = 3.54E-8 - 4.46E-8 = -9.2E-9



The change is actually a decease, so the answer is negative, so ';c'; is correct. Higher pH values mean a decrease in the hydrogen ion concentration. However, there is no such thing as an actual negative concentration. What has been calculated is a change in pH. This is sort of like working with electrical engineering where voltage is the difference between two points in a circuit, not a single value in itself.



///////////////// how it work



You can't subtract LOG values and have them behave like ordinary numbers. On a LOG scale, subtraction is equivalent to division and addition of LOGS is multiplying the numbers they represent.



For example, LOG(100) - LOG(10) = 2-1 = 1. 10^(1) = 10 and 100/10 also = 10.



LOG(100) + LOG(10) = 2 + 1 = 3. 100*10 = 1000. LOG(1000) = 3.



Not too hard once you get the haing of this sort of thing.



FYI: The natural log function is just what happens when 1/x is integrated. Nice to know what natural logs actually represent. Too bad they never actually explained this in college.