Chem Help!!I Dont Understand logarithms and pH calculations....!!!!!?

Question

I AM COMPLETELY LOST!!!
I dont even understand what a logarithm is and I am doing HARD equations using them in m chem class.
I dont understand how to answer the following problems...If someone could please EXPLAIN THE STEPS to getting the answers that would be WONDERFUL. THANK YOU!!!

Find the pH if the concentration of [H3O] is 0.25M
pH= -log[H3O]
pH= -log[0.25]
pH= 0.60
???

Find [H3O] if pH is 2.7
[H3O]= 10 to the -pH
[H3O]= 10 to the 27th power
[H3O]= 0.200
???


Find the [H] of a pH of 6.8
1.58 x 10 to the negative 7 power
???


Find the [OH] if the [H3O] is 0.000035M
2.86 x 10 to the negative 10 M
???
SOMEBODY PLEASE PLEASE EXPLAIN!!!! PLEASE!!!!!!

Answer 1

To use logs for pH calculations, all you have to do is remember what buttons to press in your calculator when given certain information. Just keep doing questions and it'll be easy soon enough.

To find the pH of a question... remember the formula. pH = -log[H]. ([H] = [H3O])
Just plug it into your formula.
Note: In calculations with logarithms only the numbers to the right of a decimal is considered significant. So your final answer should have 2 in this case.
You're right, it's 0.60

To find the concentration of [H3O], you need to use the negative log of the pH. To do the negative log you can use your calculator. (I'm using a TI-83 Plus, but it should be the same for any calculator).
1: Press [2nd] [LOG] and insert the pH value as a negative. The negative log of 2.7 is just 10^-2.7.
2: Answer should be 0.002 or 2 x 10^-3

Same thing as before, just plug it into your calculator to find the negative log.
Answer: You're right, just be careful of your significant digits.

Last question is just a little bit more complicated, but still simple nonetheless. There's multiple ways of doing it and whatever jumps out as you first should be the way you do it.
Option 1: You should know that [OH] x [H3O] = 1.00 x 10^-14
- so just manipulate the formula to figure it out
- [1.00 x 10^-14] / 0.000035M
Answer: 2.9 x 10^-10

Option 2: You should also know that pH + pOH = 14.
- find the pH since you're given the concentration of H3O by using your formula
- -log(0.000035M) = 4.4559...
- that's your pH and since pH + pOH = 14... then pOH must be 14 - 4.4559 = 9.5440...
- that's your pOH... take its negative log to find [OH] (just like pH and [H3O])
- 10^-9.5440...
Answer: 2.9 x 10^-10

Answer 2

If you have a positive number m, it can be represented by 10^-b if m is less than 1 or 10^b if m is greater than 1. Then we can write that
-b = Log m m is less than 1
b = Log m m is greater than 1.
We use the convention that "Log" is alway a base-10 logarithm.
Logs are (or were) very useful in math operations, because when you add logs, it is the same as multiplying the corresponding numbers. In other words, if you had 2 numbers, for the sake of discussion, both greater than 1, and they were m and n, Log m + Log n = Log (mn). So you could take the numbers m and n, fifnd their logarithms, add them together, and then find the "antilog" of the sum, which gave you the product mn. Doing powers became easier; once converted to logs, the logs could be multiplied or divided, and the resulting log converted back to the answer.

Chemists, being poor math people, and dealing with small numbers, used pH to describe the hydrogen ion content in water. It runs inversely to the hydrogen ion content. Thus, a pH of 2 means a [H+] of 10-2 and pH of 3 means a [H+] of 10-3. If the pH is between 2 and 3, the [H+] is larger than 10-3 but not as high as 10-2.

Hopefully, this is enough to get into pH conversions to [H+] and vice versa.
PROB 1. Looks OK to me.
PROB 2. You lost a decimal there, [H+]=10^-2.7
At this point we do a little trick, and rewrite this as
[H+]= 10^-3 x 10^0.3. Cranking this through your calculator, you will get an [H+] of 2 x 10^-3
PROB 3. This looks OK.
PROB 4. With this, you use the relation
[H+][OH-] = 10x10^-15. This is somewhat easier to work with than 1x10^-14 for problems like this. [H+] = 3.5x10-5 so [OH-] = 2.8x10^-10. This is OK.

Answer 3

If you really want to understand logs the best way to break out a college algebra book...

The short way is this.
In chemistry the only log that they use is the log base 10, which is you button on TI-83 calculator or equivalent, so I will show you those steps.

If you want to find a pH from a concentration then all you need to do is type in -log(concentration number) and it will give you the pH.
(note: (-) is the negative sign next to the decimal point in the TI-84)

pH = -log[ ]

if you need to find the concentration from a pH then you type in the reversal of log which is the 10^x button...(on the TI-84 it is the second button option for log)...

[ ] = 10^(-pH)

therefore problem two would look like 10^(-2.7) on your calculator which gives 0.001995

The third problem looks like the first just with a different notation for the hydronium ion. They sometimes use H instead of H3O:
Therefore on your calculator it would look like
10^(-6.8) = 1.58x10^7

the last problem remember that [OH] times [H3O] = 1.00x10^-14 therefore all you have to do is:
[OH]X[H3O] = 1.00x10^ -14
[OH] = (1.00x10^ -14)/[H3O]
[OH] = (1.00x10^-14)/(0.000035)
[OH] = 2.86X10^ -10

Hope that helps--
It would be better to try and find a tutor or someone who can show you in person
^ means raised to some power

Answer 4

pH can be solved through this equation.

pH = -log[H3O+]

To understand the concept of logarithm, let's replace pH with y and [H3O+] with x. So the equation then becomes

y = -log x

This equation means, to what exponent y should you raise 10 in order to get x. In equation,

10^-y = x.

Get it so far? We go about pH calculations using this theory. So for the first problem, it's just like saying

10^-0.60 = 0.25

I hope you get the concept of logarithm now. For computations using the calculator, well, that would depend on what calculator you use. You just have to study how to use it. =)