SOURCE: Find the antilog on the TI-83 Plus for finding pH in chemistry problems
I reccently had the same problem, but I have a Ti-84 Plus,and I am disgusted by how poorly this matter is covered by Texas Intruments. Esp. since the solution is rather simple.
The antilog key is [2nd] [log] or the 10^x. On your screen it should apear as 10^( and then you just input the value that you want to find the antilog of.
When used to solve pH problems remember to input the pH value as a negative number, since pH= -log[H+] then [H+]= antilog(-pH).
Hope that helped))))) ^^
SOURCE: antilog in chemistry problem
pH is minus (log to base 10) of the hydrogen ion activity of an aqueous solution, or (log to base 10) of (1/hydrogen ion activity)
To get the inverse log, i.e the hydrogen ion activity corresponding to a specified pH, simply enter the pH value and press
2nd
LOG
1/x
Answer 0.001
SOURCE: Find the antilog on the TI-84 Plus for finding pH in chemistry problems
I'm not specifically familiar with the TI83 or TI84 but I've used a lot of TI calculators in my time, so I'll give it a try. If your trying to find the antilog of a number in base 10 enter the number and hit the (10 to the X) button. If you're trying to find the antilog of a number in in base e (natural log), enter the number and hit the (e to the X) button.
SOURCE: simplifying fractions on the TI-30XA
Two ways:
1) hit the equal sign
2) hit 2nd and [d/c] (you can do this multiple times to see it as the most reduced improper fraction or as a mixed fraction)
SOURCE: How to use antilog?
Hello,
As you well know the pH is the negative of the log in base 10 of the H+/H3O+ ion concentration. If we use [H+] to represent that concentration, then pH=-log[H+].
To obtain the [H+] you need to calculate the antilog. You write the definition in the form log[H+] =-pH and then calculate 10 to the power of each member. The equality remains valid as both members are treated similarly. Thus
10^( log[H+] ) = 10^(-pH)
Since raising 10 to a power is the inverse function of taking the log in base 10, 10^(log(x))=log(10^(x)) = x (they are inverse of one another), you are left with
[H+]=10^(-pH)
Your calculator has a function [10 to x] accessed by pressing the [2nd] function key. To use it you must enter the negative value of the pH, press the [2nd] function key then the [10 to x], then the = key to get the result (concentration)
Exemple: let the pH=5.5, what is the H+ concentration?
With [(-)] being the change sign key, then
[H+]:[ (-) ] 5.5 [2nd][10 to x] [=]
The result is 0.000003162 or 3.16 x 10^(-6)
Hope it helps.
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