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Chemistry problem How do I find the antilog of -PH on MY TI30XA calculator?

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You enter a pH value and change the sign to minus, then hit 2nd LOG which is 10^x. The x value is what you're looking for. To put the answer in scientific notation, press 2nd 5 to do so.

Example:

For a pH = 4.1

-4.1 2nd LOG will return 0.0000794

Hiting 2nd 5 will express it as 7.94 -5, giving you the [H+] concentration.

Posted on May 17, 2008

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1 Answer

How do I get the antilog on the Sharp el-531x scientific calculator for doing ph calculations


Try <second function> log, which is the 10 to the power of x key.
Good luck,
Paul

May 27, 2015 | Office Equipment & Supplies

1 Answer

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.


Dec 07, 2009 | Texas Instruments TI-30XA Calculator

1 Answer

I am trying to do ph problems for chemistry and I cannot figure out how to use the log button right i keep getting an error


Hello,
To use the log function press [LOG] enter the number, close the right parenthesis and press [ENTER]
[LOG] 1 [ ) ] [ENTER] gives 0.

I have a hunch your problem is not the syntax of the log function on this calculator, but the calculation of the pH from the concentration or the calculation of a concentration from the pH. So I am inserting a copy of a previous post on the question. Use it as an exemple.

Let y=10^(x) 10 to the power of x
Take the log of both tems of the equality. You get log(y)=log[10^(x)] where I used square brackets for clarity. But from the general properties of logarithms

log(b^(a)) = a*log(b)

Applied to our expression above log(10^x)=x*log10
But since we are using log as log in base 10, log_10(10)=1 so log(y)=x
We thus have two equivalent relations

y=10^x <----> x=log(y) The double arrow stands for equivalence.

If y is the log of x, then x is the antilog of y

Your question: With log_10 standing for logarithm in base 10
pH=-log_10(c) where c= concentration. Then log_10(c)=-pH
The equivalence above translates as
log_10(c)=-pH is equivalent to c=10^(-pH)

The concentartion corresponding to a pH of 7.41 is
c=10^(-7.41)=3.89x10^(-8)

Hope it helps

Oct 28, 2009 | Texas Instruments TI-30 XIIS Calculator

1 Answer

How do i use the inv log for calculations on my sharp el-531wh


Hello,

Let y=10^(x) 10 to the power of x
Take the log of both tems of the equality. You get
log(y)=log[10^(x)] where I used square brackets for clarity. But from the general properties of logarithms

log(b^(a)) = a*log(b)
Applied to our expression above
log(10^x)=x*log10
But since we are using log as log in base 10, log_10(10)=1
so
log(y)=x
We thus have two equivalent relations
y=10^x <----> x=log(y) The double arrow stands for equivalence.

If y is the log of x, then x is the antilog of y

Your question: With log_10 standing for logarith in base 10
pH=-log_10(c) c= concentration
Then log_10(c)=-pH the equivalence above translates as
log_10(c)=-pH is equivalent to c=10^(-pH)

The concentartion corresponding to a pH of 7.41 is
c=10^(-7.41)=3.89x10^(-8)

You enter it as follows
10[^]7.41[(-)][ENTER/=]

Hope it helps

Aug 09, 2009 | Sharp EL-531VB Calculator

2 Answers

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.

May 29, 2009 | Texas Instruments TI-83 Plus Calculator

3 Answers

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))))) ^^

May 10, 2008 | Texas Instruments TI-83 Plus Calculator

1 Answer

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

Apr 22, 2008 | Texas Instruments TI-30XA Calculator

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