Question about Texas Instruments TI-30XA 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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Posted on Jan 02, 2017

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pH=-log[c]

concentration=10^(-pH)=10^(-4.32)=4.786*10^(-5) mol/L

concentration=10^(-pH)=10^(-4.32)=4.786*10^(-5) mol/L

Mar 11, 2016 | Texas Instruments TI-30Xa Scientific...

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

Good luck,

Paul

Good luck,

Paul

May 27, 2015 | Office Equipment & Supplies

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.

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

To obtain the [H+] you need to calculate the antilog. You write the definition in the form

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

Exemple: let the pH=5.5, what is the H+ concentration?

With [(-)] being the change sign key, then

[H+]:

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

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

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

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

If

Your question: With log_10 standing for logarithm in base 10

The equivalence above translates as

log_10(c)=-pH is equivalent to

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

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

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

If

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

"2nd" + "log"

May 29, 2009 | Texas Instruments TI-84 Plus Silver...

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

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))))) ^^

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

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

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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