Question about Texas Instruments TI-30XA Calculator

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Hello,

This post answers two questions

- How to obtain the concentration knowing the pH?
- How to obtain the pH knowing the concentration

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

1. Let the pH=5.5, what is the H+ concentration?

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

[H+]:

Calculating the pH

For all H+/H3O+ concentrations of the form

Concentration =10^(-3), pH=3

Concentration=10^(-11), pH=11

For other concentrations such as 3.567*10^(-8), one cannot use the shortcut above, but have to calculate the log of the concentration

[H+/H3O+] = 3.567*10^(-8)

pH= - log(3.567*10^(-8))

This is keyed in as follows (to minimize the number of parentheses)

Here you have two (-) change sign, the first is entered after the exponent of 10, the other at the end of the calculation to take the negative of the displayed result.

To verify your calculation, the result is 7.447696891 or just 7.45

If you have a problem with the first (-) try entering it before you type in 8.

Posted on Dec 10, 2009

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SOURCE: Find the antilog on the TI-83 Plus for finding pH in chemistry problems

how do i find the anilog on my calculator i have a texas instrument ti-30

Posted on Oct 15, 2008

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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.

Posted on May 29, 2009

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...

pH=-log( Concentration in mol/Liter)

May 08, 2014 | Texas Instruments TI-30XA Calculator

Don't use the - key to enter a negative value. Use the (-) key located to the right of the decimal point key.

Jan 31, 2011 | Texas Instruments TI-84 Plus Calculator

If you have the pH, and want to calculate the concentration that yields that value of the pH, you proceed as follows

c[H3O ] = 10^(-pH)

Ex pH=8.23

c[H3O ] : [2nd][LOG] [(-)] 8.23 [)] where [(-)] is the change sign key.

Here is a screen capture to show you how to calculate the pH (first example) and the concentration (2nd example)

c[H3O ] = 10^(-pH)

Ex pH=8.23

c[H3O ] : [2nd][LOG] [(-)] 8.23 [)] where [(-)] is the change sign key.

Here is a screen capture to show you how to calculate the pH (first example) and the concentration (2nd example)

Sep 03, 2010 | Texas Instruments TI-83 Plus Calculator

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

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

Basically you first have to put your answer into scientific notation by pressing the 3rd key then the 6 key. The screen display should change to one big zero and two small zeros.

I was workinggon the same problem with a -log of a chemistry problem to get pH and I knew the answer because I calculated it on it TI-85, but we are not allowed to use graphing calulators on the exam (:-((() so I had to figure out how to do it on this one!

For example, I had pH =-log(2.828947x10^4)

On the TI-36X you change to scientific notation using the instructions above. Once the three zeros enter on the screen:

[4]

[-]

[2nd]

[10^x]

1^-04 should appear on screen, then you multiply by 2.828947

[x]

[2][.][828947]

then your screen will show the answer in scientific notation.

2.828947 in big numbers and in small numbers -04

From there you simply press:

[log]

and your answer will show up as a negative number. Just take the absolute value of that and you have your answer. Took me like 30 minutes the night before a huge chemistry exam to figure that out! Hope it helps someone!

I was workinggon the same problem with a -log of a chemistry problem to get pH and I knew the answer because I calculated it on it TI-85, but we are not allowed to use graphing calulators on the exam (:-((() so I had to figure out how to do it on this one!

For example, I had pH =-log(2.828947x10^4)

On the TI-36X you change to scientific notation using the instructions above. Once the three zeros enter on the screen:

[4]

[-]

[2nd]

[10^x]

1^-04 should appear on screen, then you multiply by 2.828947

[x]

[2][.][828947]

then your screen will show the answer in scientific notation.

2.828947 in big numbers and in small numbers -04

From there you simply press:

[log]

and your answer will show up as a negative number. Just take the absolute value of that and you have your answer. Took me like 30 minutes the night before a huge chemistry exam to figure that out! Hope it helps someone!

May 09, 2009 | Texas Instruments TI-36 X Solar 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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