How do I get the inverse log of -PH on my TI-30XA calculator?

Ad

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

Posted on Jan 01, 2009

Ad

Hi there,

Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this issue over the phone in a minute or two.

Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

Here's a link to this great service

Good luck!

Posted on Jan 02, 2017

Ad

The inverse of the log function is the power function.

For log in base 10 that inverse is 10 to a power of

More generally, let b be the base of the logarithm. If y=log_b (x) then x=b^y

For your case log=log_10, to calculate the inverse you perform 10^(-2)=0.01=1/100

On calculators the log in base 10 and its inverse share the same physical key. One is accessed directly, the other is the shifted key function.

For log in base 10 that inverse is 10 to a power of

More generally, let b be the base of the logarithm. If y=log_b (x) then x=b^y

For your case log=log_10, to calculate the inverse you perform 10^(-2)=0.01=1/100

On calculators the log in base 10 and its inverse share the same physical key. One is accessed directly, the other is the shifted key function.

Sep 09, 2011 | Texas Instruments TI-30XA Calculator

The inverse of the common log function is the power function with base 10 (10^). It shares the same physical key as the log. You acees the function by pressing [2nd] [log]

Sep 09, 2011 | Texas Instruments TI-30XA Calculator

You calculate the pH with the relation

pH=-log(c) where c is the value of concentration in mol/L.

ex: c[H+]= 3.57x10^(-6)

pH: (-) [LOG] 3.57 [2nd][EE] [(-)] 6 [ ) ]

Here you have 2 change sign key press [(-)]

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

c= 10^(-pH)

Ex pH=8.23

c: [2nd][LOG] [(-) 8.23 [)]

General shortcut.

If concentration is of the form c[H+]= 1*10^(-a) where a is an integer between 0 and 14, the pH is equal to the positive value of exponent.

Ex:c[H+]=1x10^(-5) Mol/L then the pH is just the positive value of the exponent, ie 5 because pH=-log(1.x10^-5) =- log1 -log(10^-5)=0 -(-5)=5

Above I use the rule log(axb)=log(a)+log(b) , the rule log(a^b) = b*log(a)and the facts that log(1)=0 and log(10)=1, the last being true if log stands for log in base 10.

pH=-log(c) where c is the value of concentration in mol/L.

ex: c[H+]= 3.57x10^(-6)

pH: (-) [LOG] 3.57 [2nd][EE] [(-)] 6 [ ) ]

Here you have 2 change sign key press [(-)]

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

c= 10^(-pH)

Ex pH=8.23

c: [2nd][LOG] [(-) 8.23 [)]

General shortcut.

If concentration is of the form c[H+]= 1*10^(-a) where a is an integer between 0 and 14, the pH is equal to the positive value of exponent.

Ex:c[H+]=1x10^(-5) Mol/L then the pH is just the positive value of the exponent, ie 5 because pH=-log(1.x10^-5) =- log1 -log(10^-5)=0 -(-5)=5

Above I use the rule log(axb)=log(a)+log(b) , the rule log(a^b) = b*log(a)and the facts that log(1)=0 and log(10)=1, the last being true if log stands for log in base 10.

May 17, 2011 | Texas Instruments TI-30XA Calculator

Hi,

Ex: calculate -log(15.32)

Here is a picture of the change sign

Hope it helps

Ex: calculate -log(15.32)

- Calculate the log 15.32 [LOG] [=] result 1.185258765
- Then press the change sign key (-) result -1.185258765

Here is a picture of the change sign

Hope it helps

Dec 13, 2009 | Texas Instruments TI-30XA Calculator

Hello,

This post answers two questions

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)

**Example**s

1. 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)

Calculating the pH

Shortcut:

For all H+/H3O+ concentrations of the form**1.*10^(a)** where a is** an integer number between 0 and -14**, the pH is the negative value of the exponent.

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)

**8 (-) [2nd][10 to x] [*] 3.567 [LOG] [=] (-)**

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.**You may notice that it is entered in the reverse order of the defining relation **- log(3.567*10^(-8)).

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.

Hope it helps** **and thank you for using FixYa

And please, show your appreciation by rating the solution**.**

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.

Dec 07, 2009 | Texas Instruments TI-30XA 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

Hi,

Sorry to contradict you but there are many types of logarithms, the most important ones are

**[LOG]**, and the natural logarithms are labeled** [LN]**.

The inverse function of the natural log function is the exponential (e^(x)), and the inverse of the log in base ten function is the function ten to the power of. It is called (sometimes) the antilog

Ex:

Question What is the antilog of 3.5678?

Answer The antilog of 3.5678 is 10^(3.5678) = 3696.579068

Verification: log(3696.57908) =3.5678

Hope it helps.

Sorry to contradict you but there are many types of logarithms, the most important ones are

- the common logarithms (log in base 10),
- the natural logarithms (logarithms in base e)
- the binary logarithms (logarithms in base 2)

The inverse function of the natural log function is the exponential (e^(x)), and the inverse of the log in base ten function is the function ten to the power of. It is called (sometimes) the antilog

Ex:

Question What is the antilog of 3.5678?

Answer The antilog of 3.5678 is 10^(3.5678) = 3696.579068

Verification: log(3696.57908) =3.5678

Hope it helps.

Nov 29, 2009 | Texas Instruments TI-30XA Calculator

Press "2nd LN" which gives you the exponential (inverse natural log) of 7.41 = 1652.426....

Pressing LN will give you back 7.41

Same procedure to use logs to base 10, use LOG and 2nd LOG

Pressing LN will give you back 7.41

Same procedure to use logs to base 10, use LOG and 2nd LOG

Dec 07, 2008 | Texas Instruments TI-30XA Calculator

The problem is simple. You're trying to get an [H+] concentration which is obviously going to have a value of some number times ten raised to a negative power. Therefore, you have to insert the negative value of the pH into the 10^(x). When you insert said negative number you will come out with the right answer.

i.e.

The pH of a sample of human blood was calculated to be 7.41. What is the [H+] concentration of the blood?

10^(-7.41) = [H+]

[H+] = 3.9 E-8

(the answer should only have two sig. figs because the pH has two digits after the decimal.

i.e.

The pH of a sample of human blood was calculated to be 7.41. What is the [H+] concentration of the blood?

10^(-7.41) = [H+]

[H+] = 3.9 E-8

(the answer should only have two sig. figs because the pH has two digits after the decimal.

Jun 10, 2008 | Texas Instruments TI-84 Plus Calculator

Use the inverse log key sequences "2nd LN" or "2nd LOG" depending on whether you are using natural logs or logs to base 10

Dec 10, 2007 | Texas Instruments TI-30XA Calculator

3,938 people viewed this question

Usually answered in minutes!

I have the same problem. I am trying to find the hydronium ion concentration and i keep getting a crazy number. The "enter the pH then 2nd then LOG then 1/x " thing doesn't work for me.

the same thing isn't working for me either-the enter the # then 2nd log then 1/x-not right answer...

×