Question about Casio FX-260 Calculator

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Posted on Jan 02, 2017

Do you have the concentration? if YES, you CALCULATE the pH with the relation

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

ex: c= 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 [)]

Here is a screen capture to show you both calculations

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

ex: c= 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 [)]

Here is a screen capture to show you both calculations

Dec 05, 2011 | Texas Instruments TI-Nspire Graphic...

To get the concentration C of hydronium ions use the relation

C=10^ (-pH)=10^ (-7.41)=3.89045*10^(-08) or just C=3.89*10^(-8) mol / L

The inverse of the function log in base 10 is 10 to the power of.

pH=-log(C) and C=10^(-pH)

C=10^ (-pH)=10^ (-7.41)=3.89045*10^(-08) or just C=3.89*10^(-8) mol / L

The inverse of the function log in base 10 is 10 to the power of.

pH=-log(C) and C=10^(-pH)

Jul 21, 2011 | Casio FX-115ES Scientific 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

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 20, 2010 | Texas Instruments TI-30XA Calculator

What do you mean by solving for the pH?

Do you have the concentration? if YES, you CALCULATE the pH with the relation

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

ex: c= 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 [)]

Here is a screen capture to show you both calculations

Do you have the concentration? if YES, you CALCULATE the pH with the relation

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

ex: c= 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 [)]

Here is a screen capture to show you both calculations

Apr 25, 2010 | Texas Instruments TI-84 Plus 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

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

"2nd" + "log"

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

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

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