Question about Texas Instruments TI-30X-IISTK Scientific Calculator

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Posted on Jan 02, 2017

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SOURCE: chemistry problem

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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SOURCE: how do i use the inverse log operation on the TI-84

If I understand you correctly, the inverse of the log operation is exponentiation. So if your log = x, the inverse is 10 raised to the x power. On most TI calculators this is on the same key as the log function.

michael

Posted on Jun 13, 2008

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

SOURCE: 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

Posted on Jan 01, 2009

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

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

General shortcut.

If the concentration is of the form 1* 10^ (-a) where a is an integer between 0 and 14, the pH is given by a.

c= 1.* 10^(-5), pH=5

c=1.*10^(-11(, pH=11.

For other values, ex: c=3.68 *10^(-8)

pH: 3.68 [EXP] [+/-] 8 [LOG] [+/-] Result is 7.434

Above, the [+/-] key is the change sign key, just above the [7] key.

If the concentration is of the form 1* 10^ (-a) where a is an integer between 0 and 14, the pH is given by a.

c= 1.* 10^(-5), pH=5

c=1.*10^(-11(, pH=11.

For other values, ex: c=3.68 *10^(-8)

pH: 3.68 [EXP] [+/-] 8 [LOG] [+/-] Result is 7.434

Above, the [+/-] key is the change sign key, just above the [7] key.

Mar 13, 2010 | Casio FX-260 Calculator

First you have pH+pOH=14

Thus pOH=14-pH.

The problem with you question is that you do not specify the meaning of the 1.2 x 10^(-3). Is it the molar concentration of the hydrpbromic acid HBr.

I am going to answer your question but I must make some assumptions. You can imitate the reasonning to obtain your answer, once you have decided or looked up what the numbers mean.

Lets assume that HBr concentration is 1.2x10^(-3) mol/L. Since it is a strong acid, it dissociates completely, releasing 1.2x10^(-3) mol/L H+/H3O+ ion.

Thus the pH is

pH=-log(1.2x10^(-3)).

The result is 2.92

Here is a screen capture of the actual calculation. The minus signs are the change sign (-) to the right of the dot. To enter the power of 10 (represented by the E in display) use the (2nd)(EE) key sequence. To access the content of (Ans) memory, press (2nd) (change sign)

Thus pOH=14-pH.

The problem with you question is that you do not specify the meaning of the 1.2 x 10^(-3). Is it the molar concentration of the hydrpbromic acid HBr.

I am going to answer your question but I must make some assumptions. You can imitate the reasonning to obtain your answer, once you have decided or looked up what the numbers mean.

Lets assume that HBr concentration is 1.2x10^(-3) mol/L. Since it is a strong acid, it dissociates completely, releasing 1.2x10^(-3) mol/L H+/H3O+ ion.

Thus the pH is

pH=-log(1.2x10^(-3)).

The result is 2.92

Here is a screen capture of the actual calculation. The minus signs are the change sign (-) to the right of the dot. To enter the power of 10 (represented by the E in display) use the (2nd)(EE) key sequence. To access the content of (Ans) memory, press (2nd) (change sign)

Mar 07, 2010 | Texas Instruments TI-83 Plus Calculator

I suggest you reformulate the question. If you are doing pH questions, give the pH and we will show you how to get the concentration. Alternatively, give the concentration and we will show you how to obtain the pH. Keep in mind that in the context of pH questions, pH is restricred to the interval [0,14], and that imposes restrictions on the possible values of the concentration.

If c=concentartion, then

pH=-log(c)

This is equivalent to

c=10^(-pH)

Assuming that this is what you want to do (ie. calculate the concentration)

Press [2nd][log] to access the 10^ function

Screen displays 10^(

Enter the change sign [+/-]

Enter the pH value 4.62

Screen displays 10^( -4.62)

Press [=]

The result is 2.398832919E-05. To 3 significant digits the concentartion is c=2.39E-05 mol/L of H+/H3O+ ion

If c=concentartion, then

pH=-log(c)

This is equivalent to

c=10^(-pH)

Assuming that this is what you want to do (ie. calculate the concentration)

Press [2nd][log] to access the 10^ function

Screen displays 10^(

Enter the change sign [+/-]

Enter the pH value 4.62

Screen displays 10^( -4.62)

Press [=]

The result is 2.398832919E-05. To 3 significant digits the concentartion is c=2.39E-05 mol/L of H+/H3O+ ion

Mar 07, 2010 | Casio FX-115ES Scientific Calculator

Let us start with the definitions:

Let c be the concentration of H+/H30+ ions in the the solution (expressed in mol/L).

By definition the pH =-log(c)

Rewriting the equation as log(c)=-pH

Taking both members of the last equation to the power of 10

10^(log(c))=10^(-pH)

Since 10^(log(x))=log(10^x)= x (identity for inverse functions)

we get c=10^(-pH).

To handle concentartion-pH problems, you have two relations, namely

pH=-log(c) : to calculate pH from the concentrartion

and

c=10^(-pH): to calculate the concentration from the pH

A SHORTCUT WORTH REMEMBERING.

If you concentration is of the form c=1.*10^(-a) where a is between 0 and 14, the pH is simply the value of a.

Ex: c=1.*10^(-3), pH=3

c=1*10^(-8), pH =8.

Now let us take a more general example: c=6.54*10^(-9), pH=?

pH=-log(6.54*10^(-9))

You have to enter twice the change-sign [(-)], the one on the bottom row of the key pad.

Here is a screen capture from a TI84Plus calculator.

As you can see, you can use the general power key [^] to enter 10^(-9)

or the shortcut [2nd][ ' ] to access the (EE) which appears on screen as E. The second way is more efficient (left screen)

On the screen to the right you will notice that I did not close the parentheses, yet I obtained the correct result. However, I suggest you always close parentheses to avoid syntax errors.

Let c be the concentration of H+/H30+ ions in the the solution (expressed in mol/L).

By definition the pH =-log(c)

Rewriting the equation as log(c)=-pH

Taking both members of the last equation to the power of 10

10^(log(c))=10^(-pH)

Since 10^(log(x))=log(10^x)= x (identity for inverse functions)

we get c=10^(-pH).

To handle concentartion-pH problems, you have two relations, namely

pH=-log(c) : to calculate pH from the concentrartion

and

c=10^(-pH): to calculate the concentration from the pH

A SHORTCUT WORTH REMEMBERING.

If you concentration is of the form c=1.*10^(-a) where a is between 0 and 14, the pH is simply the value of a.

Ex: c=1.*10^(-3), pH=3

c=1*10^(-8), pH =8.

Now let us take a more general example: c=6.54*10^(-9), pH=?

pH=-log(6.54*10^(-9))

You have to enter twice the change-sign [(-)], the one on the bottom row of the key pad.

Here is a screen capture from a TI84Plus calculator.

As you can see, you can use the general power key [^] to enter 10^(-9)

or the shortcut [2nd][ ' ] to access the (EE) which appears on screen as E. The second way is more efficient (left screen)

On the screen to the right you will notice that I did not close the parentheses, yet I obtained the correct result. However, I suggest you always close parentheses to avoid syntax errors.

Feb 11, 2010 | Texas Instruments Office Equipment &...

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

It should be this sequence:

2.9 {EE} 4 {+ -> -} will give you the Sci.Not.

Then {=} (0.00029)

Then {LOG} (-3.537602002)

Then {+ -> -} (3.537602002)

pH=3.54

2.9 {EE} 4 {+ -> -} will give you the Sci.Not.

Then {=} (0.00029)

Then {LOG} (-3.537602002)

Then {+ -> -} (3.537602002)

pH=3.54

Feb 25, 2009 | Texas Instruments TI-30XA 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

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