Question about Texas Instruments TI-Nspire Graphic Calculator

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Posted on Jan 02, 2017

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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

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

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.

Posted on May 18, 2009

SOURCE: TI-nspire will not turn on

Yes we changed out the batteries on now the TI-84 will not turn on HELP

Posted on Nov 25, 2009

SOURCE: TI-Nspire 84 Plus keypad problem

Remove the TI84 Keyboard, insert the TI-Nspire keyboard and try to reset the calculator with a starightened paper clip.Calculator has a reset button on the back.

Posted on Feb 01, 2010

The pH scale is a logarithmic scale used to measure the concentrations of hydrogen/Hydronium ions in an aqueous solution.

If c[H] is the concentration of hydrogen ions in mol/L, the pH is defined as

pH=-log(c[H]).

Inversely, the concentration is given by

c[H]=10^(-pH)

The pOH=14-pH

If c[H] is the concentration of hydrogen ions in mol/L, the pH is defined as

pH=-log(c[H]).

Inversely, the concentration is given by

c[H]=10^(-pH)

The pOH=14-pH

Jan 19, 2012 | Office Equipment & Supplies

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...

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

To do this on the TI NSpire:

- Press [HOME] [1]
- Press [CATALOG] [1] [I] and scroll down until the integral symbol or integral( function is located. Press [ENTER] on either function.
- Input: x sin x, x, -1, 3
- Press [ENTER].

Jan 23, 2011 | Texas Instruments TI-Nspire Graphic...

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

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,

You should use the equivalence

**y=10^(x) is equivalent to x=log(y)**

You should not use word that have no meaning. x above is sometimes called antilog, not reverse log.

The relation linkin pH and pOH is as follows

** pH+pOH=14,** thus **pH=14-pOH**

Let c[H+] be the concentation (or if you use the hydronium ion, instead of H+ ) c[H3O+] the concentration of hydronium ion)

c[H+]=c[H3O+]=10^(-pH)

pOH= 6.95 thus pH=14-6.95 = 7.05

If pH=7.05 the c[H+]= 10 ^(-7.05) =8.9125E(-8)

pH=**10^ [X to the power] [(-)]7.05 **

The key/buttom to raise to the power is the one to the right of x squared

Hope it helps

You should use the equivalence

You should not use word that have no meaning. x above is sometimes called antilog, not reverse log.

The relation linkin pH and pOH is as follows

Let c[H+] be the concentation (or if you use the hydronium ion, instead of H+ ) c[H3O+] the concentration of hydronium ion)

c[H+]=c[H3O+]=10^(-pH)

pOH= 6.95 thus pH=14-6.95 = 7.05

If pH=7.05 the c[H+]= 10 ^(-7.05) =8.9125E(-8)

pH=

The key/buttom to raise to the power is the one to the right of x squared

Hope it helps

Sep 10, 2009 | Texas Instruments Office Equipment &...

Yes, the Ti nspire not only functions as a ti83 but, also operate w/ nice advance math software that any teacher would envey. It comes with the ti83 keypad plus software and great instructions to help the instructor at all levels of math learning. I don't know any better and it is a beautiful plan that ti has set up.

Apr 03, 2009 | Texas Instruments TI-84 Plus 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

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