Question about Texas Instruments TI-83 Plus Calculator

Ad

Hi,

a 6ya expert can help you resolve that 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.

New users get to try the service completely Free afterwhich it costs $6 per call and covers almost anything you can think of (from cars to computers, handyman, and even drones).

click here to download the app (for users in the US for now) and get all the help you need.

Goodluck!

Posted on Jan 02, 2017

Ad

Number of moles [H+] in first solution =concentration of [H+] in mol/L times the number of liters. **[C1]=10^(-pH)=10^(-3) mol /L**

**Number of moles **N1=10^(-3) mol/L * 1L= **10^(-3) mol [H+]**

Concentration [C2] of second solution

[C2]= 10^(-5.5)=3.16227766*10^(-6) mol/L of [H+]

**Number of moles **of H+ in second solution

**N2**=[C2]*10 L =**31.6227766*10^(-6) mol of H+**

Total number of moles H+ in the new solution

**N=10^(-3) +31.6227766*10^(-6) =1.031622777*10^(-3) mol**

Concentration of H+ in new solution

[C]=**1.031622777*10^(-3) mol** / (1+10) L=9.358388878 *10^(-5) mol/L

Resulting pH=-log( 9.358388878 *10^(-5) =4.02787

**pH is 4.03**

Concentration [C2] of second solution

[C2]= 10^(-5.5)=3.16227766*10^(-6) mol/L of [H+]

Total number of moles H+ in the new solution

Concentration of H+ in new solution

[C]=

Resulting pH=-log( 9.358388878 *10^(-5) =4.02787

Jun 15, 2014 | Office Equipment & Supplies

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

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

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

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,

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 logarith in base 10

pH=-log_10(c) 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)

You enter it as follows

10[^]7.41[(-)][ENTER/=]

Hope it helps

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

If

Your question: With log_10 standing for logarith in base 10

pH=-log_10(c) 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)

You enter it as follows

10[^]7.41[(-)][ENTER/=]

Hope it helps

Aug 09, 2009 | Sharp EL-531VB Calculator

45 people viewed this question

Usually answered in minutes!

×