Question about Texas Instruments TI-30XA Calculator

Hello,

The pH is defined as the negative log in base 10 of the H+ (H3O+) ion concentration. The concentration of your solution is 1.2x10^(-5). You calculate the log of 1.2x10^(-5) which is -4.9208188754, let us just say it is equal to -4.921.

Once result is displayed press the change sign key (-) or on some calculators [+/-] and calculator will take the negative of the result.

So your pH is about 4.921

Another less efficient way to do it is to enter

0-log(1.2x10^-5) [ENTER]. The result will be positive.

The change sign (-) is smaller than the regular MINUS sign. It is used with negative exponents and to change the sign (hence its name) of a displayed result.

Posted on Oct 06, 2009

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Posted on Jan 02, 2017

pH=-log[c]

concentration=10^(-pH)=10^(-4.32)=4.786*10^(-5) mol/L

concentration=10^(-pH)=10^(-4.32)=4.786*10^(-5) mol/L

Mar 11, 2016 | Texas Instruments TI-30Xa Scientific...

The graph of y=22log(10) is a straight line, though it should go through (0,0) and (1,22). Log(10) is a constant 1, so the function is simply y=22x .

If you want "twenty-two times log-base-ten of x", write it as 22*log(x) .

If you want "twenty-two times log-base-ten of x", write it as 22*log(x) .

Aug 11, 2012 | Texas Instruments TI-89 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

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

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,

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

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 &...

Basically you first have to put your answer into scientific notation by pressing the 3rd key then the 6 key. The screen display should change to one big zero and two small zeros.

I was workinggon the same problem with a -log of a chemistry problem to get pH and I knew the answer because I calculated it on it TI-85, but we are not allowed to use graphing calulators on the exam (:-((() so I had to figure out how to do it on this one!

For example, I had pH =-log(2.828947x10^4)

On the TI-36X you change to scientific notation using the instructions above. Once the three zeros enter on the screen:

[4]

[-]

[2nd]

[10^x]

1^-04 should appear on screen, then you multiply by 2.828947

[x]

[2][.][828947]

then your screen will show the answer in scientific notation.

2.828947 in big numbers and in small numbers -04

From there you simply press:

[log]

and your answer will show up as a negative number. Just take the absolute value of that and you have your answer. Took me like 30 minutes the night before a huge chemistry exam to figure that out! Hope it helps someone!

I was workinggon the same problem with a -log of a chemistry problem to get pH and I knew the answer because I calculated it on it TI-85, but we are not allowed to use graphing calulators on the exam (:-((() so I had to figure out how to do it on this one!

For example, I had pH =-log(2.828947x10^4)

On the TI-36X you change to scientific notation using the instructions above. Once the three zeros enter on the screen:

[4]

[-]

[2nd]

[10^x]

1^-04 should appear on screen, then you multiply by 2.828947

[x]

[2][.][828947]

then your screen will show the answer in scientific notation.

2.828947 in big numbers and in small numbers -04

From there you simply press:

[log]

and your answer will show up as a negative number. Just take the absolute value of that and you have your answer. Took me like 30 minutes the night before a huge chemistry exam to figure that out! Hope it helps someone!

May 09, 2009 | Texas Instruments TI-36 X Solar 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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