How do I covert H+ ions(2.9*10^-4) to ph? I need the keys used specifically.

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

Posted on Mar 05, 2009

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 | Calculators

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

pH is defined as the negative logarithm of the hydrogen ion activity in a solution:

The value is defined as the molar concentration of H+ Ions, multiplied by a correction factor, the activity coefficient. The activity coefficient is required because in nature there are rarely actual H+ ions in a solution. For example, in water H+ ions combine with water molecules to Oxonium and Hydronium ions:

The associated products are less reactive than H+ ions would be, so the activity coefficient is always greater than 0 and smaller than 1. It depends on plenty of factors, e.g. the temperature of the solution, or the presence of other substances, and is very difficult to determine precisely.

For this reason, in water solutions the pH is usually approximated by:

for other solutions (not in water), similar approaches exist, but the resulting "pH" equivalents are not directly comparable.

Long story short: get the H3O+ concentration in mol/L by measuring it, or from the problem text. For example, if the concentration is 0.003 mol/L, key in:

and get a pH of 2.52

The value is defined as the molar concentration of H+ Ions, multiplied by a correction factor, the activity coefficient. The activity coefficient is required because in nature there are rarely actual H+ ions in a solution. For example, in water H+ ions combine with water molecules to Oxonium and Hydronium ions:

The associated products are less reactive than H+ ions would be, so the activity coefficient is always greater than 0 and smaller than 1. It depends on plenty of factors, e.g. the temperature of the solution, or the presence of other substances, and is very difficult to determine precisely.

For this reason, in water solutions the pH is usually approximated by:

for other solutions (not in water), similar approaches exist, but the resulting "pH" equivalents are not directly comparable.

Long story short: get the H3O+ concentration in mol/L by measuring it, or from the problem text. For example, if the concentration is 0.003 mol/L, key in:

and get a pH of 2.52

Feb 02, 2011 | Texas Instruments TI-30XA 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

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 Calculators

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,

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.

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.

Oct 06, 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 Calculators

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

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

Apr 22, 2008 | Texas Instruments TI-30XA Calculator

131 people viewed this question

Usually answered in minutes!

And what about the reverse project. Which key on the calculator gives anti-log?!

×