Ph calculation If I have say 1 litre of water at pH 3.0 and add 10 litres of water at pH 5.5, what is the resultant pH? Is there a formula that I can apply? Thanks

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.

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.

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.

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

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.

Hello,
This post answers two questions

1. How to obtain the concentration knowing the pH?
2. How to obtain the pH knowing the concentration

If you are not interested in the theory, jump to the examples at the end of the post.

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)
Examples

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.

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.

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.

Sir,

Please inform me who is the authorized spares/service agent at Vashi Navi Mumbai for Braun mixie 1 litre.
I want a wet grinder 1 litre jar with new blade.

Regards,
Ranganathan

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))))) ^^