calculatin ph...

[albertlee]

-log(concentration of H+ ions)

 

the above is the formula for calculating the ph of an acid...

 

however, when the concentraion is 1M, the pH is 10.. but how could that be?? it should be less than 7, because it is an acid..

 

secondly, how do you calculate the "alkalidity" of pH??

 

thx

[Borek]

No. When the concentration of H+ is 1M, pH=0.

 

Best,

Borek

--

Chemical calculators at www.chembuddy.com

pH calculator

concentration conversion

[rakuenso]

so whats the relationship between the pKa and pH?

[albertlee]

so, Borek, how did you calculate that?????

 

oh come on, plz help...

[Tetrahedrite]

Simple:

 

get calculator, enter -log(1), answer = 0

 

A pH of ten would require a hydrogen ion concentration of 0.0000000001M

[Tetrahedrite]

so whats the relationship between the pKa and pH?

 

Where HA is a weak acid,

 

pKa = pH +([A-]/[HA])

[albertlee]

Simple:

 

get calculator' date=' enter -log(1), answer = 0

 

A pH of ten would require a hydrogen ion concentration of 0.0000000001M[/quote']

 

 

thx, but how come I get negative ph for 0.5M??

[jdurg]

Because you're calculating it wrong. -LOG(0.5) = 0.3 and not a negative number. Also remember that the pH scale doesn't stop at 0 and 14. It's very possible to have a pH higher than 14 or lower than 0.

[Dak]

-log(0.5)

 

=

 

log(0.5) X -1

 

=

 

-0.3 X -1

 

=

 

0.3

 

 

Where HA is a weak acid' date='

 

pKa = pH +([A-']/[HA])

 

just to test my memory/understanding, is the following right?

 

pKa = pH plus dissociation constant (Ka)

 

Ka = ([conk products]^no.mols)/[conk reagents]^no.mols)

 

so Ka{HA --> H⁺ + A^-} = ([H⁺]¹ + [A^-]¹)/[HA]¹

 

as it is a weak acid, [H⁺] will be small, and thus can be ingnored, giving Ka = [A-]/[HA]

 

Hence, pKa = pH + ([A-]/[HA])

 

?

[YT2095]

it`s also important to remember that PH is dependant on temperature also, IIRC all standards are taken at 298k at 101kpa.

the same condition where water has the PH of 7

 

(boil water and the PH drops, and it becomes acid with a ph of about 6.2ish )

 

just a little added extra for you

[albertlee]

thanks...

 

btw, how do you calculate pH higher than 7?? since they are for alkalis...

 

thx

[ecoli]

same way you do it for a chemical under ph lower then 7.

[Yggdrasil]

albertlee, remember the autodisociation of water:

 

H₂O <--> H⁺ + OH^-

 

Has the equilibrium constant, K_w = [H⁺][OH^-] ~ 10^-14. Therefore, you can use [OH^-] to calculate [H⁺].

 

[Edit: Kw ~10^-14, not 10¹⁴]

 

Also, pKa = pH + ([A-]/[HA]) is incorrect. The correct equation is called the Henderson-Hasselbach equation and takes the form:

 

pKa = pH + log([HA]/[A^-])

 

It's derrivation is as follows:

Ka = [H⁺][A^-]/[HA] (definition of Ka)

Now, take the -log of both sides:

-log(Ka) = -log([H⁺][A^-]/[HA])

-log(Ka) = -log([H⁺]) - log([A^-]/[HA])

Since pKa = -log(Ka) and pH = -log([H⁺]),

pKa = pH - log([A^-]/[HA]), or equivalently

pKa = pH + log([HA]/[A^-])

[albertlee]

so..

 

-log(OH-)??

but how do you turn that into a ph scale??

 

secondly, ph 10 means the solution is alkali... but why is 0.0000000001M of acid is 10??

 

plz help

[Dak]

no, alkaly = -log[H+].

 

there are free-floating H+ ions in water. acids make there be more H+ ions. alkalys introduce OH- ions that bind to H+ ions, resulting in there being less H+ ions.

 

so...

 

lots of H+ ions = acid = low pH.

 

lots of OH- = very few H+ ions = alkaly = high pH

[Borek]

so whats the relationship between the pKa and pH?

 

There is no one, simple answer.

 

To calculate pH you have to know concentration and Ka (or pKa) of acid.

 

Check these pH calculation lectures on my site.

 

Best,

Borek

--

Chemical calculators at www.chembuddy.com

pH calculator

concentration conversion

[albertlee]

so, does lots of H+ imply very few OH-?

[Yggdrasil]

Yes. Here are some example calculations:

 

For a solution at pH = 2.0

[H⁺] = 1.00x10^-2M

[OH^-] = Kw/[H⁺] = 10^-14/10^-2 = 1.00x10^-12M

 

For a solution at pH = 10

[H⁺] = 1.00x10^-10M

[OH^-] = Kw/[H⁺] = 10^-14/10^-10 = 1.00x10^-4M

 

When [H⁺] is high, [OH^-] is low and vice versa.

[Tetrahedrite]

Also' date=' pKa = pH + ([A-']/[HA]) is incorrect. The correct equation is called the Henderson-Hasselbach equation and takes the form:

 

pKa = pH + log([HA]/[A^-])

 

It's derrivation is as follows:

Ka = [H⁺][A^-]/[HA] (definition of Ka)

Now, take the -log of both sides:

-log(Ka) = -log([H⁺][A^-]/[HA])

-log(Ka) = -log([H⁺]) - log([A^-]/[HA])

Since pKa = -log(Ka) and pH = -log([H⁺]),

pKa = pH - log([A^-]/[HA]), or equivalently

pKa = pH + log([HA]/[A^-])

 

Sorry, you are quite correct, I forgot the "log" in the equation. I use the Henderson-Hasselbach equation all the time for calculating the pH of buffer solutions, so its not a mistake I should make!