Ph, Poh, and kw

Moderators: Chem_Mod, Chem_Admin

Belle Calforda3f
Posts: 67
Joined: Fri Sep 29, 2017 7:07 am

Ph, Poh, and kw

Postby Belle Calforda3f » Sun Dec 03, 2017 8:06 pm

Do we use the kw value because water is amphoteric and can act as a base or an acid or is there another reason? Also why does water and kw allow us to figure out ph and Poh?

Danah Albaaj 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: Ph, Poh, and kw

Postby Danah Albaaj 1I » Sun Dec 03, 2017 10:20 pm

We use Kw because it is has a calculated experimental constant for autoprotolysis at 25* C, which allows us to determine how the formation of hydronium and hydroxide within an equilibrium can be used to determine the pH and pOH based on their concentrations.

We use Kw to find the pH and pOH because Kw has a set value of 1.0 x 10^-14 which can be plugged into this equation to find the values of pH and pOH.
Kw = [H30+][OH-] and using the basic log rules we see that -logKw= -log([H3O+][OH-]) becomes -logKw= -log[H3O+] + -log[OH-]
In addition, -logKw = 14 which is then just plugged into the equation and used to determine either pH or pOH depending on which concentration you have found.

Julian Krzysiak 2K
Posts: 49
Joined: Fri Sep 29, 2017 7:07 am

Re: Ph, Poh, and kw

Postby Julian Krzysiak 2K » Sun Dec 03, 2017 10:39 pm

The Kw constant is another name for the autoprotolysis constant.

Autoprotolysis is a reaction where water naturally decomposes into hydronium and hydroxide, due to the amphiprotic nature of water allowing proton transfer between the same type of molecule.

The equilibrium constant would be found like any other equilibrium equation, but since water is a pure liquid, we take it out, and are left with the the equilibrium constant, Kw= (H3O+)(OH-), which is found to be 1.00 x 10^-14. Therefore, Kw=(H3O+)(OH-)=1.0x10^-14

Now that we know this equation, we can use it in many places.

If you know the value of the hydronium concentration, then you automatically know the concentration of the hydroxide solution, by simply plugging in the value into the equation.

To deal easier with this equation, we can convert these values by using logs. pH= -log(H30+), while pOH=-log(OH-). Applying this log to the whole equation (p=-log), we would get pKw=(pH)(pOH)=14, and using the log rule, we can convert this to pKw= pH+pOH=14

So if we know the pH of something, we can simply add/subtract to find the pOH.

You can use this equation in many other instances by using the constant with other known values to find unknown values.

Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”

Who is online

Users browsing this forum: No registered users and 2 guests