[The Guy]

The pH of 5.6X10^-7 g of KOH diluted in 1L of water is 6 (according to my calculation)
Now, as a base, I would assume that the pH has to be above 7. As far my understanding, the pH shouldn't fall lower than 7 even in a diluted solution.

Maybe my calculation is wrong, or I am missing something here.

[sjb]

The pH of 5.6X10^-7 g of KOH diluted in 1L of water is 6 (according to my calculation)
Now, as a base, I would assume that the pH has to be above 7. As far my understanding, the pH shouldn't fall lower than 7 even in a diluted solution.

Maybe my calculation is wrong, or I am missing something here.

How do you know that the pH has to be above 7?

[The Guy]

water is neutral. the pH of the DI water should be around 7, so, how about a neutral water with a little base?

[Burner]

https://en.wikipedia.org/wiki/Self-ionization_of_water

[AWK]

The pH of 5.6X10^-7 g of KOH diluted in 1L of water is 6 (according to my calculation)
Now, as a base, I would assume that the pH has to be above 7. As far my understanding, the pH shouldn't fall lower than 7 even in a diluted solution.

Maybe my calculation is wrong, or I am missing something here.

pH of pure water in standard temperature is equal to 7. Addition of any amount of any base increase pH over 7.

[The Guy]

Thanks everybody for the help.

AWK:
That's write. But, according to my calculation, the pH of 5.6X10^-7 g of KOH diluted in 1L of water is 6 (lower than 7) KOH is a strong base. How can that be?

[Burner]

https://en.wikipedia.org/wiki/Self-ionization_of_water

That's write. But, according to my calculation, the pH of 5.6X10^-7 g of KOH diluted in 1L of water is 6 (lower than 7) KOH is a strong base. How can that be?

Have you read the link? You missed out the OH^- produced from self-ionisation of water.

[AWK]

Thanks everybody for the help.

AWK:
That's write. But, according to my calculation, the pH of 5.6X10^-7 g of KOH diluted in 1L of water is 6 (lower than 7) KOH is a strong base. How can that be?

Wrong approximation

[The Guy]

Thanks everybody for the help.

AWK:
That's write. But, according to my calculation, the pH of 5.6X10^-7 g of KOH diluted in 1L of water is 6 (lower than 7) KOH is a strong base. How can that be?

Wrong approximation

What is the wrong here? any clue please?

[AWK]

In this case you should take into account two equilibria as Burner suggested.
The concentration of OH^- from KOH diminishes concentration OH^- from water, but for calculations both concnetration should be taken into account.
The same reasoning concerns a very diluted acids eg 10^-8 M HCl for which pH is below 7.

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base