Question

What is the equilibrium constant for water?

Answer 1

`color(orange)(K_eq=([H_2O])/([H_2]^2[O_2])`

Explanation:

`color(red)(a)`A+`color(red)(b)`B `rightleftharpoons` `color(blue)(c)`C+`color(blue)(d)`D

`K_eq=` `([C]^color(blue)(c)[D]^color(blue)(d))/([A]^color(red)(a)[B]^color(red)(b))` `(larrProducts)/(larrReactants)`

`2H_2+O_2->2H_2O`
So, let's put it this way:
`color(orange)(K_eq=([H_2O])/([H_2]^2[O_2])`

Answer 2

Water undergoes autoprotolysis.....

Explanation:

`2H_2O(l) rightleftharpoons H_3O^+ + HO^-`

The hydroxide and hydronium ions are labels of convenience. We could conceive of them as clusters of water molecules with ONE proton MORE....i.e. the acid principle, `H_3O^+`, or ONE PROTON LESS, the base principle, `HO^-`...

Very careful measurement establishes that the equilibrium LIES to the left as we face the page...

`K_w=[H_3O^+][HO^-]=10^-14`...under standard conditions of temperature and pressure. And this is an equation which we can divide, multiply, subtract from, PROVIDED that we do it to BOTH sides of the equation.... One thing that we can do is to take `log_10` of both sides....

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)=-14`

And so `14=underbrace(-log_10[H_3O^+])_"pH by definition"underbrace(-log_10[HO^-])_"pOH by definition"`

And so our working relationship...

`14=pH+pOH`...which is obeyed by aqueous solutions... And thus under BASIC conditions...`pH` is high, and `pOH` is LOW....and under acidic conditions, `pH` is low to negative, and `pOH` is high...

`pK_w=10^-14` for water at `298*K`. How do you think `K_w` would evolve under non-standard conditions....i.e. say at `100` `""^@C`?