WHEN WATER IS THE SOLVENT

• The water ionization or hydrolysis reaction is ALWAYS present.
• H₂O(ℓ) ⇌ H⁺(aq) + OH^-(aq).
• K_w = [H⁺][OH^-] = 1.01x10^-14 at 25 °C.

When an acid or a base is added to water, the contribution of [H⁺] or [OH^-] on the pH from the natural dissociation of water (hydrolysis or ionization) is very small and can often be neglected in most problems (99.9%).

Therefore, the pH in water is mostly "driven" by the acid or base added.

However, if the concentration of OH^- added (here KOH = 1.0x10^-8 M) is close to the one of the water hydrolysis concentration, you need to do an ICE table to know the exact pH.

ICE Table

	H2O(ℓ)	⇌	H+(aq)	+	OH-(aq)
I	(ℓ)		1.0x10-7		1.0x10-7 + 1.0x10-8
C	+ x		-x	+	-x
E	(ℓ)		1.0x10-7 - x	+	1.1x10-7 -x

The equilibrium runs backward since adding a product will shift the equilibrium to the left (leChatelier)

K_w = 1.0x10^-14 = (1.0x10^-7 - x)(1.1x10^-7 - x)

0 = x² - 2.1x10^-7x + 1.0x10^-15

Then, you need to solve this quadratic equation to get the answer.

The solution is x = 4.88x10^-9, therefore, the [H⁺] concentration is 1.00x10^-7 - 4.88x10^-9 = 9.51x10^-8

pH = -log(9.51x10^-8) = 7.02.

Note: This problem is similar to the one in Zumdahl 14-59c with HI and 14-89b with NaOH