Answer to Question #60042 in General Chemistry for kevin from 1st

Question

If the concentration of hydroxide, [OH-], of a solution is 4.98×10-6 then determine the missing variables.

Answer 1

Solution:
The self-ionization of water (also autoionization of water, and autodissociation of water) is an ionization reaction in pure water or an aqueous solution, in which a water molecule, H2O, deprotonates (loses the nucleus of one of its hydrogen atoms) to become a hydroxide ion, OH−. The hydrogen nucleus, H+, immediately protonates another water molecule to form hydronium, H3O+.
H2O + H2O = H3O+ + OH-
Kw = [H3O+][OH-] = 1.0×10−14
where [H3O+] is the concentration of hydrogen or hydronium ion, and [OH−] is the concentration of hydroxide ion.
We can also define pKw = −log10 Kw = 14 at 25 °C. This is analogous to the notations pH and pKa for an acid dissociation constant, where the symbol p denotes a cologarithm. The logarithmic form of the equilibrium constant equation is pKw = pH + pOH.
p[OH-] = -log10[OH-]=5.3
pH = pKw – p[OH] = 14 – 5.3 = 8.7
[H3O+] = 10-8.7 = 1.99×10^-9