..................C6H5OH ==> C6H5O^- + H^+
I....................0.1......................0.............0
C....................-x........................x.............x
E....................0.1-x....................x.............x
pH = 5.43 = -log(H^+)
(H^+) = 3.72E-6 but you should confirm that.
Ka = (x)(x)/(0.1-x)
x = 3.72E-6. Substitute and solve for Ka.
% ionization = 100[(x)/0.1]
Sore-throat medications sometimes contain the weak
acid phenol, HC6H5O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C.
What is the acid-dissociation constant Ka for this acid at 25°C?
Hence, what is its degree of ionisation (percent ionisation)?
1 answer
1 answer