Calculate pOH of a .1540 M solution of phenol (C6H5OH) Ka=1.61x10^-10?

1 Answer
Dec 17, 2016

Answer:

#sf(pOH=8.7)#

Explanation:

This expression can be used for a weak acid:

#sf(pH=1/2(pK_(a)-log[acid])#

#sf(pK_a=-logK_a=-log(1.61xx10^(-10))=9.79)#

#:.##sf(pH=1/2(9.79-(-0.812))=5.3)#

#sf(pH+pOH=14" "at" "25^@C)#

#:.##sf(pOH=14-5.3=8.7)#