4. (15) A saturated solution of phenol, C6H5OH, a very weak acid with pKa=10.00, has pH=4.90. What is the molarity of phenol in this solution?
use:
pKa = -log Ka
10.0 = -log Ka
Ka = 1*10^-10
Let the concentration of C6H5OH be c
use:
pH = -log [H+]
4.9 = -log [H+]
[H+] = 1.259*10^-5 M
C6H5OH dissociates as:
C6H5OH -----> H+ + C6H5O-
c 0 0
c-x x x
Ka = [H+][C6H5O-]/[C6H5OH]
Ka = x*x/(c-x)
1*10^-10 = 1.259*10^-5*1.259*10^-5/(c-1.259*10^-5)
c-1.259*10^-5 = 1.585
c=1.585 M
Answer: 1.58 M
4. (15) A saturated solution of phenol, C6H5OH, a very weak acid with pKa=10.00, has pH=4.90....
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