Here is the relevant chemical equation for the above problem:
C6H5COO- + H2O <===> C6H5COOH + OH-
kb = [C6H5COOH][OH-] / [C6H5COO-]
5.6 x 10-10 = x . x / 0.10
x2 = 0.56 x 10-10
x = 0.74 x 10-5 where x = [OH-]
pOH = -log[OH-]
pOH = -log(0.74 x 10-5) = 5.13
pH = 14 - pOH
pH = 14 - 5.13 = 8.87
Please help me find the pH , please show work as well. Thank you . Please...
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