Calculate the pH at 25°C of a 0.54M solution of sodium benzoate NaC6H5CO2. Note that benzoic acid HC6H5CO2 is a weak acid with a pKa of 4.20.Round your answer to 1 decimal place.
Given C6H5COONa is the salt of a strong base (NaOH) and weak acid (C6H5COOH), and thus the salt in aqueous solution will have a pH.
The hydrolysis of the cyanide ion is as follows,
C6H5COO- + H2O ==> C6H5COOH + OH-
pKa value of C6H5COOH is given, that is, pKa = 4.20, to find Ka
Ka = 10-pKa
Ka = 10-4.20
Ka = 6.3096 x 10-5
Using the formula Kw = KaKb
Kb = Kw/kb
Kb = 1 x 10-14/ 6.3096 x 10-5
Kb = 0.15849 x 10-9
Kb = 0.15849 x 10-9= [C6H5COOH][OH-]/[ C6H5COO-]
0.15849 x 10-9= (x)(x)/0.54M where, [C6H5COOH] = x, [OH-] = x, C6H5COO- = 0.54 M
x2 = 2.935 x 10-10
x =[OH-]= 1.7132 x 10-5
pOH = -log[OH-]
pOH = 4.7662
14 = pH + pOH
pH = 14 – pOH
pH = 14 – 4.7662
pH = 9.2338
pH = 9
Calculate the pH at 25°C of a 0.54M solution of sodium benzoate NaC6H5CO2. Note that benzoic...
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