Homework SourseHow many grams Benzoic acid HC7H5O2 ka63x105 must be dissolv
How many grams Benzoic acid (HC7H5O2), (ka=6.3x10^-5) must be dissolved in 255mL of water to mae a solution with a pH=3.6?
Solution
F = Initial [C6H5CO2H] = (0.56 g)/(122.13 g/mol) = 4.6 x 10-3 mol/L
Ka = x2/F = x2/4.6 x 10-3 and x = 5.4 x 10-4 M
Percent dissociation = x/F = 12%
Percent dissociation is greater than 5% so approximation is not appropriate
Ka = x2/(F-x), solve quadratic equation to get x = [C6H5CO2-] = [H+] = 1.4 x 10-4 M
[OH-] = 1.0 x 10-14/1.4 x 10-4 = 7.1 x 10-11 M
Fraction of dissociation = x/F = 0.030
[C6H5CO2H] = F - x = 4.6 x 10-3 M - 1.4 x 10-4 M = 4.5 x 10-3 M
pH = - log(1.4 x 10-4) = 3.85
![How many grams Benzoic acid (HC7H5O2), (ka=6.3x10^-5) must be dissolved in 255mL of water to mae a solution with a pH=3.6?SolutionF = Initial [C6H5CO2H] = (0.56](/WebImages/35/how-many-grams-benzoic-acid-hc7h5o2-ka63x105-must-be-dissolv-1104164-1761584033-0.webp "How many grams Benzoic acid (HC7H5O2), (ka=6.3x10^-5) must be dissolved in 255mL of water to mae a solution with a pH=3.6?SolutionF = Initial [C6H5CO2H] = (0.56")
