Question

What mass of benzoic acid, HC7H5O2, would you dissolve in 400.0 mL of water to produce a solution with a pH = 2.80?
HC7H5O2+H2O⇌H3O++C7H5O−2Ka=6.3×10−5

Answer 1

Apply:

Ka = [H+][C7H5O-] / [C7H5OH]

initially

[H+]= [C7H5O-] = 0

[C7H5OH]= M

in equilbirium

[H+]= [C7H5O-] = 0 - x

[C7H5OH]= M + x

substitute in Ka

Ka = [H+][C7H5O-] / [C7H5OH]

6.3*10^-5 = x*x/(M-x)

note that

x = [H+] = 10^-pH = 10^-2.8 = 0.001584

so

6.3*10^-5 = 0.001584*0.001584/(M-0.001584)

solve for M

M = ( 0.001584*0.001584)/(6.3*10^-5) + 0.001584

M = 0.0414102 M of acid

V = 400 mL = 0.4 L

so

mol = MV = 0.0414102*0.4 = 0.01656408 mol of benzoic acid

mass = mol*MW = 0.01656408*122.123 = 2.022855 g of benzoic acid required