At 25°C, a saturated solution of benzoic acid (C6H5CO2H; Ka = 6.4 10-5) has a pH of 2.80. Calculate the water?
At 25°C, a saturated solution of benzoic acid (C6H5CO2H; Ka = 6.4 10-5) has a pH of 2.80. Calculate the water solubility of benzoic acid in moles per liter and grams per 100. milliliters.
_______mol/L
_______g per 100. mL
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BenzH <-----> Benz- + H+
Ka = [Benz-][H+] / [BenzH]
If pH = 2.80, then [H+] = 10^-pH = 10^-2.80
[H+] and [Benz-] have a 1:1 stoichiometric ratio, so [Benz-] = [H+] = 10^-2.80
6.4E-5 = (10^-2.80)^2 / [BenzH]
[BenzH] = 0.0392 M
0.0392 moles / L (122.21 g/mole) = 4.79 g / L = 4.79 g / 1000 mL = 47.9 g / 100 mL