What mass of the salt sodium benzoate (NaC_6 H_5 COO) should be added to 150 mL
ID: 963768 • Letter: W
Question
What mass of the salt sodium benzoate (NaC_6 H_5 COO) should be added to 150 mL of a 0.125 M benzoic acid (C_6 H_5 COOH) solution to obtain a buffer solution with a pH of 4.00. Data: K_a (benzoic acid, a monoprotic acid) = 6.5 Times 10^-5 at 25 degree C. Assume there is not change in volume of the solution when the salt is added. Data: Molar mass of sodium benzoate = 144.10 gm/mol, Molar mass benzoic acid - 122.12 gm/mol If 0.20 mL of 3.0 M HCl is added to the solution of part (i), what is the pH of the resulting solution?Explanation / Answer
Ka of benzoic acid 6.5 x 10-5
pKa = -log Ka
pKa = 4.187
Henderson Hasselbach equation
pH = pKa + log base/acid
4.00 = 4.187 + log base/acid
log base/acid = -0.187
base/acid = 10-0.187
base/acid = 0.65
base/0.125 = 0.65
base = 0.65 x 0.125
base = 0.0815 M
0.0815 M in 150 mL is 0.0815 x 0.15 = 0.0122 moles
0.0122 x 144.1 g/mol = 1.76 g of sodium benzoate.
2) 0.2 mL of 3 M HCl is 0.0002 x 3M = 0.0006 moles
sodium benzoate is consumed 0.0122 - 0.0006 = 0.0116 moles of sodium benzoate in 150.2 mL = 0.077 M
benzoic acid increases from 0.125 x 0.15 = 0.01875 moles + 0.0006 = 0.01935 moles in 150.2 mL = 0.1288 M
pH = pKa + log base/acid
pH = 4.187 + log 0.077/0.1288
pH = 4.187 - 0.222
pH = 3.964
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.