Calculate the pH for 300. mL of a 0.850 M solution of the benzoic acid, (C6H5COO
ID: 491046 • Letter: C
Question
Calculate the pH for 300. mL of a 0.850 M solution of the benzoic acid, (C6H5COOH), Ka=6.30x10-5 ) being titrated with 0.850 M NaOH at the following positions in the titration.
a) The initial pH (before any NaOH has been added). a. 1.436 b. 12.864 c. 5.435 d. 8.534 e. 2.136
b) The pH of the solution after 125.0 mL of 0.850 M NaOH has been added. a. 10.034 b. 4.054 c. 3.543 d. 8.534 e. 2.132
c) The pH of the solution at the equivalence point (i.e. after 300. mL of 0.850 M NaOH has been added). a. 8.914 b. 5.423 c. 7.000 d. 5.086 e. 2.365
d) The pH of the solution after 400. mL of 0.850 M NaOH has been added. a. 0.916 b. 4.578 c. 13.084 d. 4.201 e. 9.800
Explanation / Answer
mmol of acid = MV = 300*0.85 = 255 mmol of acid...
a)
initially, there is only acid
so
HA <-> H+ + A-
Ka = [H+][A-]/[HA]
in equilbirium
[H+] = [A-] = x
[HA] = M-x = 0.85-x
substitute
6.3*10^-5 = x*x/(0.85-x)
x = H+ = 0.00728
pH = -log(0.00728) = 2.1378
b)
pH for V = 125 mL of base M = 0.85
mmol of base MV = 125*0.85 = 106.25
mmol of weak acid left = 255-106.25 = 148.75
mmol of conjugte formed = 0 + 106.25= 106.25
this is a buffer so
pH = pKa + log(A-/HA)
pH= -log(6.3*10^-5) + log(106.25/148.75) = 4.054
c)
in equivalce point
mmol of acid = mmol of base
255 = M*V
V = 255/0.85 = 300 mL
so total V = 300 + 300 = 600 mL
recaculate benzoate ion in solution
[A-] = 300*0.85/600 = 0.425
so
A- + H2O <-> HA + OH- Kb
hydolysis will occur
Kb = [HA][OH-]/[A-]
Kb = Kw/Ka = (10^-14)/(6.3*10^-5) =1.587*10^-10
1.587*10^-10 = x*x/(0.425-x)
x = OH = 8.21*10^-6
pOH = -log(8.21*10^-6
pOH = 5.0856
pH = 14-pOH = !4-5.0856 = 8.9144
d)
finally..
NaOH left over = 400*0.85 = 340 mmol of base
mmol of HA = 255
OH left = 340-255 = 85
Vtotal = 400+300 = 700
[OH-] =85/700 = 0.121428
´pH = 14 + log(0.121428) = 13.084