Suppose 25.0 mL of 0.100 M benzoic acid (a weak acid be abbreviated HBz) is titr
ID: 501504 • Letter: S
Question
Suppose 25.0 mL of 0.100 M benzoic acid (a weak acid be abbreviated HBz) is titrated by 0.200 M NaoH. (K_a of benzoic acid = 6.3 times 10^-5) The balanced equation representing the neutralization would be: HBz (aq) + NaOH (aq) rightarrow NaBz (aq) H_2O (l) Calculate the pH after addition of 4.0 mL of 0.200 M NaOH. show your work. Calculate the pH at the equivalence point of this titration. Show your work. Calculate the pH when 5.00 mL of 0.200 M NaOH has been added beyond the equivalence point. Show your work.Explanation / Answer
millimoles of benzoic aicd = 25 x 0.1 = 2.5
pKa = - log Ka = - log [6.3 x 10-5]
pKa = 4.20
a) millimoles of NaOH added = 4.0 x 0.2 = 0.8
2.5 - 0.8 = 1.7 millimoles acid left
0.8 moles salt formed
total volume = 25 + 4 = 29 mL
[salt] = 0.8 / 29 = 0.0276 M
[acid] = 1.7 / 29 = 0.0586 M
pH = pKa + log [salt] / [acid]
pH = 4.20 + log [0.0276] / [0.0586]
pH = 3.87
b) at equivalence point 2.5 millimoles NaOH must beaaded
2.5 = V x 0.2
V = 12.5 mL NaOH must beadded
total volume = 25 + 12.5 = 37.5 mL
[salt] = 2.5 / 37.5 = 0.067 M
pH = 1/2 [pKw + pKa + log C]
pH = 1/2 [14 + 4.20 + log 0.067]
pH = 8.51
c) exess NaOH = 5 x 0.2 = 1.0 millimoles
total volume = 37.5 + 5 = 42.5 mL
[OH-] = 1.0 / 42.5 = 0.023 M
pOH = - log [OH-]
pOH = -log [0.023]
pOH = 1.64
pH = 14 - 1.64
pH = 12.36
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