Find the pH during the titration of 20.00 mL of 0.1450 M benzoic acid, C6H5COOH
ID: 1025244 • Letter: F
Question
Find the pH during the titration of 20.00 mL of 0.1450 M benzoic acid, C6H5COOH (Ka = 6.3 10-5), with 0.1450 M NaOH solution after the following additions of titrant.
(a) 0 mL.... (2.52) Correct
(b) 10.00 mL... (4.20) Correct
(c) 15.00 mL... (3.00) Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations.
(d) 19.00 mL... (3.81) Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations.
(e) 19.95 mL... (4.98) Incorrect: Your response differs from the correct answer by more than 10%. Double check your calculations.
Explanation / Answer
c) No of moles of NaOH added = (0.1450mol/1000ml)×15ml = 0.002175
Initial no of moles of C6H5COOH = (0.1450mol/1000ml)×20ml = 0.0029
remaining moles of C6H5COOH = 0.0029 - 0.002175 =0.000725
No of moles of C6H5COO- formed = 0.002175M
Total volume = 35ml
[ C6H5COOH] = (0.000725mol/35ml)×1000ml = 0.0207M
[C6C5COO-] = (0.002175mol/35ml)×1000ml = 0.0621M
Henderson - Hasselbalch equation is
pH = pKa + log([A-]/[HA])
pH = 4.20 + log(0.0621M/0.0207M)
pH = 4.20 + 0.48
pH = 4.68
d)
No of moles of NaOH added = (0.1450mol/1000ml)×19.00ml = 0.002755
0.002755mol of NaOH react with 0.002755 mole of C6H5COOH to produce 0.002755moles of C6H5COO-
After addition
No of moles of C6H5COOH = 0.0029 -0.002755 = 0.000145
No of moles of C6H5COO- = 0.002755
Total volume = 20 + 19 = 39
[ C6H5COO-] = (0.002755mol/39ml)×1000ml = 0.0706M
[C6H5COOH] = (0.000145mol/39ml)×1000ml =0.0037M
Therefore,
pH = 4.20 + log ( 0.0706M/0.0037M)
pH = 4.20 + 1.28
pH = 5.48
e) No of moles NaOH added = (0.1450mol/1000ml)×19.95ml=0.002893
0.002893 moles of NaOH react with 0.002893 moles of C6H5COOH
After addition
No of moles of C6H5COOH = 0.0029 - 0.002893 = 0.000007
No of moles of C6H5COO- = 0.002893
Total volume = 39.95ml
[C6H5COOH]=( 0.000007mol/1000ml)×39.95ml = 0.00001752M
[C6H5COO-] = (0.002893mol/39.95ml)×1000ml =0.0724M
Therefore,
pH = 4.20 + log(0.0724M/0.00001752M)
pH =4.20 + 2.62
pH = 6.82
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