5. 0/10 points | Previous Answers | My Notes ·Ask Your Teacher Find the pH durin

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5. 0/10 points | Previous Answers | My Notes ·Ask Your Teacher Find the pH during the titration of 20.00 mL of 0.2650 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10-5), with 0.2650 M NaOH solution afterthefollowing additions of titrant. (a) 0 mL 2.91 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. (b) 10.00 mL (c) 15.00 mL (d) 20.00 mL (e) 25.00 mL Submit Answer

Explanation / Answer

Let HA be the shorthand for butanoic acid

At 0 ml NaOH:
0.2650 M HA
Ka = 1.54x10^-5 = [H+][A-]/[HA]
let X = [H+]
1.54x10^-5 = [H+][A-]/[HA] = X^2 / (0.2650 - X)
assume X << 0.2650 M
X = 0.00202
pH = 2.69

At 10.0 and 15.00 mL NaOH added you have a buffer solution composed of HA and A^-1. Use the Henderson-Hasselbalch equation-
pH = pKa + log(A-/HA)

at 10.00 mL NaOH added HA=A- so pH = pKa
pH = -log(1.54x10^-5) = 4.81

at 15.00 mL NaOH added, 3/4 of HA has been converted to A-
pH = pKa + log(0.75/0.25) = 5.29

at 20.00 mL NaOH added you have a 40.00 mL of a solution containing 5.3 millimoles of A-
A- + H2O <--> HA + OH-
Kb = [HA][OH-]/[A-]
[A-] = 5.3 mmol / 40.00 mL = 0.1325 M
Kb = Kw/Ka = 1x10^-14/1.54x10^-5 = 6.49x10^-10
let X = [OH-]
Kb = 6.49x10^-10 = X^2 / (0.1325 - X)
X = 9.27x10^-6
pOH = 5.03
pH = 14.00 - 5.03 = 8.97

at 25.00 mL of NaOH the excess NaOH is the main source of OH- and not the equilibrium A- + H2O <--> HA + OH-

At 25 mL NaOH added, you have added 5 mL excess NaOH
5 x 0.2650 M = 0.01325 millimoles
0.01325 millimoles / 45 mL = 0.000294 M
pOH = 3.53
pH = 14.00 - 3.53 = 10.47.

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