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Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka...

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10^-5), with 0.1000 M NaOH solution after the following additions of titrant (a) 14.00 mL (b) 20.30 mL (c) 26.00 mL

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Expert Solution

Ka = 1.54 x 10^-5

pKa = -log Ka

pKa = 4.81

a)

millimoles of acid = HA =20 x 0.1 = 2

millimoles of OH- = 14 x 0.1 = 1.4

HA + OH- -----------------> H2O + A-

2          1.4                           0        0

0.6          0                                      1.4

pH = pKa + log [A-/ HA]

pH = 4.81 + log (1.4 / 0.6)

pH = 5.18

(b) 20.30 mL

base millimoles dominent

[OH-] = 20.30 x 0.1 - 20 x 0.1 / (20 + 20.30)

            = 7.44 x 10^-4 M

pOH = -log [OH-] = 3.13

pH + pOH = 14

pH = 10.87

(c) 26.00 mL

[OH-]   = 2.6 - 2.1 / (46) = 0.0109M

pOH = 1.88

pH = 12.12

queen_honey_blossom answered 2 years ago
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