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Determine the pH during the titration of 56.0 mL of 0.389 M hydrofluoric acid (Ka =...

Determine the pH during the titration of 56.0 mL of 0.389 M hydrofluoric acid (Ka = 7.2×10-4) by 0.389 M NaOH at the following points. (Assume the titration is done at 25 °C.)

(a) Before the addition of any NaOH

(b) After the addition of 14.0 mL of NaOH

(c) At the half-equivalence point (the titration midpoint)

(d) At the equivalence point

(e) After the addition of 84.0 mL of NaOH

Solutions

Expert Solution

pKa = -log Ka

pKa = 3.14

(a) Before the addition of any NaOH

pH = 1/2 [pKa - log C]

pH = 1/2 [3.14 - log 0.389]

pH = 1.78

(b) After the addition of 14.0 mL of NaOH

millimoles of HF = 56.0 x 0.389 = 21.8

millimoles of NaOH = 0.389 x 14 = 5.45

HF    +    NaOH -------------------------> NaF + H2O

21.8           5.45                                    0         0

16.4           0                                        5.45   

pH = pKa + log [NaF/ HF]

pH = 3.14 + log (5.45 / 16.4)

pH = 2.66

(c) At the half-equivalence point (the titration midpoint)

here pH = pKa

pH = 3.14

(d) At the equivalence point

salt concentration =- 56 x 0.389 / (56 +56)

                             = 0.504 M

pH = 7 + 12/ [pKa + logC]

pH = 7 + 1/2 [3.14 + log 0.504]

pH = 8.42

(e) After the addition of 84.0 mL of NaOH

answer : 12.9

queen_honey_blossom answered 1 year ago
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