Consider a 0.20 M solution of NaF (20.0 mL) that is being titrated with 0.10 M H

Question

Consider a 0.20 M solution of NaF (20.0 mL) that is being titrated with 0.10 M HCl. a) write the reaction that is happening. b) Determine the pH after the addition of i) 0 mL ii) 20 mL iii) 40 mL iv) 60 mL Consider a 0.20 M solution of NaF (20.0 mL) that is being titrated with 0.10 M HCl. a) write the reaction that is happening. b) Determine the pH after the addition of i) 0 mL ii) 20 mL iii) 40 mL iv) 60 mL a) write the reaction that is happening. b) Determine the pH after the addition of i) 0 mL ii) 20 mL iii) 40 mL iv) 60 mL

Explanation / Answer

a)

F-   (aq)   +   H+ (aq)    ----------> HF (aq)

b)

i)

NaF is a salt of strong base and weak acid. so pH > 7

pKa of HF = 3.20

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (3.20 + log 0.20)

pH = 8.25

ii)

mmoles of acid = 20 x 0.1 = 2

mmoles of F- = 20 x 0.2 = 4

F-   (aq)   +   H+ (aq)    ----------> HF (aq)

4                    2                               0

2                     0                               2

this is half - equivalence . so pH = pKa

pH = 3.20

iii)

mmoles of acid = 40 x 0.1 = 4

F-   (aq)   +   H+ (aq)    ----------> HF (aq)

4                       4                             0

0                        0                           4

concentration of HF = 4 / (20 + 40) = 0.067 M

pH = 1/2 (pKa - log C)

     = 1/2 (3.20 - log 0.067)

pH = 2.19

iv)

mmoels of acid = 6

F-   (aq)   +   H+ (aq)    ----------> HF (aq)

4                  6                                  0

0                   2                                   4

here strong acid remains.

[H+] = 2 / (20 + 60) = 0.025

pH = -log (0.025)

pH = 1.60

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