Consider a 020 M solution of NaF 200 mL that is being titrat

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

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

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

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 addi

Consider a 020 M solution of NaF 200 mL that is being titrat

Solution

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