Question

50.0ml of 0.125M HF is titrated with 0.100M NaOH. Calculate the pH after the addition of

a) 0.00ml of NaOH

b) 15.00ml of NaOH

c) 31.25ml of NaOH

d) 60.00ml of NaOH

e) 62.50ml of NaOH

f) 70.00ml of NaOH

Answer 1

a) 0.00 ml NaOH added

HF dissociates as

HF + H₂O   F^- + H₃O⁺

K_a = [F^- ][H₃O⁺] / [HF]

but  [F^- ] = [H₃O⁺] = x

K_a = [x][x] / [HF]

Ka of HF = 6.6 10^-4

Substitute the value in equation

6.610^-4 = [x]²/ 0.125

[x]² = 6.610^-4   0.125 = 8.2510^-5

[x] = 0.00908

Concentration of H₃O⁺ = 0.00908 M

pH = - log[H⁺]

pH = - log (0.00908)

pH = 2.04

b)

nutrilization reaction between HF and NaOH take place as given below

HF + NaOH   NaF + H₂O

calculate mole of NaOH and HF

no. of mole = Molarity volume of solution in liter

no. of mole of NaOH = 0.100 0.015 = 0.0015 mole

no. of mole of HF = 0.125 0.050 = 0.00625

0.0015 mole of NaOH react with 0.0015 mole of HF

no. of mole of HF remain in solution = 0.00625 - 0.0015 = 0.00475 mole

total volume of solution = 15 ml + 50 ml = 65 ml = 0.065 liter

Molarity = no. of mole / volume of solution in liter

Molarity of HF = 0.00475 / 0.065 = 0.0730 M

HF dissociates as

HF + H₂O   F^- + H₃O⁺

K_a = [F^- ][H₃O⁺] / [HF]

but  [F^- ] = [H₃O⁺] = x

K_a = [x][x] / [HF]

Ka of HF = 6.6 10^-4

Substitute the value in equation

6.610^-4 = [x]²/ 0.0730

[x]² = 6.610^-4   0.0730 = 4.81810^-5

[x] = 0.00694

Concentration of H₃O⁺ = 0.00694 M

pH = - log[H⁺]

pH = - log (0.00694)

pH = 2.16

c)

nutrilization reaction between HF and NaOH take place as given below

HF + NaOH   NaF + H₂O

calculate mole of NaOH and HF

no. of mole = Molarity volume of solution in liter

no. of mole of NaOH = 0.100 0.03125 = 0.003125 mole

no. of mole of HF = 0.125 0.050 = 0.00625

0.003125 mole of NaOH react with 0.003125 mole of HF

no. of mole of HF remain in solution = 0.00625 - 0.003125 = 0.003125 mole

total volume of solution = 31.25 ml + 50 ml = 81.25 ml = 0.08125 liter

Molarity = no. of mole / volume of solution in liter

Molarity of HF = 0.003125 / 0.08125 = 0.0385 M

HF dissociates as

HF + H₂O   F^- + H₃O⁺

K_a = [F^- ][H₃O⁺] / [HF]

but  [F^- ] = [H₃O⁺] = x

K_a = [x][x] / [HF]

Ka of HF = 6.6 10^-4

Substitute the value in equation

6.610^-4 = [x]²/ 0.0385

[x]² = 6.610^-4   0.0385 = 2.54110^-5

[x] = 0.00504

Concentration of H₃O⁺ = 0.00504 M

pH = - log[H⁺]

pH = - log (0.00504)

pH = 2.98

d)

nutrilization reaction between HF and NaOH take place as given below

HF + NaOH   NaF + H₂O

calculate mole of NaOH and HF

no. of mole = Molarity volume of solution in liter

no. of mole of NaOH = 0.100 0.060 = 0.0060 mole

no. of mole of HF = 0.125 0.050 = 0.00625

0.0060 mole of NaOH react with 0.0060 mole of HF

no. of mole of HF remain in solution = 0.00625 - 0.0060 = 0.00025 mole

total volume of solution = 60 ml + 50 ml = 110 ml = 0.110 liter

Molarity = no. of mole / volume of solution in liter

Molarity of HF = 0.00025 / 0.110 = 0.00227 M

HF dissociates as

HF + H₂O   F^- + H₃O⁺

K_a = [F^- ][H₃O⁺] / [HF]

but  [F^- ] = [H₃O⁺] = x

K_a = [x][x] / [HF]

Ka of HF = 6.6 10^-4

Substitute the value in equation

6.610^-4 = [x]²/ 0.00227

[x]² = 6.610^-4   0.00227 = 1.498210^-6

[x] = 0.001224

Concentration of H₃O⁺ = 0.001224 M

pH = - log[H⁺]

pH = - log (0.001224)

pH = 2.91

e)

nutrilization reaction between HF and NaOH take place as given below

HF + NaOH   NaF + H₂O

calculate mole of NaOH and HF

no. of mole = Molarity volume of solution in liter

no. of mole of NaOH = 0.100 0.0625 = 0.00625 mole

no. of mole of HF = 0.125 0.050 = 0.00625

0.00625 mole of NaOH react with 0.00625 mole of HF

no. of mole of HF remain in solution = 0.00625 - 0.00625 = 0.00 mole

nuetrilization reaction take place complete and product salt and water formed and pH of reaction = 7

f)

nutrilization reaction between HF and NaOH take place as given below

HF + NaOH   NaF + H₂O

calculate mole of NaOH and HF

no. of mole = Molarity volume of solution in liter

no. of mole of NaOH = 0.100 0.070 = 0.0070 mole

no. of mole of HF = 0.125 0.050 = 0.00625

0.00625 mole of NaOH react with 0.00625 mole of HF

no. of mole of NaOH remain in solution = 0.0070 - 0.00625 = 0.00075 mole

total volume of solution = 70 ml + 50 ml = 120 ml = 0.120 liter

Molarity = no. of mole / volume of solution in liter

Molarity of NaOH = 0.00075 / 0.120 = 0.00625 M

NaOH is strong base dissociate comletly

NaOH Na⁺  +OH^-

[NaOH] = [OH^-] = 0.00625 M

pOH = - log[OH^-] = -log[0.00625] = 2.20

pH = 14 - 2.20 = 11.80