Beaker 1 Beaker 2 Beaker 3 0.100 M NaOH 0.100 M NaF 0.100 M HF 25.0 mL 50.0 mL 5
ID: 229837 • Letter: B
Question
Beaker 1 Beaker 2 Beaker 3 0.100 M NaOH 0.100 M NaF 0.100 M HF 25.0 mL 50.0 mL 50.0 mL Beaker 1 contains a 25.0 mL solution of NaOH. Whereas, Beaker 2 & 3 have 50.0 mL solutions of NaF and HF respectively. Each solution is at 25°C. 1. a) Determine the pH of the solution in beaker 1. b) The K, for HF is 2.6x105. What is the hydronium ion concentration in beaker 3? c) Calculate the pH of the solution in beaker 3. d) Calculate the pH of the solution in beaker 2. e) The contents of beaker 2 are poured into beaker 3 and the resulting solution is stirred. Assume the volumes are additive. i. Calculate the pH of the resulting solution. f) The contents of beaker 1 are then added to the solution made in part d i. Write the reaction that takes place? ii. Calculate the pH of the resulting solution.Explanation / Answer
a)
[OH-] = 0.100M
pOH = - log[OH-] = - log(0.100) = 1
pH + pOH = 14
pH = 14 - 1
pH = 13
b)
HF(aq) + H2O(l) < - - - - > H3O+(aq) + F-(aq)
Ka = [H3O+] [F-] /[HF] = 2.6×10-5
x2/0.1 - x = 2.6×10-5
x = 0.00160
Therefore,
[H3O+] = 0.0016M
c)
pH = - log[H3O+]
pH = - log(0.0016)
pH = 2.80
d)
NaF is partly hydrolysed by water
F- + H2O <------> HF + OH-
Kb = [HF] [OH-] /[F-]
Kb = Kw/Ka
Kw = ionic product of water, 1.00×10-14
Ka = Dissociation constant of HF, 2.6×10-5
Kb = 1.00×10-14/2.6×10-5 = 3.85×10-10
x2/0.100 - x = 3.85×10-10
we can assume, 0.100 - x ~ 0.100
x2/0.100 = 3.85×10-10
x2 = 3.85×10-11
x= 6.20×10-6
Therefore,
[OH-] = 6.20×10-6M
pOH = - log[OH-] = - log(6.20×10-11)=5.21
pH = 14 - pOH
= 14 - 5.21
= 8.79
e) Tota volume after mixing two solution = 100ml
[NaF] = [F-] = 0.100M/2 = 0.050M
[HF] = 0.100M/2 = 0.050M
According to Henderson - Hasselbalch
pH = pKa + log([A-] /[HA])
pKa = - logKa = - log(2.6×10-5) = 4.59
Therefore,
pH = 4.59 + log(0.05M/0.05M)
pH = 4.59
f)
No of moles of OH- in beaker 1 =(0.100mol/1000ml)×25ml = 0.0025
No of moles of F- present in solution made in part e = (0.05mol/1000)×100ml = 0.0050
No of moles of HF present in solution made in part e = (0.05mol/1000ml)×100ml =0.0050
OH- react with acid HF to form F-
HF + OH- <------> F- + H2O
After addition
No of moles of F- = 0.0050 + 0.0025 = 0.0075
No of moles of HF = 0.0050 - 0.0025 = 0.0025
Total volume = 100ml + 25ml = 125ml
[HF] =(0.0025mol/125ml)×1000ml =0.02M
[F-] = (0.0075mol/125ml)×1000ml =0.06M
Applying Henderson-Hasselbalch equation
pH = 4.59 + log(0.06M/0.02M)
= 4.59 + 0.48
= 5.07
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