1. A 20.0 ml sample of hydrofluoric acid ( ka= 7.1 x 10^-4 ) is titrated to equi
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Question
1. A 20.0 ml sample of hydrofluoric acid ( ka= 7.1 x 10^-4 ) is titrated to equivalence with 37.12 ml of 0.1211 M NaOH. What is the pH of the resulting solution2. What is the solubility of barium fluoride ( ksp = 1.7 x 10^-6) in water ? a) what is the solubility in 1 M NaF?
3. 3H2 ( g) + N2 (g) <—> 2 NH3 This reaction has initial concentration of [H2] = 1.60 M [N2] = 1.20 M [NH3] =0.00 M At equilibrium [H2] = 0.23 Determine Kc for reaction 1. A 20.0 ml sample of hydrofluoric acid ( ka= 7.1 x 10^-4 ) is titrated to equivalence with 37.12 ml of 0.1211 M NaOH. What is the pH of the resulting solution
2. What is the solubility of barium fluoride ( ksp = 1.7 x 10^-6) in water ? a) what is the solubility in 1 M NaF?
3. 3H2 ( g) + N2 (g) <—> 2 NH3 This reaction has initial concentration of [H2] = 1.60 M [N2] = 1.20 M [NH3] =0.00 M At equilibrium [H2] = 0.23 Determine Kc for reaction
2. What is the solubility of barium fluoride ( ksp = 1.7 x 10^-6) in water ? a) what is the solubility in 1 M NaF?
3. 3H2 ( g) + N2 (g) <—> 2 NH3 This reaction has initial concentration of [H2] = 1.60 M [N2] = 1.20 M [NH3] =0.00 M At equilibrium [H2] = 0.23 Determine Kc for reaction
Explanation / Answer
1)
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1211 M * 37.12 mL = 4.4952 mmol
At. equivalence point, equal mol of both acid and base would have reacted.
We have:
mol(HF) = 4.4952 mmol
mol(NaOH) = 4.4952 mmol
4.4952 mmol of both will react to form F- and H2O
F- here is strong base
F- formed = 4.4952 mmol
Volume of Solution = 37.12 + 37.12 = 74.24 mL
Kb of F- = Kw/Ka = 1*10^-14/7.1*10^-4 = 1.408*10^-11
concentration ofF-,c = 4.4952 mmol/74.24 mL = 0.0605M
F- dissociates as
F- + H2O -----> HF + OH-
0.0605 0 0
0.0605-x x x
Kb = [HF][OH-]/[F-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.408*10^-11)*6.055*10^-2) = 9.235*10^-7
since c is much greater than x, our assumption is correct
so, x = 9.235*10^-7 M
[OH-] = x = 9.235*10^-7 M
use:
pOH = -log [OH-]
= -log (9.235*10^-7)
= 6.0346
use:
PH = 14 - pOH
= 14 - 6.0346
= 7.9654
Answer: 7.97
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