Determine the pH of a a solution formed by mixing 25.0 mL of0.44 M HF and 25.0 m

Question

Determine the pH of a a solution formed by mixing 25.0 mL of0.44 M HF and 25.0 mL of 0.44 M NaOH at 25 degree C. The K_a of HF is 3.5 times 10^-5 M. 5.10 11.44 10.20 8.90 2.56 A solution contains 0.012 M in Al^3+ and 0.24 M in NaF. K_f for Al F_6^3- is 7 times 1019 M-6. What is [AI^3] when the system achieves equilibrium? 1.3 times 10^-17M 7.6 times 10^-18M 3.1 times 10^-22 M 1.9 times 10^-21 M Calculate delta S degree_rxn for the following reaction. The S degree for each species shown below the is reaction. C_2H_2(g) + 2H_2(g) rightarrow C_2 H_6(g) -303.3 J/K -229.2 J/K -233.1 J/K +560.8 J/K +102.4 J/K

Explanation / Answer

12)

Given:

M(HF) = 0.44 M

V(HF) = 25 mL

M(NaOH) = 0.44 M

V(NaOH) = 25 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.44 M * 25 mL = 11 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.44 M * 25 mL = 11 mmol

We have:

mol(HF) = 11 mmol

mol(NaOH) = 11 mmol

11 mmol of both will react to form F- and H2O

F- here is strong base

F- formed = 11 mmol

Volume of Solution = 25 + 25 = 50 mL

Kb of F- = Kw/Ka = 1*10^-14/3.5*10^-5 = 2.857*10^-10

concentration ofF-,c = 11 mmol/50 mL = 0.22M

F- dissociates as

F- + H2O -----> HF + OH-

0.22 0 0

0.22-x x x

Kb = [HF][OH-]/[F-]

Kb = x*x/(c-x)

since kb is small, x will be small and it can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.857*10^-10)*0.22) = 7.928*10^-6

So [OH-] = 7.928*10^-6

[OH-] = x = 7.928*10^-6 M

use:

pOH = -log [OH-]

= -log (7.928*10^-6)

= 5.1008

use:

PH = 14 - pOH

= 14 - 5.1008

= 8.80

Answer: D

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