0.350 mol NaF in sufficient water 22. An aqueous solution of NaF is prepared by

Question

0.350 mol NaF in sufficient water 22. An aqueous solution of NaF is prepared by dissolving to yield 1.0 L of solution. The pH of the solution was 8.93 at 25°C. The Ks of P is (A) 2.8x10-12 (B) 1.2x10 (C) 6.9x10 (D) 2.1x10 (E) 9.9x102 23. Which of the following pairs cannot be mixed together to form a buffer solution? (A) KOH, HF (B) NHs, NH&C; (C) NaCaHO2, HC2H3O2 (E) RbOH, HBr 24. Which solution has the greatest buffering capacity? (A) 0.520 M HC2Hs02 and 0.116 M NaC,H,O2 (B) 0.120 M HC2H,02 and 0.115 M NaC-H,O2 (C) 0.820 M HC,Hs02 and 0.715 M NaC,H02 (D) 0.335 M HC2H,02 and 0.497 M NaC2H,02 (E) These buffer solutions will provide the same buffering capacity. 25. The Ka of benzoic acid is 6.20x109. The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium benzoate and 50.0 mL of 1.00 M benzoic acid is (A) 4.201 (B) 0.851 (C) 3.406 (D) 2.383 (E) 1.705

Explanation / Answer

22)

22)

we have below equation to be used:

pH = -log [H+]

8.93 = -log [H+]

log [H+] = -8.93

[H+] = 10^(-8.93)

[H+] = 1.175*10^-9 M

we have below equation to be used:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(1.175*10^-9)

[OH-] = 8.511*10^-6 M

Lets write the dissociation equation of F-

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

0.35 0 0

0.35-x x x

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

Kb = x*x/(c-x)

Kb = 8.511*10^-6*8.511*10^-6/(0.35-8.511*10^-6)

Kb = 2.1*10^-10

Answer: D

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