0.350 mol NaF in sufficient water 22. An aqueous solution of NaF is prepared by
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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) 1.705Explanation / 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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