What is the pH of the solution prepared by adding 3.343 g of solid NH_4CI to 250

Question

What is the pH of the solution prepared by adding 3.343 g of solid NH_4CI to 2500 mL of 0.25 M aqueous ammonia ? Assume the volume does not change (A) 11.33 (B) 9.26 (C) 4.93 (D) 4.74 An aqueous solution is 0.10 M in both Ca^2+ and Pb^2+. As Na_2CO_3(aq) is added, which will precipitate first, Ca^2+ or Pb^2+? Assume the volume does not change. (A) Pb^2+ (B) Ca^2+ (C) They will precipitate at the same time, since they both have the same concentration (D) Neither; Na_2CO_3 will precipitate first. What is the pH at the equivalence point of a titration of HCI and the weak base pyridine, C_5H_5N? (A) below 7.0 (B) 7.0 (C) above 7.0 (D) The K_b of pyridine is required. Given this reaction at equilibrium, which change in reaction conditions will favor the production of D? 2A(g) + B(g) C(g) + D(g) Delta H = + 495 kJ (A) increase the temperature (B) increase the volume of the system (C) increase the concentration of C D) decrease the pressure of the system

Explanation / Answer

Q47.

fin pH for:

mol of NH4Cl = mass/M W= 3.343/53.49 = 0.06249 mol of NH4+

now

mol of NH3 = MV = 0.25*0.250 = 0.0625 mol

apply buffer equation

pH = pKa + log(NH4+/NH3)

pKa = 14-pKb = 14-4.75 = 9.25

pH = 9.25+ log(0.06249 /0.0625 )

pH = 9.24993

best answer shown is B

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