Question

APP-17 A 4 Problem Set 13 CHEM 1252 October 21, 2019 Textbook Reading Assignment: 14.6-10 Additional Practice Problems: 1466 -68.20.73.75.27.79.83.86-87,89,101,107,111-115,119 1. Ammonia and hydrogen sulfide can both be formed directly from their elements. At 90 "C... Nale) + 3H2(e) -2NH3G) * = 120 Hale) + SC) - H256) K = 0.068 Find Kat 90 °C for the following reaction 2NH3(e) + 35() - 3H.SG) + Nale) K? 2. The equilibrium constant for the following reaction is called the solubility product of calcium fluoride: CaF (s) - Ca2+(aq) + 2F-(ay) K = K - 3.2 x 10-11 (a) Write an expression for the equilibrium constant of this reaction in terms of concentrations Why do you suppose we call this a solubility product instead of a solubility quotient or ratio? (b) Calculate the equilibrium concentrations of Ca2+ and F if excess solid CaF is placed in water. (c) In which direction will the equilibrium shift if some NaF is added to the saturated CaF2 solution from part b? (Hint: is NaF soluble in water?] (d) Calculate the equilibrium concentrations of Ca2+ and F- if enough NaF is added to the saturated CaF2 solution from part b to make the total concentration (prior to re-equilibration) = 0.250 M. (Hint: look for a simplifying approximation!)

Answer 1

p criven that, Temperature = 90°C. N2 (9) +3+3(9) = 2 NHS (9) KR=120 . KRIE Pana Proz.PARE H2(9)+S(s) = H2SC) ; KP2 = 0.068. KP2 = PH2S PH2 Now, Cubing on both sides KP3-PINS 2N1 H3 (9) +3515) = 31535(g) + N2 (9) KP= P42S XPne Prviz? x 3 KP23 ₂ PBA's/Pith KPI Phng. PN 2 P KR 3 RP = KP (0.06839 = KP 120 Kpa 2:67156