Two equations can be written for the dissolution of in
Chapter 19, Problem 50(choose chapter or problem)
Two equations can be written for the dissolution of in acidic solution. (a) Explain why these two equations have different values. (b) Will K for these two equations be the same or different? Explain. G G = -47.8 kJ mol-1 1 2 Mg1OH221s2 + H+1aq2 1 2 Mg2+1aq2 + H2O1l2 G = -95.5 kJ mol-1 Mg1OH221s2 + 2 H+1aq2 Mg2+1aq2 + 2 H2O1l2 Mg1OH221s2 Kc = 9.14 * 10-6 at 25 C 2 Fe2+1aq2 + 2 Hg2+1aq2 2 Fe3+1aq2 + Hg2 2+1aq2 Kc = 4.61 * 10-3 at 25 C N2O41g2 2NO21g2 Kc = 1.7 * 10-13 at 25 C N2O1g2 + 1 2 O21g2 2NO1g2 H21g2 + I21g2 2 HI1g2 Kc = 50.2 at 445 C G SO2 , 0.18 mol O2 , and 0.72 mol SO3 G Kp ? Kc = 2.8 * 102 2 SO21g2 + O21g2 2 SO31g2, H2O? CO2 , H2 , G (c) Will the solubilities of in a buffer solution at depend on which of the two equations is used as the basis of the calculation? Explain.
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer