(c) 100.0 mL of 0.0800 molar Ba(NO3)2 is mixed with 100.0 ml of 0.0600 molar KCl
ID: 917879 • Letter: #
Question
(c) 100.0 mL of 0.0800 molar Ba(NO3)2 is mixed with 100.0 ml of 0.0600 molar KCl. How many grams of BaCl2 precipitate? Calculate the equilibrium concentrations of Ba+, K+, NO3?, and Cl? in the supernatant aqueous solution. Assume all activity coefficients equal 1.00.
iPad 1:10 AM 56% chegg.com Step 1 of 2 We have to calculate the value of HR and G; for the reaction between barium nitrate with KCI Problem: 10P Ba (NO,), (aq) +2KCI(aq) BaCl2 (s) +2KNO3(aq) Now, the enthalpy change values of them are shown in the table below: Chapter: CH10 show all steps AH (kJ mol Ba* (aq Cr(aq BaCl, (s 855.0 CH10 90 CH10 10Q )-167.2 ()537.6 So, the enthalpy change in the reaction will be Putting the values we get -855.0k/ mori-(-537.6) kJ mol-2x (-167.2)k/ mol l -855.0kJ mol+(537.6)kJ mol+(334.4)kJ mol = | 17.0 kJ mol-1 Hence, the value of the enthalpy change is 17.0kJ mol Provide feedback(0) Step 2 of2 2. Consider the following reaction between barium nitrate with KCI Ba (NO,), (aq)+2KCI(aq) >BaCI2 (s)+2KNO, (aq) If we write the ionic form of this reaction, then this will be as Ba2 (aq)+2CI (a)BaCI, (s) Now, the Gibb's free energy values of them are shown in the table below: AG? (kJ mol Ba2(aq-560.8 CI (a)-131.2 BaCl, (s -806.7 So, the Gibb's free energy in the reaction will be Putting the values we get 806.7k/mol.'-(-560.8)k/mol-I-2x(-131.2)kJmol ' 806.7kJmol+(560.8)kJ mol+(262.4)kJ mol 16.5 kJ mol-1 Hence, the value of the Gibb's free energy is 16.5 kJmol Provide feedback(0) Was this solution helpful? Was this solution helpful?1Explanation / Answer
The relation between Keq and Delta G is
Delta Go = -2.303RTlog Keq
Given free enrgy change = 16.5 kJ= 16500J, R = 8.314 , t= 298 K substituting these values
we have Keq = 1.282x 10-3
c) Ba(NO3)(aq) + KCl (aq) -----> BaCl2(aq) + 2KNO3(aq)
Since the solutions are very dilute and barium chloride is water soluble , no precipitation occurs.
Thus the concentrations of all the ions are
[Ba+2] = 100x0.08/200= 0.04M
[K+] = 100x0.06/200 = 0.03M
[NO3-] = 100 x 2x 0.08/200 = 0.08M
[Cl-] = 100x 0.06/200 = 0.03M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.