Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry277Practice Exercise 6.9To help us in determining what ions are in solution, let’s write down allthe reaction leading to the preparation of the solutions and the equilibriawithin the solutions. These reactions are the dissolution of two solublesalts+ −KH PO () s → K ( aq) + H PO ( aq )2 4 2 4+ 2−Na HPO () s → 2Na ( aq) + HPO ( aq )2 4 4and the acid–base dissociation reactions for H 2 PO 4 – , HPO 4 2– , andH 2 O.− + 2−HPO ( aq) + H O() l H O ( aq) + HPO ( aq )2 4 2 34−−HPO ( aq) + H O() l OH ( aq) + H PO ( aq )2 42 3 42− + 3−HPO ( aq) + H O() l H O ( aq) + PO ( aq )42 32H O() l H O + ( aq) + OH− ( aq )2 3Note that we did not include the base dissociation reaction for HPO 42–because we already have accounted for its product, H 2 PO 4 – , in another reaction.The mass balance equations for K + and Na + are straightforward[ K + ] = 010 . M [ Na + ] = 0.10 Mbut the mass balance equation for the phosphate takes a little bit ofthought. Both H 2 PO 4 – andHPO 4 2– produce the same ions in solution.We can, therefore, imagine that the solution initially contains 0.15 MKH 2 PO 4 , which gives the following mass balance equation.− 2− 3−0.15 M= [ H PO ] + [ HPO ] + [ HPO ] + [ PO ]3 4 2The charge balance equation is+ + +[ HO ] + [ K ] + [ Na ] =34 4−2−[ HPO ] + 2× [ HPO ] + 3×[ PO2 4 44Click here to return to the chapter.Practice Exercise 6.10443−−] + [ OH ]In determining the pH of 0.050 M NH 3 , we need to consider two equilibriumreactions—the base dissociation reaction for NH 3NH ( ) HO () −OH ( ) +aq + l aq + NH ( aq )and water’s dissociation reaction.3 242H O() l H O + ( aq) + OH− ( aq )2 3