17.4% K_(2)SO_(4) solution at 27^(@)C is isotonic with 4% NaOH solution at the same temperature. If NaOH is 100% ionized, what is the degree of ionization of K_(2)SO_(4) in aqueous solution?

17.4% K_(2)SO_(4) solution at 27^(@)C is isotonic with 4% NaOH solutio

1.	17.4% K_(2)SO_(4) solution at 27^(@)C is isotonic with 4% NaOH solution at the same temperature. If NaOH is 100% ionized, what is the degree of ionization of K_(2)SO_(4) in aqueous solution?
Answer» Solution :As the solution are isotonic, the molar concentration of particles in the two solution must be same. Molar concentration of `K_(2)SO_(4)=(17.4)/(174)molxx(1)/("100 mL")xx"1000 mL L"^(-1)=1.0M` `""(because "molar mass of "K_(2)SO_(4)="174 g mol"^(-1))` Molar concentration of NaOH `=(4)/(40)molxx(1)/(100mL)xx1000"mL L"^(-1)=1.0M` `""(because" molar mass of NaOH = 40 g mol"^(-1))` As NaOH completely ionizes as `NaOH rarr Na^(+) +OH^(-)`, molar COCENTRATION of particles = 2.0 M If `alpha` is the degree of ionization of `K_(2)SO_(4)`, then `{:(,K_(2)SO_(4)rarr2K^(+)+SO_(4)^(-)),("INITIAL conc.","1 M"),("Conc. after ionization",""1-alpha""2alpha""alpha):}` Thus, the total concentration of the particles after ionization `=(1-alpha)+2alpha+alpha =1+2alpha` As two solution are isotonic, `1+2alpha=2 or alpha=0.5` Thus, degree of ionization of `K_(2)SO_(4)=0.5`, i.e., it is `50%` IONIZED.