17.4 % of K2SO4 at 27°C isotonic …

In case of  isotonic electrolytes ,

* the concentrations of both the electrolytes are equimolecular

*  their  osmotic pressures are equal

*  both the electrolytes have same temperature 

Therefore , considering the given electrolytes ,  

{tex}\pi{/tex}_{(K2SO4)   =}  {tex}\pi{/tex}_(NaOH ) 

i_(K2SO4) C_(K2SO4) RT  =  i_(NaOH)  C (_NaOH)RT

but  ,  C_{(K2SO4)  =}  C_(NaOH)

 {tex}\therefore{/tex}   i _{(K2SO4)  =   i ( NaOH )}

Now,  we know that  i (ie .  vant Hoff' 's  factor )

 =   No. of particles in dissociated molecule ( ie. after dissociation ) /   No. of particles in undissociated molecule  ( ie before dissociation )

ie.  =   degree of dissociation  ({tex}\alpha{/tex})

So, let us calculate degree of dissociation for both the given electrolytes-

K₂ SO₄----------------->  2K⁺    +    SO₄^2-                    ;        NaOH ----------->   Na⁺     +    OH^- 

( 1 - {tex}\alpha{/tex} ).........................( 2 {tex}\alpha{/tex} )...........({tex}\alpha{/tex}  ) -----------------------(1- {tex}\alpha{/tex} )..................({tex}\alpha{/tex}).............({tex}\alpha{/tex})

Therefore total number of  particles ( ie ions / molecules )  after dissociation in both cases  are -

( 1 - {tex}\alpha{/tex}  +  2{tex}\alpha{/tex} +  {tex}\alpha{/tex} ).............................................................................( 1 - {tex}\alpha{/tex}  +  {tex}\alpha{/tex}   +    {tex}\alpha{/tex}  )

=  (1  +  2{tex}\alpha{/tex} )................................. ...............................................................=  (  1  +    {tex}\alpha{/tex}  )

..........................................substituting the given value for degree of dissociation ({tex}\alpha{/tex}) in case of NaOH as = 1

.......................................................................................................................= 2

Further  ,   i( K₂ S O₄ )   =  [ (1 + 2{tex}\alpha{/tex} ) /  1  ] ........................&    i (NaOH )   =  ( 2 / 1 )    

.......................................= [ ( 1  +  2{tex}\alpha{/tex} (K₂ SO₄ )  ] .......................................= 2....................................

           Equating i_(K2SO4)   and  i _{( NaOH)} ,   as inferred above we get -

1  +  2{tex}\alpha{/tex} _(K2SO4)    =   2

{tex}\therefore{/tex}   2 {tex}\alpha{/tex} _{(K2SO4)   =  1}

&     {tex}\alpha{/tex}  _(K2SO4)   =  1/ 2

...........................=  0.50

Hence , the degree of dissociation of  given 17.4% K₂ SO₄  is  0.50  and is ,therefore ,50%  dissociated  when it is isotonic with  given 4% NaOH solution .

1