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17.4%(w/v) K2SO4 solution at 27℃ is an isotonic with 5.85%(w/v) NaCl solution at 27℃. if NaCl is 100% ionised what will be the percent of ionisation K2SO4

17.4%(w/v) K2SO4 solution at 27℃ is an isotonic with 5.85%(w/v) NaCl solution at 27℃. if NaCl is 100% ionised what will be the percent of ionisation K2SO4

Grade:12

3 Answers

Abhiraj Pathak
82 Points
5 years ago
Isotonic solutions have same osmotic pressure. Osmotic pressure (p) =icRTSince solutions are isotonic,(ic)NaCl = (ic)K2SO4For NaCl, i=2, since it dissociates 100%.c for NaCl = 5.85c for K2SO4 = 17.84Sunstiuting the values we get i= 0.655i.e K2SO4 is 65.5% ionised
T_chauhan
13 Points
4 years ago
As Isotonic solutions have same osmotic preesure so,
i1c1RT = i2c2RT    so, Let V1=V2=100ml
put i1=m=2                  c1=5.85/58.5*100ml = 0.1/100ml
c2=17.4/174*100ml = 0.1/100ml     so,
i2 upon substituting is 2,    Therefore 
Alpha=Degree of disassociation2 = i2-1/m-1 = 2-1/3-1 = 0.5
m = 3 because k2so4 is disassociated into 3 ions i.e. 2K+ + 1SO42- 
is D.of.disasso. = 0.5 then K2so4 is 50% disassociated !!
Hope it helps
ankit singh
askIITians Faculty 614 Points
3 years ago

For K2SO4,

             K2SO42K++SO42
At eqm,   1α             2α           α       [α= degree of dissociation of K2SO4]
 
Vant Hoff factor (i)=11α+α+2α=1+2α
 
Let the concentration be C of K2SO4
 
Concentration at equilibrium =C(1+2α)
 
For NaOH,
              NaOHNa++OH
At eqm,   1β            β            β
i=1β+β+β=1+β
 
Let the concentration also be C (since both solution are isotonic)
Concentration at eqm= C(1+β)
 
Both solution are isotonic
C(1+2α)=C(1+β)
1+2α=1+β=1+1    [β=1 i.e. 100% ionized]
1+2α=2
α=0.5
Degree of ionisation of K2SO4 in aqueous solution is 50% .

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