wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The freezing point of an aqueous solution of NaHSO4 is 0.372 oC. The dissociation constant for the reaction
HSO4H++SO24 is 0.4. Find the concentration of NaHSO4 in mol/kg.

[Kf for H2O=1.86 Kkgmol1]
(Assume dilute solution and take molality = molarity)

Open in App
Solution

Let the molality of NaHSO4 solution be m, then assuming molarity = molality.
HSO4 H+ + SO24
m(1α) mα mα

i=1+1+α=2+α

Effective molality =[Na+]+i[HSO4]
=m+m(1+α)=m(2+α)
Tf=Kfim
0.372=1.86×m(2+α)
0.2=m(2+α)

Also, Ka=mα21α=0.4
0.22+α×α21α=0.4
α2=0.2(2+α)(1α)
So,
1.2α2+0.2α0.4=0
6α2+α2=0
α=0.5
So, m=0.22.5=0.08 mol/kg


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Depression in Freezing Point
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon