Question
For a solution of (a certain type of) locust bean gum in water at room temperature,
the Mark–Houwink parameters areK¼ 8? 10 ^4 m^3 ?kg^1 anda¼ 0 :79. What would
be the zero shear rate viscosity of a 0.01%solution, assuming thatMequals (a) 10^5
or (b) 10^6 Da?ðZs¼1 mPa?s:ÞAre the results obtained by applying Eq. (6.6)
reliable?
Answer
Equation (6.6) directly gives for (a), and (b) that ½Z 0 ¼ 7 :1 and 44 m^3 ?kg^1 ,
respectively. Taking Eqs. (5.9) and (5.10), and assuming dZ=dcto be independent of
c, it is derived (check this) that
Z¼Zsð 1 þc½Z 0 Þat very small shear rate. 0.01%corresponds toc& 0 :1kg?m^3 , and it follows that
for (a), and (b) the viscosity would be 1.7 and 5:4 mPa?s, respectively.
To check whether these results are reasonable, it is useful to invoke the
Einstein equation (5.6)
Z¼Zsð 1 þ 2 : 5 jÞand compare it with the relation above; it follows thatjwould equal½Zc= 2 :5. This
gives for sample (a)j¼ 0 :28 and for (b) 1.76. The latter value is clearly impossible,
and the derivation given thus is invalid; the actual viscosity would be very much
higher than calculated above. Actually, also 0.28 is much too high ajvalue to
ensure that dZ=dcis independent ofc (the viscosity actually would be about
2 :4 mPa?s in this case).
6.3 POLYELECTROLYTES
In this section some specific aspects of polymers bearing electric charge will
be discussed. The reader is referred to Section 2.3, Electrolyte Solutions, for
basic aspects.