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155
C
Na
= C
mono
= c
alg
/M
0
= 5.05 mM = 5.05E-3 M (500 x that of the polymer
chains)
The total molar concentration of ‘osmotically active’ particles is thus:
C
total
= C
alg
+ C
Na
= 1.0E-5 M + 5.05E-3 M = 5.06E-3 M
≈
can
Hence, the number of ions totally dominates over the number of polymer
molecules, which generally can be ignored in such calculations. Using the
van’t Hoff equation we obtain:
Some textbooks use the (average) number of charges per molecule (z) in the
equations:
C
Na
=
zC
a
lg
∏
tot
= ∏
a
lg
+∏
Na
=
RTC
a
lg
+
RTC
Na
=
RTC
a
lg
+
RT zC
a
lg
(
)
=
1
+
z
(
)
RTC
a
lg
≈
zRTC
a
lg
By further calculation we find that the result is in fact independent of the
molecular weight of the alginate:
∏
tot
≈ ∏
Na
=
RT z c
a
lg
M
⎛
⎝⎜
⎞
⎠⎟
Inserting
z
=
M
M
0
∏
tot
≈ ∏
Na
=
RT M
M
0
c
a
lg
M
⎛
⎝⎜
⎞
⎠⎟ =
RT c
a
lg
M
0
⎛
⎝⎜
⎞
⎠⎟
If no ions dissociated, C
total
would be 1.0E-5 M, i.e. 0.2% of the value above.
Hence, the osmotic pressure would be roughly 500 times lower.
∏
tot
= ∏
a
lg
+∏
Na
=
RTC
a
lg
+
RTC
Na
=
0.12
atm