﻿14 Mr. W. Sutherland on the Molecular 



Solutions of Na 2 S0 4 may be taken as typical o£ those in 

 which p d c + JO4C4 — c' becomes a negative at a certain concen- 

 tration. The experimental data are taken from the Tabdlen 

 of Landolt and Bornstein, 3rd edit., as are all the subsequent 

 data for specific heats. I find ar/^ = 0'49 and b=l'7, with 

 the following comparison : — 



1OOOj0 4 ... 303 193 108 19 



1000c' exp. ... 

 ,, calc... 



781 

 784 



8)3 

 835 



892 

 892 



977 

 977 



I000(p 3 c+p^-c')... 



-14 



+9 



25 



8 



1000{p. 3 c +/> 4 c 4 - c')/p#4 exp. . . . 

 „ „ „ calc. ... 



-7 

 -8 



+6 

 + U 



26 

 26 



43 

 42 



The next table contains the values of ar and be for those 

 electrolytes for which the necessary data are available. 



Table II L 



NaCl. NaBr. Nal. NaN0 3 . Na 2 S0 4 . NaOH. KC1. KBr. KT. 



ar 29 32 31 32 32 51 41 39 41 



be 99 56 51 124 125 122 91 36 43 



K 2 S0 4 . NH 4 C1. MgCl 2 . CaCl 2 . SrCl 2 . BaCl 2 . MgS0 4 . ;Pb(N0 3 ) 2 . 



ar "39 34 41 40 44 44 42 21 



be 104 57 55 64 79 112 105 43 



The first point to arrest attention in this table is that ar 

 for NH 4 C1 has a positive value 34, although r is 1*29 - 1*39 

 or —0*1. The second is that ar for all the Na compounds 

 except NaOH has a value near to 31, and for all the K com- 

 pounds a value near to 40. For MgCl 2 and MgS0 4 the values 

 of ar are nearly the same. From these facts it follows that 

 the ordinary negative ions contribute no part to t in ar. The 

 reason for this is fairly obvious. A rise of temperature 

 makes trihydrol more unstable, causing more of it to become 

 dihydrol in pure water, whereas a rise of temperature pro- 

 duces no appreciable effect on dihydrol, the predominant 

 ingredient of water. Now the negative ions attack dihydrol 

 to make trihydrol out of it ; but as a rise of temperature does 

 not make (H 2 0) 2 more unstable, the negative ions contribute 

 a negligible part to (D + dh/dpz^dpz'/dt, and therefore also a 

 negligible part to ar. We can test this reasoning by applying 

 it to the case of NaOH. The OH ion dissociates O86 H 2 6 

 into 086 OH and 0*86 H ; but the OH contributes no part 

 to ar; so that the value 51 for NaOH is the sum of 31 

 for Na and 0*86 times the part of ar due to H, which is 

 (51-31)/0'86 or 23. We have then the result that the 



