765 
another way still, viz. by measuring the electrical conductivity of 
the solutions. 
If to a solution containing C-- and Cl’ ions, C, 0," and Na’ ions 
be added, then the product Ca X C,O, cannot exceed the square 
of the solubility of CaC,0O,. If too many C, 0, ions have been 
added, then undissociated CaC,O, must be formed. How much 
CaC,O, will be formed, if the product is exceeded by a fixed 
quantity of C, O,? 
To a binary electrolyte with a solubility A, a salt which has an anion in 
common with the first is added in a concentration x. Through this the solubility 
of the first salt is changed to A’. The total concentration of the anion then 
amounts to A’+ 2, that of the kation to A’. The solubility product is therefore ; 
A‘(A’+ x), and because this is constant we have: 
A' (A! + 2) = A? 
or 
A 
— wt WAA Ee? 
5 8 
The quantity of undissociated salt which results when x Mol salt 
that has 1 ion in common with the first is added, therefore is: 
=~ se VAR dE 
A culls SII Lice 4 Ae maar ena (| 
5 (1) 
if A represents the solubility of the first salt. 
We have now e.g. 5c.c. of an aqueous solution of CaCl, .6aq., free of COs, con- 
taining per litre 0.56 millimol Ca: and (2 Cl’). To this there is added several times 
successively 0.0050 ce. of a 0.05 N solution of Na,C,O,. After every addition the 
conduetivity is measured. The Na,C,;0, may here be added in solution, for here 
the solution may be supersaturated. 
By means of the first method the value now found for the solubility product is 
0.055. From this it follows that a C,O, concentration of a magnitude 1 millimol 
corresponds to the 0.56 millimol Ca. Upon every addition of 0.0050 c.c. 0.05 N. 
Na,0,0, to 5 ce, of a solution of CaCl, 6 aq. the C,0, concentration increases 
by 0.25 mm. After 4 additions therefore the solubility product is reached. What is 
the relation between the total concentrations of ions during these additions? 
For the first addition the total ion concentration is 
0,56 millimol Ca” + 0,56 m.m. (2 Cl”) = 1,12 m.m. 
After the first addition of 0.025 mm. NasC,0, 
0,56 Ca + 0,56 (2 CI”) + 0,025 40, + 0,025 (2 Na”) = 1,17 m.m. 
Thus the total ion concentration after the 2nd addition is 1.22 m.m., after the 
Srd 1.27 mm. and after the 4th 1.32 m.m. Upon the 5th addition the solubility 
product is exceeded. According to the deduced formula (1) the amount of undis- 
sociated CaC,0, formed = 
0,025 + V4 « 0,055 +4- 0.025? 
VY 90,055 0 ONO A Ke 
, 
“ 
The total ion concentration becomes thus after the 5th addition 
