510 Proceedings of Royal Society of Edinburgh. [sess. 
Haying found by means of this equation two pairs of correspond- 
ing values of x and A, tbe required value of x , when A is equal to 
0‘259, was calculated by the ordinary interpolation formula, viz.: — 
X — X j A o1te — A-^ 
X ibg AqJjs. Ag 
where x x and A 15 x 2 and A 2 , x and A obs- are pairs of corresponding 
values, and A obs . is equal to 0’259. 
acid is easily obtained, being equal to 
From x tbe percentage of free 
(B — a)100 . In the follow . 
ing tables are exhibited the calculated values of x and of the 
percentage of free acid in all the solutions, together with the 
observed values of the same. 
First Series . 
h 2 so 4 +k 2 so 4 
X . 
Percentage Free Acid. 
Observed. 
Calculated. 
Observed. 
Calculated. 
0-025 + 0-1 
*00960 
•00938 
61-6 
62'5 
0-05 „ 
•01705 
•01646 
65-9 
67*1 
o-i „ 
•0282 
•0271 
71 ’ 8 
72-9 
0-2 „ 
•0400 
•0415 
80-0 
79-3 
0-35 „ 
•0528 
*0543 
84;9 
84-5 
Second Series. 
h 2 so 4 +k 2 so 4 
X . 
Percentage Free Acid. 
Observed. 
Calculated. 
Observed. 
Calculated. 
0-1 + 0-4 
•0606 
•0638 
39-1 
36*2 
„ 0-2 
•0457 
•0442 
54-3 
55-8 
„ o-i 
•0282 
•0271 
71-8 
72-9 
„ 0-05 
•0154 
•01545 
84*6 
84'6 
„ 0'025 
•0081 
•00825 
91-8 
91-8 
