654 LABORATORY HINTS ' 



Deduct the weight of dish, remainder equals total solids in 200 c.c. 

 Simply calculate how much in 100,000 c.c. 

 Example : — 



Dish + total solids — 27*848 grm. 

 Dish alone = 27797 ,, 



Weight of residue = o'05i grm. x 100,000 



= 25'5 parts 



200 per 100,000 



(2) To Estimate the Hardness. 



The hardness of water is largely due to the presence of bi- 

 carbonates, sulphates and chlorides of calcium and magnesium. 



(i) Total Hardness. 



In a stoppered bottle of 150 c.c. capacity place 100 c.c. of sample. 



Run in from a burette standard soap solution (i c.c. = o"ooi grm. 

 CaCOg), shaking vigorously until a permanent lather of \ in. is 

 obtained. 



Exaynple. 



13*8 c.c. of soap solution is required. 



Hence I3'8 x cooi is contained in 100 c.c. of water. 



I3'8 X 001 X 100,000 

 = I3'8 parts per 100,000. 



10 



If 100 c.c. are used this latter calculation is avoided as the c.c. used, 



i3'8 in this case, indicates the parts per 100,000. 



(2) Temporary Hardness. 



This can be removed by boiling. 

 It is due to the bicarbonates of lime and magnesia. 

 On boiling the COo is driven off and the carbonates are precipitated. 

 Commercially it is removed by adding lime. 



The degree of temporary hardness is calculated by subtracting the 

 permanent from the total hardness. 



(3) Permanent Hardness. 



This is due to the sulphates and chlorides of calcium and 

 magnesium. 



They are soluble in water and cannot be precipitated by boiling. 

 To estimate them : — 



Boil 100 c.c. in a beaker for fifteen minutes. 



Filter when cold and make up the filtrate with distilled water to 

 100 c.c. Shake well. 



Run in soap solution as described. 

 Example : 4*2 c.c. required. 

 Hence 4*2 parts per 100,000. 



But the total hardness was 138 



= 9'6 parts per 100,000 temporary hardness. 



The permanent „ „ 4-2 



