The Calculations of Analytical Chemistry
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— for example suppose we have two solutions of NH4SCN (a) I c.c. = 0.01275 gram of silver (b) I c.c. = 0.00447 gram of silver and we wish to make one liter of a solution equivalent to 10 mgs. of silver per c.c. How much of each shall we mix. Let x be the number of c.c. of (a) ; then 1000 — X = number of c.c. of (b) : and we have 64 EXCESS NECESSARY TO AFFECT THE INDICATOR.
X (0.01275) + Ciooo — x) 0.00447 = 10 (as one liter will contain 10 grams if i c.c. contains lomgs.) This gives 667.9 c.c. ...of (a) and 332.1 c.c. of (b).
If we have a solution of KMn04 containing 2 -grams per liter and a ^/s solution of the same, to make two liters of a -"^/lo KMn04. Let x be the volume of the 2 grams solution in liters and (2 — x) the volume of the ^/j solution ; then 2 x + (2 -x) 6.326 = 2 (3.163) and x = 1.4623 liters, (2 -x) = 0.5377 liter.
SUBTRACTION OF EXCESS NECESSARY TO AFFECT THE INDICATOR.
In some titrations, conspicuously that of zinc by potas- sium ferrocyanide and that of lead by ammonium molyb- date, the end point is not very sharp and a considerable excess of the standard solution is used before the indica- tor is affected.
What to read after The Calculations of Analytical Chemistry?
You can find similar books in the "Read Also" column, or choose other free books by Edmund H Edmund Howd Miller to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
— for example suppose we have two solutions of NH4SCN (a) I c.c. = 0.01275 gram of silver (b) I c.c. = 0.00447 gram of silver and we wish to make one liter of a solution equivalent to 10 mgs. of silver per c.c. How much of each shall we mix. Let x be the number of c.c. of (a) ; then 1000 — X = number of c.c. of (b) : and we have 64 EXCESS NECESSARY TO AFFECT THE INDICATOR.
X (0.01275) + Ciooo — x) 0.00447 = 10 (as one liter will contain 10 grams if i c.c. contains lomgs.) This gives 667.9 c.c. ...of (a) and 332.1 c.c. of (b).
If we have a solution of KMn04 containing 2 -grams per liter and a ^/s solution of the same, to make two liters of a -"^/lo KMn04. Let x be the volume of the 2 grams solution in liters and (2 — x) the volume of the ^/j solution ; then 2 x + (2 -x) 6.326 = 2 (3.163) and x = 1.4623 liters, (2 -x) = 0.5377 liter.
SUBTRACTION OF EXCESS NECESSARY TO AFFECT THE INDICATOR.
In some titrations, conspicuously that of zinc by potas- sium ferrocyanide and that of lead by ammonium molyb- date, the end point is not very sharp and a considerable excess of the standard solution is used before the indica- tor is affected.
What to read after The Calculations of Analytical Chemistry?
You can find similar books in the "Read Also" column, or choose other free books by Edmund H Edmund Howd Miller to read onlineMoreLess
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