A Course in Qualitative Chemical Analysis
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When freshly precipitated, A1(OH) 3 is readily soluble in acids : A1C1 3 + 3 NH 4 OH = | A1(OH) 3 + 3 NH 4 C1; A1(OH) 3 + 3 HC1 = A1C1 8 + 3 H 2 O.
Aluminum hydroxide is also soluble in caustic alkalies (see 2).
2. Potassium or Sodium Hydroxide precipitates A1(OH) 3, soluble in excess with the formation of alkali aluminate : A1(OH) 3 + 3 NaOH = Na 3 AlO 3 + 3 H 2 O.
On carefully neutralizing the alkaline solution of sodium aluminate with hydrochloric acid, A1(OH) 3 is reprecipitated : (a) Na 3 ...A10 3 + 3 HC1 = 3 NaCl + 1 A1(OH) 3 .
If an excess of acid is added, the precipitate which first forms is dissolved and we obtain : (b) A1(OH) 3 + 3 HC1 = AlClg + 3 H 2 O.
It is evident that if the original sodium aluminate solution is at once acidified with HC1, we shall get the net result of (a) and (b). Adding (a) and (), and eliminating Al(OH)g, which ap- pears on opposite sides, we get: Na 3 A10 3 + 6 HC1 = A1C1 3 + 3 NaCl + 3 H 2 O.
If, now, we heat this solution to boiling and add ammonium hydroxide in faint excess, all the aluminum will be precipitated as the hydroxide : AlClg + 3 NH 4 OH = | A1(OH) 3 + 3 NH 4 C1.
What to read after A Course in Qualitative Chemical Analysis?
You can find similar books in the "Read Also" column, or choose other free books by Baskerville, Charles, 1870-1922, Ed to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
When freshly precipitated, A1(OH) 3 is readily soluble in acids : A1C1 3 + 3 NH 4 OH = | A1(OH) 3 + 3 NH 4 C1; A1(OH) 3 + 3 HC1 = A1C1 8 + 3 H 2 O.
Aluminum hydroxide is also soluble in caustic alkalies (see 2).
2. Potassium or Sodium Hydroxide precipitates A1(OH) 3, soluble in excess with the formation of alkali aluminate : A1(OH) 3 + 3 NaOH = Na 3 AlO 3 + 3 H 2 O.
On carefully neutralizing the alkaline solution of sodium aluminate with hydrochloric acid, A1(OH) 3 is reprecipitated : (a) Na 3 ...A10 3 + 3 HC1 = 3 NaCl + 1 A1(OH) 3 .
If an excess of acid is added, the precipitate which first forms is dissolved and we obtain : (b) A1(OH) 3 + 3 HC1 = AlClg + 3 H 2 O.
It is evident that if the original sodium aluminate solution is at once acidified with HC1, we shall get the net result of (a) and (b). Adding (a) and (), and eliminating Al(OH)g, which ap- pears on opposite sides, we get: Na 3 A10 3 + 6 HC1 = A1C1 3 + 3 NaCl + 3 H 2 O.
If, now, we heat this solution to boiling and add ammonium hydroxide in faint excess, all the aluminum will be precipitated as the hydroxide : AlClg + 3 NH 4 OH = | A1(OH) 3 + 3 NH 4 C1.
What to read after A Course in Qualitative Chemical Analysis?
You can find similar books in the "Read Also" column, or choose other free books by Baskerville, Charles, 1870-1922, Ed to read onlineMoreLess
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