Question

What happens when each inorganic pigments is placed in each of the following solutions? Details below.

4 Inorganic Pigments: barium white, zinc yellow, chromium oxide green, prussian blue

3 Solutions at room temp: 3M NaOH, 3M HCl, 10% H2SO4

Please explain what happens to each pigment in each solution. Maybe even include a chemical equation?

Answer 1

Answer on Question #66701, Chemistry / Inorganic Chemistry

What happens when each inorganic pigments is placed in each of the following solutions? Details below.

4 Inorganic Pigments: barium white, zinc yellow, chromium oxide green, prussian blue

3 Solutions at room temp: 3M NaOH, 3M HCl, 10% H2SO4

Please explain what happens to each pigment in each solution. Maybe even include a chemical equation?

Answer

1. $\mathrm{BaSO}_4$ - barium white:

$\mathrm{BaSO}_4 + 2 \mathrm{NaOH} = \mathrm{Ba(OH)}_2 + \mathrm{Na}_2\mathrm{SO}_4$; dissolution of sediment;

$\mathrm{BaSO}_4 + 2 \mathrm{HCl} = \mathrm{BaCl}_2 + \mathrm{H}_2\mathrm{SO}_4$; dissolution of sediment;

$\mathrm{BaSO}_4$ and $\mathrm{H}_2\mathrm{SO}_4$ do not react

2. $\mathrm{ZnCrO}_4$ - zinc yellow:

$\mathrm{ZnCrO}_4 + 2 \mathrm{NaOH} = \mathrm{Zn(OH)}_2 + \mathrm{Na}_2\mathrm{CrO}_4$; we received a yellow white;

$\mathrm{ZnCrO}_4 + 2 \mathrm{HCl} = \mathrm{ZnCl}_2 + \mathrm{H}_2\mathrm{CrO}_4$; bleaching;

$\mathrm{ZnCrO}_4 + \mathrm{H}_2\mathrm{SO}_4 = \mathrm{ZnSO}_4 + \mathrm{H}_2\mathrm{CrO}_4$; bleaching;

3. $\mathrm{Cr}_2\mathrm{O}_3$ - chromium oxide green:

$\mathrm{Cr}_2\mathrm{O}_3 + 2 \mathrm{NaOH} = 2 \mathrm{NaCrO}_2 + \mathrm{H}_2\mathrm{O}$; bleaching;

$\mathrm{Cr}_2\mathrm{O}_3 + 6 \mathrm{HCl} = 2\mathrm{CrCl}_3 + 3\mathrm{H}_2\mathrm{O}$; green became purple;

$\mathrm{Cr}_2\mathrm{O}_3 + 3\mathrm{H}_2\mathrm{SO}_4 = \mathrm{Cr}_2(\mathrm{SO}_2)_3 + 3\mathrm{H}_2\mathrm{O}$; green became pink;

4. $\mathrm{CoO*4Al_2O_3}$ - prussian blue:

CoO and NaOH do not react; $\mathrm{Al}_{2}\mathrm{O}_{3} + 2\mathrm{NaOH} = 2\mathrm{NaAlO}_{2} + \mathrm{H}_{2}\mathrm{O}$; without changes;

$\mathrm{CoO} + 2 \mathrm{HCl} = \mathrm{CoCl}_2 + \mathrm{H}_2\mathrm{O}$; $\mathrm{Al}_{2}\mathrm{O}_{3} + 6 \mathrm{HCl} = 2\mathrm{AlCl}_3 + 3\mathrm{H}_2\mathrm{O}$; blue became purple red or pink;

$\mathrm{CoO} + \mathrm{H}_2\mathrm{SO}_4 = \mathrm{CoSO}_4 + \mathrm{H}_2\mathrm{O}$; $\mathrm{Al}_{2}\mathrm{O}_{3} + \mathrm{H}_2\mathrm{SO}_4 = \mathrm{Al}_2(\mathrm{SO}_4)_3 + \mathrm{H}_2\mathrm{O}$; blue became pink.

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