Question

what regent(s) would distinguish between
a) barium nitrate and lead nitrate solution?
b) copper (ii) nitrate and barium nitrate solution ?
c) silver nitrate and sodium nitrate solution ?
d) sodium nitrate and sodium sulfate solution ?

Answer 1

what regent(s) would distinguish between

a) barium nitrate and lead nitrate solution?

b) copper (ii) nitrate and barium nitrate solution?

c) silver nitrate and sodium nitrate solution?

d) sodium nitrate and sodium sulfate solution?

Answer:

a) We use potassium iodide (KI) to distinguish lead nitrate from barium nitrate. Bright yellow precipitate of lead iodide forms when lead nitrate solution is mixed with $a$ solution. The ionic equation is:

$$\mathrm{Pb}^{2+}_{(aq)} + 2\mathrm{I}^{-}_{(aq)} = \mathrm{PbI}_{2(s)}$$

Bright yellow precipitate of $\mathrm{PbI}_2$ is obtained from hot solution after fast cooling.

The presents of barium nitrate in solution $a$ we can verify by reaction with sodium sulfate:

\mathrm{Ba}^{2+}_{(aq)} + \mathrm{SO}_4^{2-}_{(aq)} = \mathrm{BaSO}_4 (white precipitate) which is insoluble in acid conditions.

b) We use sodium sulfate $(\mathrm{Na}_2\mathrm{SO}_4)$ to distinguish barium nitrate from copper (II) nitrate. White precipitate of barium sulfate forms when sodium sulfate solution is mixed with $b$ solution. The ionic equation is:

\mathrm{Ba}^{2+}_{(aq)} + \mathrm{SO}_4^{2-}_{(aq)} = \mathrm{BaSO}_4_{(s)}

It is note the barium sulfate is insoluble in acid conditions.

The presents of copper (II) nitrate in solution $b$ we can verify by reaction with sodium sulfide.

Black precipitate of copper (II) sulfide will form by the ionic equation:

$$\mathrm{Cu}^{2+}_{(aq)} + \mathrm{S}^{2-}_{(aq)} = \mathrm{CuS}$$

(black precipitate).

c) We use sodium chloride (NaCl) to distinguish silver nitrate from sodium nitrate. White precipitate of silver chloride forms when sodium chloride solution is mixed with $c$ solution: The ionic equation of this reaction is:

$$\mathrm{Ag}^{+}_{(aq)} + \mathrm{Cl}^{-}_{(aq)} = \mathrm{AgCl}_{(s)}$$

Silver chloride is insoluble in acid conditions but it is soluble in strong solution of $\mathrm{NH}_4\mathrm{OH}$.

Sodium nitrate can verify when solution of $(\mathrm{Zn(UO_2)_3(CH_3COO)_8 + CH_3COOH})$ is mixed with $c$ solution. The pale yellow crystals of $\mathrm{Na_2(UO_2)_3(CH_3COO)_8}$ will form by the ionic equation:

$$\mathrm{Na}^{+} + \mathrm{Zn(UO_2)_3(CH_3COO)_8 + CH_3COOH + 9H_2O = NaZn(UO_2)_3(CH_3COO)_9 \cdot H_2O \downarrow + H^+}$$

d) We use barium chloride $(\mathrm{BaCl}_2)$ to distinguish sodium sulfate from sodium nitrate. White precipitate of barium sulfate forms when sodium sulfate solution is mixed with $d$ solution: The ionic equation is:

\mathrm{SO}_4^{2-}_{(aq)} + \mathrm{Ba}^{2+}_{(aq)} = \mathrm{BaSO}_4_{(s)}

Barium sulfate is insoluble in acid conditions.

Sodium nitrate in solution $d$ we identify by means of diphenylamine test. The solution which includes $\mathrm{NO}_3^-$ ions is painted in black blue color.