Insoluble sulfide compounds are generally black in color.

Which of the following combinations could yield a black precipitate?

Na2S (aq) + KCl (aq)

Li2S (aq) + Pb (NO3) 2 (aq)

Pb (ClO3) 2 (aq) + NaNO3 (aq)

AgNO3 (aq) + KCl (aq)

K2S (aq) + Sn (NO3) 4 (aq

Li₂S (aq) + Pb (NO₃) ₂ (aq) → PbS (s) + LiNO₃ (aq)

K2S (aq) + Sn (NO₃) ₄ (aq→ SnS (s) + KNO₃ (aq)

Explanation: According to solubility rules, metal sulfides are insoluble except Calcium sulfide (CaS), Magnesium sulfide (MgS), Barium sulfide (BaS), and those of potassium, sodium, and ammonium. Therefore, when all the other sulfides are formed during a reaction they form a precipitate that is shown to be in solid-state. Precipitates are formed during precipitation reactions when cations and anions combine to form a compound that is insoluble in water.

In our case; In the equations given, some reactions will take place to form a precipitate while others will not occur.

That is;

Na₂S (aq) + KCl (aq) → No reaction Li₂S (aq) + Pb (NO₃) ₂ (aq) → PbS (s) + LiNO₃ (aq) Pb (ClO₃) ₂ (aq) + NaNO₃ (aq) → No reaction AgNO₃ (aq) + KCl (aq) →AgCl (s) + KNO₃ (aq) K2S (aq) + Sn (NO₃) ₄ (aq→ SnS (s) + KNO₃ (aq)

The first reaction will not occur.

The second equation is a precipitation reaction that forms lead (ii) sulfide which is a black precipitate.

The third reaction will not take place.

The fourth reaction will be a precipitation reaction forming silver chloride precipitate.

The fifth equation is also a precipitation reaction that forms tin (ii) sulfide, SnS which is a black precipitate.

Therefore; the equations that answers our question are;

Li₂S (aq) + Pb (NO₃) ₂ (aq) → PbS (s) + LiNO₃ (aq) K2S (aq) + Sn (NO₃) ₄ (aq→ SnS (s) + KNO₃ (aq)