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3 years ago

Insoluble sulfide compounds are generally black in color.

Chemistry

1 answer:

AfilCa [17]3 years ago

5 0

<h3>Answer:</h3>

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

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

<h3>Explanation:</h3>

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)

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3 years ago

A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.

seropon [69]

<u>Answer:</u> The p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

<u>Explanation:</u>

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

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Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

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1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, $M_1=(2\times 0.0031)=0.0062M$

Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, $M_2=(2\times 0.0042)=0.0084M$

To calculate the concentration of nitrate ions in the solution, we use the equation:

$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$

Putting values in above equation, we get:

$M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M$

Calculating the p-function of zinc ions and nitrate ions in the solution:

<u>For zinc ions:</u>

$\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]$

$\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51$

<u>For nitrate ions:</u>

$\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]$

$\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14$

Hence, the p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

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