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Karolina [17]
3 years ago
11

What are the equilibrium concentrations of pb2+ and f– in a saturated solution of lead(ii) fluoride if the ksp of pbf2 is 3.20×1

0–8?
Chemistry
2 answers:
erica [24]3 years ago
6 0
Answer is:  the equilibrium concentrations fluorine anion are 0.004 M and lead cation are 0.002 M.<span>
Chemical reaction: PbF</span>₂(aq) → Pb²⁺(aq) + 2F⁻(aq).<span>
Ksp = 3,2·10</span>⁻⁸.
[Pb²⁺] = x.
[F⁻] = 2[Pb²⁺] = 2x<span>
Ksp = [Pb²</span>⁺] · [F⁻]².
Ksp = x · 4x².
3,2·10⁻⁸ = 4x³.
x = ∛3,2·10⁻⁸ ÷ 4.
x = [Pb²⁺] = 0,002M = 2·10⁻³ M.
[F⁻] = 2 · 0,002M = 0,004 M = 4·10⁻³ M.

murzikaleks [220]3 years ago
3 0

The equilibrium concentrations of  {\text{P}}{{\text{b}}^{2+}} and {{\text{F}}^-}  in a saturated solution of {\text{Pb}}{{\text{F}}_{\text{2}}}  are \boxed{{\text{0}}{\text{.002 M}}}  and \boxed{{\text{0}}{\text{.004 M}}}  respectively.

Further explanation:

The equilibrium constant between the compound and its ion, when dissolved in solution, is known as solubility product constant. It is denoted by {{\text{K}}_{{\text{sp}}}} . The solubility product constant is used to calculate the product of the concentration of ions at equilibrium. Higher the solubility product constant more will be the solubility of the compound.

The general reaction is as follows:

{\text{AB}}\left({aq}\right)\to{{\text{A}}^+}\left({aq}\right)+{{\text{B}}^-}\left( {aq}\right)

The expression to calculate the solubility product for the general reaction is as follows:

{{\text{K}}_{{\text{sp}}}}=\left[{{{\text{A}}^+}}\right]\left[{{{\text{B}}^-}}\right]

Here,

{{\text{K}}_{{\text{sp}}}}  is the solubility product constant.

\left[{{{\text{A}}^+}}\right]  is the concentration of {{\text{A}}^+} ions.

\left[{{{\text{B}}^-}}\right] is the concentration of {{\text{B}}^-} ions.

The dissociation of {\text{Pb}}{{\text{F}}_2}  occurs as follows:

{\text{Pb}}{{\text{F}}_2}\left({aq}\right)\rightleftharpoons{\text{P}}{{\text{b}}^{2+}}\left({aq}\right)+2{{\text{F}}^-}\left({aq}\right)

Consider the molar solubility of {\text{Pb}}{{\text{F}}_2} to be S. Therefore, after dissociation, the concentration of {\text{P}}{{\text{b}}^{2+}}  and {{\text{F}}^-}  are S and 2S respectively.

The formula to calculate the molar solubility of {\text{Pb}}{{\text{F}}_2} is as follows:

{{\text{K}}_{{\text{sp}}}}=\left[{{\text{P}}{{\text{b}}^{{\text{2+}}}}}\right]{\left[{{{\text{F}}^-}}\right]^2}                                        …… (1)

Substitute S for \left[{{\text{P}}{{\text{b}}^{{\text{2+}}}}}\right] , 2S for \left[{{{\text{F}}^-}}\right] and {\text{3}}{\text{.20}}\times{10^{-8}} for {{\text{K}}_{{\text{sp}}}} in equation (1).

\begin{aligned}{\text{3}}{\text{.20}}\times {10^{ - 8}}&=\left[{\text{S}}\right]{\left[ {{\text{2S}}}\right]^2}\hfill\\{\text{3}}{\text{.20}}\times{10^{-8}}&=\left({\text{S}}\right)\left( {4{{\text{S}}^2}}\right)\hfill\\4{{\text{S}}^3}&={\text{3}}{\text{.20}}\times{10^{-8}}\hfill\\\end{aligned}

Simplifying the above equation for S,

\begin{aligned}4{{\text{S}}^3}&={\text{3}}{\text{.20}}\times{10^{-8}}\hfill\\{{\text{S}}^3}&=\frac{{{\text{3}}{\text{.20}}\times{{10}^{ - 8}}}}{4}\hfill\\{\text{S}}&=\sqrt[3]{{\frac{{{\text{3}}{\text{.20}}\times{{10}^{-8}}}}{4}}} \hfill\\\end{aligned}

Solving for S,

{\text{S}}=0.00{\text{2 M}}

Therefore the equilibrium concentration of {\mathbf{P}}{{\mathbf{b}}^{{\mathbf{2+}}}}  is 0.002 M.

The equilibrium concentration of {{\text{F}}^-}  is calculated as follows:

\begin{aligned}\left[ {{{\text{F}}^-}}\right]&=2\left({{\text{0}}{\text{.002 M}}}\right)\\&=0.00{\text{4 M}}\\\end{aligned}

Therefore the equilibrium concentration of  {{\mathbf{F}}^{\mathbf{-}}} is 0.004 M.

Learn more:

1. Sort the solubility of gas will increase or decrease: <u>brainly.com/question/2802008</u>.

2. What is the pressure of the gas?: <u>brainly.com/question/6340739</u>.

Answer details:

Grade: School School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: Molar solubility, saturated solution, PbF2, Pb2+, F-, S, 2S, 0.002 M, 0.004 M, equilibrium concentration, Ksp, 3.2*10^-8.

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Answer:

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

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Given the following reactions and their standard enthalpy changes:

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(3) 2 NO₂(g) → N₂O₄(g) ΔH o rxn = −57.2 kJ

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

You need to get the heat of reaction from: N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)

Hess's Law states: "The variation of Enthalpy in a chemical reaction will be the same if it occurs in a single stage or in several stages." That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when verified in a single stage.

This law is the one that will be used in this case. For that, through the intermediate steps, you must reach the final chemical reaction from which you want to obtain the heat of reaction.

Hess's law explains that enthalpy changes are additive. And it should be taken into account:

  • If the chemical equation is inverted, the symbol of ΔH is also reversed.
  • If the coefficients are multiplied, multiply ΔH by the same factor.
  • If the coefficients are divided, divide ΔH by the same divisor.

Taking into account the above, to obtain the chemical equation

N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)  you must do the following:

  • Multiply equation (3) by 2

(3) 2*[2 NO₂(g) → N₂O₄(g) ] ΔH o rxn = −57.2 kJ*2

<em>4 NO₂(g) →  2 N₂O₄(g)  ΔH o rxn = −114.4 kJ</em>

  • Reverse equations (1) and (2)

(1) <em>N₂O₃(g)  → NO(g) + NO₂(g) ΔH o rxn = 39.8 kJ</em>

(2) <em>N₂O₅(g) →  NO(g) + NO₂(g) + O₂(g)  ΔH o rxn = 112.5 kJ</em>

Equations (4) and (5) are maintained as stated.

(4) <em>2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ </em>

(5) <em>N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ </em>

The sum of the adjusted equations should give the problem equation, adjusting by canceling the compounds that appear in the reagents and the products according to the quantity of each of them.

Finally the enthalpies add algebraically:

ΔH= -114.4 kJ + 39.8 kJ + 112.5 kJ -114.2 kJ + 54.1 kJ

ΔH= -22.2 kJ

<u><em>The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ</em></u>

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