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umka2103 [35]
1 year ago
15

The concentration of iodide ions in a saturated solution of lead (ii) iodide is ________ m. The solubility product constant of p

bi2 is?
Chemistry
1 answer:
alexandr402 [8]1 year ago
7 0

The concentration of iodide ion in a saturated solution of lead (ii) iodide is 3.04 × 10⁻³ M.

<h3>What is a Chemical Reaction ?</h3>

A chemical reaction is a process in which chemical bonds between atoms to break and reorganize to form other new substances.

Now write the equation:

PbI₂ (s) ↔ Pb⁺² (aq) + 2I⁻ (aq)

_                x                 2x

<h3>How to find the concentration when solubility product is given ? </h3>

It is expressed as:

K_{sp} = [A^{+}]^{a} [B^{-}]^{b}

where,

K_{sp}<em> = </em>Solubility product constant

A⁺ = Cation in an aqueous solution

B⁻ = Anion in an aqueous solution

a, b = Relative concentration of A and B.

Now put the values in above formula we get

K_{sp} = [A^{+}]^{a} [B^{-}]^{b}

1.4 × 10⁻⁸ = [Pb⁺²]¹ [I⁻]²

1.4 × 10⁻⁸ = [x]¹ [2x]²

1.4 × 10⁻⁸ = 4x³

x³ = 3.5 × 10⁻⁹

x = 1.52 × 10⁻³

So,

Pb⁺² = x = 1.52 × 10⁻³

I⁻ = 2x = 2 × 1.52 × 10⁻³

          = 3.04 × 10⁻³ M

Thus from the above conclusion we can say that The concentration of iodide ion in a saturated solution of lead (ii) iodide is 3.04 × 10⁻³ M.

Disclaimer: Th question was given incomplete on the portal. Here is the complete question.

Question: The concentration of iodide ions in a saturated solution of lead (ii) iodide is ________ m. The solubility product constant of PbI₂ is 1.4 × 10⁻⁸.

Learn more about the Solubility Product here: brainly.com/question/1419865

#SPJ4

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The number of atoms of each element :

C : 1 atom

H : 3 atoms

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<h3>Further explanation</h3>

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Required

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The empirical formula is the smallest comparison of atoms of compound forming elements.  

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Calculate the oxidation number for carbon in CH4
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Dolomite is a mixed carbonate of calcium and magnesium. Calcium and magnesium carbonates both decompose upon heating to produce
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Answer:

72.03 %

Explanation:

Total mass of dolomite = 9.66 g

Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol

According to the reaction shown below:-

MgCO_3\rightarrow MgO+CO_2

CaCO_3\rightarrow CaO+CO_2

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = \frac{x}{84.3139}\ mol

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = \frac{x}{84.3139}\times 40.3044 \ g

Moles of CaO = \frac{9.66 - x}{100.0869}\ mol

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = \frac{9.66 - x}{100.0869}\times 56.0774 \ g

Given that total mass of the oxide = 4.84 g

Thus,

\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84

\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84

-694.1618435x+45673.48749\dots =40843.38968\dots

x=\frac{4830.09780\dots }{694.1618435}

x=6.9582

Thus, the mass of Magnesium carbonate = 6.9582 g

\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100

\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%

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