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3 years ago

Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed.

Chemistry

1 answer:

wolverine [178]3 years ago

6 0

Answer:

a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. NO REACTION

Explanation:

For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.

a. Potassium carbonate = K₂CO₃

Lead(II) nitrate = Pb(NO₃)₂

Products = KNO₃ and PbCO₃.

According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:

K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Lithium sulfate = Li₂SO₄

Lead(II) acetate = Pb(C₂H₃O₂)₂

Products = Li(C₂H₃O₂) and PbSO₄

All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:

Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Copper(II) nitrate = Cu(NO₃)₂

Magnesium sulfide = MgS

Products = CuS and Mg(NO₃)₂

All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:

Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. Strontium nitrate = Sr(NO₃)₂

Potassium iodi = KI

Products = K(NO₃)₂ and SrI₂

All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.

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What is a Boitic factor

svetoff [14.1K]

Answer:

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Explanation:

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3 years ago

Consider the solutions, 0.04 m urea [(NH_2)_2C=O)], 0.04 m AgNO_3 and 0.04 m CaCl_2. Which has

kaheart [24]

Answer:

Explanation:

All the colligatives properties are modified by the Van't Hoff factor. This is shown as i, in the formula of the colligative properties freezing point depression, boiling point elevation, and osmotic pressure.

The Van't Hoff factor shows the number of particles into which the solute dissociates. In organic compounds we use 1, in inorganic compounds we have to think the dissociation. For example in aquous solutions, inorganic salts as AgNO₃ and CaCl₂ dissociate like this:

AgNO₃ → Ag⁺ + NO₃⁻

CaCl₂ → Ca²⁺ + 2Cl⁻

AgNO₃ has 2 moles of ions and CaCl₂ has 3 moles.

In urea: i = 1

In silver nitrate: i = 2

In calcium chloride: i = 3

The i = 3 for calcium chloride, makes that this salt has the highest osmotic pressure, highest vapor pressure and highest boiling point.

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3 years ago

The chemical formula is different from the empirical formula in

Alik [6]

Answer:be careful and relax

Explanation:

7 0

3 years ago

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4) Balance the following redox reaction in an acidic solution. What are the coefficients in front of H⁺ and Fe3+ in the balanced

Kryger [21]

Answer:

- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5

- There are transferred 5 moles of e-

Explanation:

This is the reaction:

Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Let's think the oxidation numbers:

Fe2+ changed to Fe3+

It has increased the oxidation number → OXIDATION

Mn in MnO₄⁻ acts with +7 and it changed to Mn²⁺

It has decreased the oxidation number → REDUCTION

Let's make the half reactions:

Fe²⁺ → Fe³⁺ + 1e⁻ (it has lost 1 mol of e⁻)

MnO₄⁻ + 5e⁻ → Mn²⁺ (it has gained 5 mol of e⁻)

Now we have to ballance the O. In acidic medium we complete with water as many oxygens we have, in the opposite side. We have 4 O in reactant side, so we fill up with 4 H2O in products side.

MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O

Now we have to ballance the H, so as we have 8 H in products side, we complete with 8H⁺ in reactants, this is the complete half reaction:

8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O

Notice that have 1e⁻ in oxidation and 5e⁻ in reduction. We must multiply (x5) the half reaction of oxidation, so the electrons can be cancelled.

(Fe²⁺ → Fe³⁺ + 1e⁻ ) .5

5Fe²⁺ → 5Fe³⁺ + 5e⁻

8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O

We sum both half reactions:

5Fe²⁺ + 8H⁺ + MnO₄⁻ + 5e⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O

The electrons are cancelled, so the ballanced reaction is this:

5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O

3 0

3 years ago

We will abbreviate malonic acid CH2(CO2H)2, a diprotic acid, as H2A (pK1 = 2.847 and pK2 = 5.696). Find the pH in (a) 0.200 M H2

larisa [96]

Answer:

a. pH → 1.77

b. pH → 4.27

Explanation:

Malonic acid is a dyprotic acid. It releases two protons:

H₂A + H₂O → H₃O⁺ + HA⁻ Ka1

HA⁻ + H₂O → H₃O⁺ + A⁻² Ka2

Let's find the first pH:

We expose the mass balance:

Ca = [HA] + [HA⁻] + [A⁻²] = 0.2 M

We can not consider the [A⁻²] so → Ca = [HA] + [HA⁻] = 0.2M

As the acid is so concentrated, we can not consider the HA- so:

Ca = [HA] = 0.2 M

Charge balance → [H⁺] = [HA⁻] + [OH⁻]

H₂A + H₂O → H₃O⁺ + HA⁻ Ka1

Ka = H₃O⁺ . HA⁻ / H₂A

We need the HA⁻ value to put on the charge balance. We re order the Ka expression:

HA⁻ = Ka . H₂A / H₃O⁺ (notice that H₃O⁺ = H⁺)

We replace: H⁺¨ = Ka . H₂A / H⁺

(H⁺)² = Ka . Ca

H⁺ = √(Ka . Ca) We determine Ka from pKa → 10²'⁸⁴⁷ = 1.42×10⁻³

H⁺ = √(1.42×10⁻³ . 0.2) = 0.016866

- log [H⁺] = pH → - log 0.016866 = 1.77

b. NaHA is the salt from the weak acid, where the HA⁻ works as an amphoterous, this means that can be an acid or a base:

HA⁻ + H₂O ⇄ A⁻² + H₃O⁺ Ka₂

HA⁻ + H₂O ⇄ H₂A + OH⁻ Kb2

There is a formula than can predict the pH, so now we need to compare the Ka₂ and Kb₂ data.

Ka₂ = 10⁻⁵'⁶⁹⁶ = 2.01×10⁻⁶

Kb₂ = 1×10⁻¹⁴ / 1.42×10⁻³ = 7.04×10⁻¹²

So Ka₂ > Kb₂. In conclussion the pH will be acid.

[H⁺] = √(Ka1 . Ka2) → [H⁺] = √ 1.42×10⁻³ . 2.01×10⁻⁶ = 5.34×10⁻⁵

- log 5.34×10⁻⁵ = pH → 4.27

3 0

3 years ago