Consider the following chemical equation: NH4NO3(s)⟶NH+4(aq)+NO−3(aq) What is the standard change in free energy in kJmol at 298.15K? The heat of formation data are as follows: ΔH∘f,NH4NO3(s)=-365.6kJmolΔH∘f,NH+4(aq)=-132.5kJmolΔH∘f,NO−3(aq)=-205.0kJmol The standard entropy data are as follows: S∘NH4NO3(s)=151.1Jmol KS∘NH+4(aq)=113.4Jmol KS∘NO−3(aq)=146.4Jmol K Your answer should include two significant figures.

Answers

Answer 1

Answer:

[tex]\Delta _rG=-4.3\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, for the given dissociation reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):

[tex]\Delta _rH=\Delta _fH_{NH^{4+}}+\Delta _fH_{NO_3^-}-\Delta _fH_{NH_4NO_3}\\\\\Delta _rH=-132.5+(-205.0)-(-365.6)=28.1kJ/mol[/tex]

Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):

[tex]\Delta _rS=S_{NH^{4+}}+S_{NO_3^-}-S_{NH_4NO_3}\\\\\Delta _rS=113.4+146.4-151.1=108.7J/mol*K[/tex]

Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the given temperature (298.15 K), we finally obtain (two significant figures):

[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=28.1kJ/mol-(298.15 K)(108.7\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\\Delta _rG=-4.3\frac{kJ}{mol}[/tex]

Best regards.


Related Questions

Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ

Answers

Answer:

CO2 will diffuse more rapidly.

Explanation:

From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:

Rate (R) & 1/√Density (d)

R & 1/√d

But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.

Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:

Rate (R) & 1/√Molar mass (M)

R & 1/√M

From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.

Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.

This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71 g/mol

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Summary

Gas >>>>>> Molar mass

Cl2 >>>>>> 71 g/mol

CO2 >>>>> 44 g/mol

From the illustration above, we can see that CO2 is lighter than Cl2.

Therefore, CO2 will diffuse more rapidly.

Answer: CO2

Explanation:

How do covalent bonds form? A Sharing valence electrons between atoms. B Donating and receiving valence electrons between atoms. C Opposite slight charges attract each other between compounds. D Scientists are still not sure how they form.

Answers

Answer:

A. Sharing valence electrons between atoms.

Explanation:

This is the definition of a covalent bond. Option B describes ionic bonds, Option C describes intermolecular forces, and Option D is wrong because then there wouldn't be any mention of them in our high school chemistry textbooks :).  

write any four characties of vertebratas?​

Answers

Answer:

1. to bend

2. to sit

3. to walk

4. To stand

Explanation:

Complete the following equation of nuclear transmutation.
23892U + 126C → 24498Cf + 6 ______
Complete the following equation of nuclear transmutation.
U + C → Cf + 6 ______
A) 1n
B) 0 e
C) 0 e
D) 1H
E) 0g 0 -1 +1 1 0

Answers

Answer:

Option A. 1 0n

Explanation:

Details on how to balanced the equation for the reaction given in the question above can be found in the attached photo.

The missing part of the transmutation equation as it has been shown is 1/o n. Option A

What is nuclear transmutation?

Nuclear transmutation is the process of shifting the number of protons in an atom's nucleus to change one element into another. Nuclear processes that change one atomic nucleus into another with a different atomic number are involved.

The production of nuclear energy, radioactive decay, and the creation of new isotopes for use in science and industry all depend on nuclear transmutation, a fundamental idea in nuclear physics.

We have the equation as;

238/92 U + 12/6 C  ----> 244/98 Cf + 6 1/0 n

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Why can long chain fatty acids can form micelles in solutions with pH > 7 but are insoluble in pH < 5

Answers

Answer:

In basic conditions that is ( pH > 7 ), the equilibrium shifts towards right and produces a lot of (-ve) negatively charged fatty acids which are polar, In water, since they have long hydrophobic hydrocarbon part, this form micelles where the hydrocarbon part  remain inside the  sphere and -coo- group remain outside the sphere due to H-bonding interaction with water.

At ( pH < 5 ) I.e acidic conditions, the equilibrium shift to the left giving neutral molecules which can not have stronger H-bonding interaction .

So micelles cant form as they become insoluble.

The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Cal­culate the value of Ksp for silver carbonate from this data.

Answers

Answer:

2.3 × 10⁻⁹

Explanation:

Step 1: Write the reaction for the solution of calcium oxalate

CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

Step 2: Make an ICE chart

We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.

        CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

I                                0                 0

C                              +S               +S

E                                S                 S

The solubility product constant is:

Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹

An individual was injected with 80 mg of inulin and 960,000 counts per min (cpm) of tritium-labeled water (3H20) to determine the volume of various body fluid compartments. After equilibration a blood sample was obtained and the plasma inulin concentration was 0.5 mg% and the plasma activity (concentration) of tritium was 20 cpm/ml. The volumes of which body compartments can be determined?

Answers

The measurement of body fluid compartments can be achieved by the dilution of chemical compounds that only circulate and disperse in the region of selected areas in the body. The dilution process is dependent on how the concentration is defined.

Given that:

the concentration of plasma insulin after equilibrium = 0.5 mg %

Concentration C  = 0.5 mg/100

Concentration C = 0.005 mg/ml

The mass of insulin = 80 mg

Since the mass amount of the chemical compound(i.e. insulin) and the concentration is known.

The volume of the body fluid compartment can be calculated as:

[tex]\mathbf{volume = \dfrac{\text{mass of the marker }}{concentration }}[/tex]

[tex]Volume = \dfrac{80 \ mg}{0.005 \ mg/ml}[/tex]

Volume = 16000 ml

Thus, it is known that insulin is generally utilized for the measurement of the extracellular fluid volume and serves as a cell impermeant marker.

As a result;

The volume of the extracellular fluid compartment is 16000 ml.

However, the tritium-labeled water is a good marker for the entire body fluid compartment due to the fact that:

its diffusion occurs throughout the entire body,it is identical to water and;the equilibrium concentration is typically easy to measure due to the radioactive characteristics of tritium.

Given that:

plasma activity of tritium = 20 cpm/ml

i.e.

In 1 ml of plasma, 20 cpm of tritium is present.

As such, in 960,000 counts per min (cpm) of tritium-labeled water, the volume of the whole body compartment is:

[tex]\mathbf{= \dfrac{960000}{20} ml \plasma}[/tex]

= 48000 ml of plasma

Therefore, we can conclude that the volumes of the body compartment that can be determined are:

The volume of the extracellular fluid compartment, which is 16000 ml.The volume of the whole body compartment, which is 48000 ml

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As a reaction proceeds, the ratio between the rate of consumption of reactant and the rate of formation of product:

Answers

Answer:

Depends on the reaction.

Explanation:

Hello,

In this case, the answer is depends on the reaction since the ratios between the rates of both consumption and formation depend upon the stoichiometric coefficients in the chemical reaction. For instance, for the reaction:

A -> 2B

The relationship is:

[tex]\frac{1}{-1}r_A =\frac{1}{2} r_B[/tex]

Therefore, we can see that the rate of consumption of A half the rate of formation of B, but is we consider the following chemical reaction:

2A -> B

The relationship is:

[tex]\frac{1}{-2}r_A =\frac{1}{1} r_B[/tex]

Therefore we can see that the rate of consumption of A doubles the rate of consumption of B.

Best regards.

If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?

Answers

Answer:

[tex]m=0.127g[/tex]

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]

Best regards.

Write a balanced chemical equation for the base hydrolysis of methyl butanoate with NaOH. (Use either molecular formulas or condensed structural formulas, but be consistent in your equation.)

Answers

Explanation:

C5H10O2 + NaOH = C2H5COONa + C2H5OH

your result are : sodium propanoate and ethanol

A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.

What is hydrolysis?

Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.

The base hydrolysis of methyl butanoate is shown as,

C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH

Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).

Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.

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A 1 L container originally holds 0.4 mol of N2, 0.1 mol of O2, and 0.08 mole of NO. If the volume of the container holding the equilibrium mixture of N2, O2, and NO is decreased to 0.5 L without changing the quantities of the gases present, how will their concentrations change

Answers

Answer:

a)  [tex]N_2=0.8mol/L[/tex]

b)  [tex]O_2=0.2mol/L[/tex]

c)  [tex]NO=0.16mol/L[/tex]

Explanation:

From the question we are told that:

Moles 0f Nitrogen [tex]N_2=0.4[/tex]

Moles 0f Oxygen [tex]O_2=0.1[/tex]

Volume Decrease [tex]V_2=0.5L[/tex]

Generally, the equation for Concentration is mathematically given by

[tex]C=\frac{moles}{V}[/tex]

For Nitrogen

[tex]N_2=\frac{0.4}{0.5}[/tex]

[tex]N_2=0.8mol/L[/tex]

For Oxygen

[tex]O_2=\frac{0.1}{0.5}[/tex]

[tex]O_2=0.2mol/L[/tex]

For Nitrogen

[tex]NO=\frac{0.08}{0.5}[/tex]

[tex]NO=0.16mol/L[/tex]

Oxygen condenses into a liquid at approximately 90 K. What temperature, in degrees Fahrenheit, does this correspond to?

Answers

Answer:

-297.67 °F

Explanation:

Oxygen condenses into a liquid at approximately 90 K. We can convert any temperature in the Kelvin scale (absolute scale) to the Fahrenheit scale using the following expression.

°F = (K − 273.15) × 9/5 + 32

°F = (90 − 273.15) × 9/5 + 32

°F = (-183.15) × 9/5 + 32

°F = -329.67 + 32

°F = -297.67 °F

A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). Find the ratio of the rms speed of the argon molecules to that of the hydrogen. Assume hydrogen molecule has only translational degree of freedom.

Answers

Answer:

Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1

Explanation:

The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:  

Vrms = [tex]\sqrt{3RT/M}[/tex]

where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol

For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R

Vrms = √(3 * R *4T)/0.04 = √300RT

For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R

Vrms = √(3 * R *T)/0.001 = √3000RT

Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316

Ratio of Vrms of argon to that of hydrogen = 0.316 : 1

Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.

Answers

Explanation:

To calculate [H3O+] in the solution we must first find the pH from the [ OH-]

That's

pH + pOH = 14

pH = 14 - pOH

To calculate the pOH we use the formula

pOH = - log [OH-]

And [OH-] = 5.5 × 10^-5 M

So we have

pOH = - log 5.5 × 10^ - 5

pOH = 4.26

Since we've found the pOH we can now find the pH

That's

pH = 14 - 4.26

pH = 9.74

Now we can find the concentration of H3O+ in the solution using the formula

pH = - log H3O+

9.74 = - log H3O+

Find the antilog of both sides

H3O+ = 1.8 × 10^ - 10 M

The solution is basic since it's pH lies in the basic region.

Hope this helps you

Consider the acid H3PO4. This acid will react with water by the following equation. H3PO4+H2O↽−−⇀H2PO−4+H3O+ What will be true of the resulting conjugate base H2PO−4? Select the correct answer below: H2PO−4 can act as an acid.

Answers

Answer:

H+/PO-4^-2

Explanation:

hydrogen has dissolved completely

In the given reaction conjugate base is H₂PO₄⁻, it also behave as a weak acid.

What is acid - conjugate base pair?

An acid and conjugate base pairs are those pairs in which they are differentiated by the one atom of hydrogen atom.

Given chemical reaction is:

H₃PO₄ + H₂O → H₂PO₄⁻ + H₃O⁺

In the above reaction H₃PO₄ is an acid as it gives H⁺ ion to the solution and formed H₂PO₄⁻, which is a conjugate base of H₃PO₄ acid. H₂PO₄⁻ will also behave as an acid because it have H⁺ ion to gives in the solution but nature of this acid is weak as they not readily dissociates.

Hence, H₂PO₄⁻ is a conjugate base.

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A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of the titrant was 38.2 mL. The starting volume of the HCl solution was 20 mL. What's the concentration of the HCl? answer options: A) 0.788 M B) 0.284 M C) 3.34 M D) 0.191 M

Answers

Answer: it is A

Explanation: I am sure

Answer:

0.191 M

Explanation:

i took the test.

What is the compound formed from the combination of the base and a hydrogen ion

Answers

Answer:

Water

Explanation:

When a base react to and hydrogen ion, we can produce water.

According to these equation

H⁺  +  OH⁻   ⇄   H₂O       Kw: 1×10¹⁴

Remember that OH⁻ is determined by a strong base.

This reaction is called neutralization. You can also produce water with a weak base, because OH⁻ are released. For example, let's mention ammonia which is a weak base, it takes protons from water (H⁺)

NH₃  +  H₂O  ⇄  OH⁻  +  NH₄⁺          Kb

When the ammonium ion (acid), reacts to a base, you produce water.

NH₄⁺   +  NaOH →   NH₃ +  H₂O  +  Na⁺



If you have 2.4L of SO2 gas (at STP) how many moles of sulfur dioxide do you have?

Answers

Answer:

0.107 mole of SO2.

Explanation:

1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).

With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.

This can be obtained as follow:

22.4 L contains 1 mole of SO2.

Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.

Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.

Methylamine is a base because it can bond to H+. Draw Lewis structures to show how methylamine reacts with water to form a hydroxide ion.

Answers

Attached are the lewis structures,

The proton released by water bonds to the nitrogen, because it can hold charges relatively better than carbon because of it's bigger size

In a reversible reaction, the endothermic reaction absorbs ____________ the exothermic reaction releases. A. less energy than B. None of these, endothermic reactions release energy C. the same amount of energy as D. more energy than

Answers

Answer: C. the same amount of energy as

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.

Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

[tex]A+B\rightleftharpoons C+D[/tex]

Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.

The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.

Answer:

C.) the same amount of energy as

Explanation:

I got it correct on founders edtell

Calculate the energy required to heat of 1.50 kg silver from -7.8 C to 15.0 C . Assume the specific heat capacity of silver under these conditions is .0235 J*g^-1*K^-1 . Be sure your answer has the correct number of significant digits.

Answers

Answer:

804 J

Explanation:

Step 1: Given data

Mass of silver (m): 1.50 kgInitial temperature: -7.8 °CFinal temperature: 15.0 °CSpecific heat capacity of silver (c): 0.0235J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]

Q = 804 J

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia

Answers

Answer:

0.327 M

Explanation:

Step 1: Write the balanced equation

2 NH₃(g) ⇄ N₂(g) + 3H₂(g)

Step 2: Make an ICE chart

        2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

I              x             0            0

C          -2y            +y         +3y

E         x-2y             y           3y

Step 3: Find the value of y

The concentration of N₂ at equilibrium is 0.0161 M. Then,

y = 0.0161

Step 4: Find the value of x

The concentration of NH₃ at equilibrium is 0.295 M. Then,

x-2y = 0.295

x-2(0.0161) = 0.295

x = 0.327

A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.

Answers

Answer:

[tex]\Delta G=-97.14kJ[/tex]

Explanation:

Hello,

In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

[tex]\Delta G=-RTln(K)[/tex]

Hence, we compute it as required:

[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]

And for 2.37 moles of hydrogen bromide, we obtain:

[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]

Best regards.

Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Answers

Answer:

D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Explanation:

Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).

All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.

PLEASE HELP!!
this is on USAtestprep
a)
b)
c)
d)

Answers

The answer is D!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

5. Calcule las concentraciones cuando se alcanza el equilibrio si partimos de unas concentraciones iniciales [A]=[B]=1M ; [C]=[D]=0M y una constante de equilibrio de 5.

Answers

Las concentraciones en el equilibrio para la reacción química presentada son:

[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]

Consideremos la siguiente reacción química genérica:

A + B ⇄ C + D

Para calcular las concentraciones en el equilibrio, debemos construir una Tabla ICE. Cada fila representa una instancia (Inicial, Cambio, Equilibrio) y la completamos con la concentración o cambio de concentración ("x" para concentraciones desconocidas). Como inicialmente no hay productos, la reacción se desplazará hacia la derecha para alcanzar el equilibrio.

          A + B ⇄ C + D

I          1      1      0    0

C       -x    -x     +x    +x

E      1-x    1-x    x     x

La constante de equilibrio, Kc, es:

[tex]Kc = 5 = \frac{[C][D]}{[A][B]} = \frac{x^{2} }{(1-x)^{2} } \\\sqrt{5} = x/1-x\\x = 0.69[/tex]

Las concentraciones en el equilibrio son:

[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]

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What are the conditions that are favorable for extensive solid solubility of one element in another (Hume-rothery rules)

Answers

Answer:

Atomic radius less than 15%, similar structure and same valency.

Explanation:

The conditions that are favorable for extensive solid solubility of one element in another are the following.

The atomic radius of the solute and solvent atoms must be less than 15%. The structure of both solute and solvent are similar. Solubility completes when both have same valency. Valency means number of electrons in the outermost shell. If both solute and solvent has same number of electrons so it will be completely soluble in each other.

The conditions that are favorable for extensive solid solubility of one element in another is the same size, electrongativity and valency.

What is Hume - Rothery rules?

Hume - Rothery rules are the sets of some important rules which gives idea about the desired condition for the formation of solid solution.

Following main points are described in this rule:

Difference between the size of the solute and the solvent should be less than 15%.Electronegativity difference between the solute and solvents should be small.And they both should have same valency, means same no. of electrons in the outermost shell.

Hence size, electronegativity and valency are the conditions.

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Could someone help with this? Much appreciated!

Answers

Answer:

The 3rd answer down.

Na²O (sodium oxide) will be a base when exposed to water H²O

Explanation:

Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).

The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.

An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint.

Required:
a. What is the concentration of the diluted HCI solution?
b. If 7.93 mL of the diluted HCI solution was required to reach the endpoint, what is the concentration of OH- in solution?
c. What is the concentration of Ag+ in solution?
d. What is the Ksp expression for the dissolution of AgOH?

Answers

Answer:

a. 3.8856x10⁻³M HCl

b. 1.23x10⁻⁴M OH⁻

c. 1.23x10⁻⁴M Ag⁺

d. Ksp = [Ag⁺] [OH⁻]

Explanation:

a. The reaction that you are studying is:

HCl(aq) + AgOH(aq) → H₂O(l) + AgCl(s)

The HCl solution is diluted from 10.00mL to 250.00mL, that is:

250.00mL / 10.00mL = 25 -The solution is diluted 25 times-

As original concentration of HCl is 0.09714M, the concentration of the diluted solution is:

0.09714M / 25 =

3.8856x10⁻³M HCl

b. 1 mole of HCl reacts per mole of AgOH, moles of HCl that reacts are:

7.93mL = 7.93x10⁻³L × (3.8856x10⁻³mol HCl / L) = 3.0813x10⁻⁵ moles of HCl.

Based on the reaction, you have in solution

3.0813x10⁻⁵ moles of AgOH = Ag⁺ = OH⁻

The AgOH solution was 250.0mL = 0.2500L, its concentration is:

3.0813x10⁻⁵ moles OH⁻ / 0.2500L =

1.23x10⁻⁴M OH⁻

c. In solution, AgOH produce Ag⁺ and OH⁻ in equals proportions, that means:

1.23x10⁻⁴M OH⁻ =

1.23x10⁻⁴M Ag⁺

d. The solubility product reaction of AgOH(s) is:

AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)

Where Ksp for this reaction is defined as:

Ksp = [Ag⁺] [OH⁻]

How many moles of NaF must be dissolved in 1.00 liter of a saturated solution of PbF 2 at 25°C to reduce the [Pb 2+] to 1.0 × 10 –6 M? The K sp for PbF 2 at 25 °C is 4.0 × 10 –8.

Answers

Answer:

0.1957 moles of NaF

Explanation:

The Pb²⁺ and F⁻ are in equilibrium with PbF₂ as follows:

PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)

Where Ksp expression is:

Ksp = 4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

A saturated solution contains the maximum possible amount of Pb²⁺ and F⁻. That is:

PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)

PbF₂(s) ⇄ X + 2X

Where X is amount of ions presents in solution

4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

4.0x10⁻⁸ = [X] [2X]²

4.0x10⁻⁸ = 4X³

4.0x10⁻⁸/4 = X³

1.0x10⁻⁸ = X³

2.15x10⁻³M = X

That means initial concentration of Pb²⁺ is = X = 2.15x10⁻³M and [F⁻] = 2X = 4.30x10⁻³M

Now, using again Ksp, if you want a [Pb²⁺] = 1.0x10⁻⁶M, the [F⁻] you need is:

4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

4.0x10⁻⁸ = [1.0x10⁻⁶M] [F⁻]²

0.04M = [F⁻]²

0.2M = [F⁻]

You need a final concentration of 0.2M of F⁻. As initial concentration was 4.30x10⁻³M and volume of the buffer is 1.00L, the moles of F⁻ = moles of NaF you must add are:

0.2M - 4.30x10⁻³M =

0.1957 moles of NaF

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