The literature value for the Ksp of Ca(OH)2 at 25 °C is 4.68E−6. Imagine you ran the experiment and got a calculated value for Ksp which was too high. Select all of the possible circumstances which would cause this result.

A. The HCl was more concentrated than the labeled molarity (0.0500 M).

B. The Ca[OH]2 solution may have been supersaturated.

C. The HCl was less concentrated than the labeled molarity (0.0500 M).

D. The Ca[OH]2 solution may have been unsaturated.

E. The titration flask may have not been clean and had a residue of a basic solution.

F. The titration flask may have not been clean and had a residue of an acidic solution.

Answers

Answer 1

Answer:

D. The Ca[OH]2 solution may have been unsaturated

Explanation:

The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.

If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.


Related Questions

If an individual proton has mass 1.007825 amu, and an individual neutron has mass 1.008665 amu, what's the calculated mass of a neptunium-236 nucleus? options: A) 237.92482 amu B) 236.99873 amu C) 237.96682 amu D) 237.04817 amu

Answers

Answer:

C) 237.96682 amu

Explanation:

The symbol for neptunium-236 is given as;

²³⁶₉₃Np

This element has 93 protons and (236 - 93 = 143) neutrons.

Mass Number =Total mass of Protons + Total mass of neutrons

Total Mass pf protons = 93 * 1.007825 amu, = 93.727725 amu

Total mass of Neutrons = 143 * 1.008665 amu = 144.239095 amu

Mass = 144.239095 + 93.727725  = 237.96682 amu

Correct option is option C.

A 5-column table with 2 rows. Column 1 is labeled number of protons, with entries 20 and 9; column 2 is number of neutrons, with entries 20 and D; Column 3 is atomic number, with entries A and E; Column 4 is Mass Number, with entries B and 19, and Column 5 is Element (symbol) with entries C and F. Using the periodic table, complete the table to describe each atom. Type in your answers

Answers

Answer:

A    ⇒ 20

B    ⇒ 40

C    ⇒ Ca

D    ⇒ 10

E     ⇒ 9

F     ⇒ F

Explanation:

edge 2021

Answer:

the person above is correct

Explanation:

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

Answers

Answer:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]

The equilibrium expression is:

[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:

[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]

Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Regards.

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

The dissociation of calcium fluoride has been given by:

[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]

The solubility constant, ksp has been given as:

[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]

From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.

The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.

The concentration of Ca in the solution has been resulted as x + 0.01 M.

The solubility product can be given as:

[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

For more information about molar solubility, refer to the link:

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Consider the bond dissociation energies listed below in kcal/mol. CH3-Br 70 CH3CH2-Br 68 (CH3)2CH-Br 68 (CH3)3C-Br 65 These data show that the carbon-bromine bond is weakest when bromine is bound to a __________.

Answers

Answer:

The answer is "Tertiary carbon".

Explanation:

Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is=  68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.

The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.

g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4

Answers

Answer:

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]

Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.

Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:

    0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M

Name the following compound from the concise formula:______.
CH3CH(CH3)CHCHCH(CH3)CH2CH3
A. 2,4-dimethyl-3-heptene
B. 2,5-dimethyl-3-heptene
C. 3,5-dimethyl-3-heptene
D. 2,5-dimethyl-4-heptene

Answers

Answer:

B. 2,5-dimethyl-3-heptene

Explanation:

Answer:

B. 2,5-dimethyl-3-heptene

Explanation:

Identify the term that matches each electrochemistry definition. The electrode where oxidation occurs Cathode The electrode where reduction occurs Choose... An electrochemical cell powered by a spontaneous redox reaction Choose... An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction Choose... A chemical equation showing either oxidation or reduction Choose...

Answers

Answer:

An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction

When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration

Answers

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

Using the standard reduction potentials Ni2+(aq) + 2 e‑Ni(s) ‑0.25 volt Fe3+(aq) + e‑Fe2+(aq) +0.77 volt Calculate the value of E°cell for the cell with the following reaction. Ni2+(aq) + 2 Fe2+(aq) →Ni(s) + 2 Fe3+(aq)

Answers

Answer:

The correct answer is - 1.02 V

Explanation:

From the reduction-oxidation reaction:

Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:

Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s)                    Eº= ‑0.25 V

Oxidation (anode) :  2 x (Fe²⁺ → Fe³⁺ + e-)(aq)                Eº= -0.77 V

                                -------------------------------------

                     Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):

Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V

After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step in balancing a redox reaction under basic conditions is to: ______

Answers

Explanation:

After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step is the  last step when balancing a redox reaction under basic condition.

The last step is to cancel all common terms that arises as a result of the formation of the water molecule. Usually, there's need to balance the water molecules in the reactant and product side of the reaction.

What will be formed when 2,2,3-trimethylcyclohexanone reacts with hydroxylamine?

Answers

Answer:

Following are the solution to this equation:

Explanation:

In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:

In the given question "Option (iii)" is correct, which is defined in the attachment file.

When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.

Predict the most likely bond type for the following.

a. Cu (Copper)
b. KCl (Potassium Chloride)
c. Si (Silicon)
d. CdTe (Cadmium Telluride)
e. ZnTe (Zinc Telluride)

Answers

Answer:

The three major types of bond are ionic, polar covalent, and covalent bonds. Ionic occurs majorly between metals and non-metals, which allows sharing of electrons to form an ionic compound. Whereas covalent bonding calls for complete transfer of electrons between atoms. Polar covalent bonds have unequaly shared electron-pair between two atoms.

Explanation:

a. Cu (Copper)- ionic bonding

b. KCl (Potassium Chloride) - ionic bonding

c. Si (Silicon) - covalent bonding

d. CdTe (Cadmium Telluride) - polar covalent bonding

e. ZnTe (Zinc Telluride)- polar covalent bonding

The condition that a reaction takes place without outside help Choose... Solution in which no more solute can be dissolved in the solvent Choose... Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature Choose... The extent of randomness in a system Choose... Sum of the internal energy plus the product of the pressure and volume for a reaction

Answers

Answer:

Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature

Explanation:

The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).

∆G is given by;

∆G= ∆H -T∆S

Where;

∆H= change in enthalpy of the system

T= absolute temperature of the system

∆S= change in entropy

Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.

g Which ONE of the following molecules and ions has trigonal planar molecular geometry? (NOTE: You must first determine what the Lewis structure of each substance is.) A) PCl3 B) HCN C) CO3 2– D) H3O+ E) NF3

Answers

Answer: C) [tex]CO^{2-}_{3}[/tex]

Explanation: The Valence Shell Electron Pair Repulsion Model (VSEPR Model) shows bonding and nonbonding electron pairs present in the valence, outermost, shell of an atom connecting to other atoms. It also gives the molecular geometric shape of a molecule.

To determine molecular geometry:

1) Draw Lewis Structure, i.e., a simplified representation of the valence shell electrons;

2) Count the number of electron pairs (count multiple bonds as 1 pair);

3) Arrange electron pairs to minimise repulsion;

4) Position the atoms to minimise the lone pair;

5) Name the molecular geometry from the atom position;

Trigonal planar molecular geometry is a model which molecule's shape is triangular and in one plane. Such molecule has three regions of electron density extending out from the central atom and the repulsion will be at minimum when angle between any two is 120°.

The Lewis structure of each molecule is shown in the attachment.

Analysing each one, it can be concluded that molecule with trigonal planar geometry is [tex]CO^{2-}_{3}[/tex]

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

A simplified version of photosynthesis can be represented as carbon dioxide combining with water to form glucose and oxygen: 6CO2+6H20 C6H12O6+6O2 In this reaction, ________ is oxidized. 1.Carbon dioxide 2.Hydrogen 3.Carbon 4.Oxygen

Answers

Answer:

2, hydrogen

Explanation:

i think

Answer:

Answer is not hydrogen

Explanation:

did the test and got it wrong

How many grams of H2O will be formed when 32.0 g H2 is mixed with 73.0 g of O2 and allowed to react to form water

Answers

hope this helps u

pls mark as brainliest .-.

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.

a sample of gas occupies a volume of 2.62 liters at 25 C and 1.00 atm. what will be the volume at 50 C and 2 atm

Answers

Answer:2.62 L

Explanation:

A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

What is ideal gas law ?

The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.

An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.

By using ideal gas equation,

P₁ V ₁ ÷ T = P₂V₂ ÷ T

1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50

V₂ = 1 × 2.62 × 50 ÷ 25 × 2

V₂ = 2.62 liters.

Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

To learn more about ideal gas law follow the link below;

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The concentration of glucose, C6H12O6, in normal spinal fluid is 75 mg/100g. What is the molality of the solution

Answers

Answer:

4.16x10⁻³m

Explanation:

Molality is defined as the ratio between moles of a solute, in this case glucose, and kg of solvent.

As there are 100g of solvent, the kg are 0.1. Thus, we only need to calculate from the mass of glucose its moles to solve the molality of the solution.

Moles glucose:

There are 75mg = 0.075g of glucose. To conver mass to moles it is necessary molar mass.

Molar mass glucose:

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

72.06 + 12.10 + 96 = 180.16g/mol

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Molality of the solution:

4.16x10⁻⁴ moles of glucose / 0.1kg of solvent =

4.16x10⁻³m

The molarity of the solution is 4.16x10⁻³m

Calculation of the molarity:

We know that the molarity refers to the ratio that arise between the moles of a solute.

Since there are 100 g of solvent so here the kg should be 0.1.

Likewise there is 75 mg so it should be 0.075g

Now the Molar mass glucose should be

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

So,

= 72.06 + 12.10 + 96

= 180.16g/mol

Now

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Now finally

Molality of the solution:

= 4.16x10⁻⁴ moles of glucose / 0.1kg of solvent

=4.16x10⁻³m

learn more about molarity here: https://brainly.com/question/13011592

What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0×10–14), with a concentration of hydroxide ions of 2.21×10–6 M? answer options: A) 2.8×10–8 M B) 4.52 ×10–9 M C) 1.6×10–9 M D) 3.1×10–6 M

Answers

Answer:

ITS NOT D. ITS B. 4.52x10^-9 M

Explanation:

Answer:

4.52 ×10–9 M

Explanation:

Both chlorine and fluorine are represented by a green modeling piece that has 4 holes. Is using the same piece for two different atoms acceptable? Why or why not

Answers

Answer:

Yes, same piece can be used.

Explanation:

The same piece can be used for two different atoms are acceptable because both atoms has 7 electrons in their outermost valance shell. Both atoms belong to same group i. e. halogens so same piece can be used for both atoms. If the atoms belong to different groups and they have different number of electrons in their outermost shell so using same piece will be a problem so it is recommended to use different pieces for different atoms.

The use of the same modeling piece for chlorine and fluorine has been accepted as it has consisted of the same properties and belongs to the same group.

Chlorine and fluorine have been the elements of group 17. The elements are halogens with the presence of 7 valence electrons.

The elements have been belonging to the same group and have the same number of valence electrons thus resembling each other in the chemical properties.

Since both the elements are similar to each other, the use of the same piece for two different atoms has been acceptable.

For more information about the modeling piece, refer to the link:

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Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.

Answers

Answer:

whats the ph  ofpoh=9.78

Explanation:

For the following set of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain constant.

V=2.91 L at 23.0 °C
V= 4.20 L at ? °C

Answers

Answer:

155 °C

Explanation:

Step 1: Given data

Initial volume (V₁): 2.91 LInitial temperature (T₁): 23.0°CFinal volume (V₂): 4.20 LFinal temperature (T₂): ?

Step 2: Convert the initial temperature to Kelvin

We will use the following expression.

K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K

Step 3: Calculate the final temperature

Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 4.20 L × 296.2 K/2.91 L

T₂ = 428 K

Step 4: Convert the final temperature to Celsius

We will use the following expression.

°C = K - 273.15 = 428 - 273.15 = 155 °C

Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
For ethylenediamine, use (en) in the formula.
a) sodium hexachloroplatinate(IV)
b) dibromobis(ethylenediamine)cobalt(III) bromide
c) pentaamminechlorochromium(III) chloride

Answers

Answer:

sodium hexachloroplatinate(IV)- Na2[PtCl6]

dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br

pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2

Explanation:

The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.

The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.

What are the products in the following chemical reaction Pb(NO3)+KCI

Answers

Answer:

The products are KNO3 + PbCl2.....

Espero que te sirva.

The entropy of a substance above absolute zero will always be:

a. Negative
b. Positive
c. Neither Negative nor positive

Answers

i will be positive. just because it’s positive

A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to

Answers

Answer:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

Explanation:

The reaction of a weak acid (HOOH) with NaOH is as follows:

HCOOH + NaOH → HCOONa + H₂O

Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).

The initial moles of both species are:

HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH

NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH

After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).

Final moles:

HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles

HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles

As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:

0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa

Thus, the initial mixture is equivalent to:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

What is pH of a buffer made by combining 45.0mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate

Answers

Answer:

3.11

Explanation:

Any buffer system can be described with the reaction:

[tex]HA~->~H^+~+~A^-[/tex]

Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:

[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]

With all this in mind, we can write the reaction for our buffer system:

-) Nitrous acid: [tex]HNO_2[/tex]

-) Sodium nitrate: [tex]NaNO_2[/tex]

[tex]HNO_2~->~H^+~+~NO_2^-[/tex]

In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).

We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:

-) moles of [tex]HNO_2[/tex]:

[tex]mol=0.150~M*0.045~L=0.00657[/tex]

-) moles of [tex]NO_2^-[/tex]:

[tex]mol=0.175~M*0.020~L=0.0035[/tex]

The total volume would be:

0.020 L + 0.045 L = 0.065 L

With this in mind, we can calculate the molarity of each compound:

-) Concentration of [tex]HNO_2[/tex]

[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]

-) Concentration of [tex]NO_2^-[/tex]

[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]

The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:

[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]

The final pH value would be 3.11

I hope it helps!

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.

Nitrous acid is a weak acid and nitrate ion is its conjugate base.

What is a buffer?

It is a solution used to resist abrupt changes in pH when acids or bases are added.

Step 1: Calculate the moles of each species.

We do so by multiplying the molar concentration by the volume in liters.

HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol

NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol

Step 2: Calculate the total volume of the mixture.

The total volume will be the sum of the volumes of each solution.

V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L

Step 3: Calculate the molar concentration of each species in the mixture.

HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M

NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M

Step 4: Calculate the pH of the buffer.

We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.

pH = pKa + log [NaNO₂]/[HNO₂]

pH = 3.16 + log 0.0538/0.104 = 2.87

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

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A voltaic cell is set up as follows: Anode: Zn electrode in a solution of 0.050 M Zn(NO 3 ) 2 Cathode: Pt electrode with 0.500 atm H 2 (g) in 0.010 M HNO 3 a) Write the overall balanced cell reaction.

Answers

Answer:

[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]

Explanation:

Given that,

Anode : Zn electrode in a solution of 0.050 M Zn(NO₃)₂

Cathode : Pt electrode with 0.500 atm H₂(g) in 0.010 M HNO₃

Anode :

[tex]Zn(s)\Rightarrow Zn^{2+}(aq) + 2e^{-}[/tex]

Cathode :

[tex]2H+(aq)+2e^{-}\Rightarrow H_{2}(g)[/tex]

We need to write the overall balanced cell reaction

Using anode and cathode

[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]

Hence, This is required answer.

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