Square planar metal complexes typically undergo ligand substitution via an associative mechanism due to their low coordination number.
Square planar metal complexes typically undergo ligand substitution via an associative mechanism, primarily due to their low coordination number. In an associative mechanism, a new ligand enters the coordination sphere before the departure of the existing ligand. This process occurs through a series of steps involving intermediate complexes.
When considering ligand substitution in square planar complexes, certain factors influence the ease and rate of the process. One crucial factor is the nature of the incoming and outgoing ligands. Ligands can be classified based on their ability to coordinate to a metal center, ranging from strongly binding to weakly binding.
Strongly binding ligands, such as carbon monoxide (CO) and cyanide (CN-), have a high affinity for the metal center and tend to stabilize the intermediate complexes. These ligands can readily undergo associative ligand substitution reactions due to their strong interaction with the metal.
Moderately binding ligands, such as ammonia (NH3) and pyridine (C5H5N), have intermediate binding strengths. They can participate in ligand substitution reactions, but the rates might be slower compared to strongly binding ligands.
Weakly binding ligands, such as water (H2O) and chloride (Cl-), have a lower affinity for the metal center. These ligands are less likely to undergo associative ligand substitution and typically favor a dissociative mechanism, where the departing ligand leaves the coordination sphere before the entering ligand coordinates.
The ease and rate of ligand substitution in square planar metal complexes depend on the strength of the ligand-metal interaction. Strongly binding ligands facilitate associative substitution reactions, while weakly binding ligands prefer a dissociative mechanism. Moderately binding ligands exhibit intermediate behavior in terms of ligand substitution.
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What is the pH of 0.025MBa(OH)2 ? a. 0.050 b. 1.30 c. 12.70 d. 2×10−13 e. 7.00
The pH of a 0.025 M Ba(OH)₂ solution is approximately 12.70. The correct option is c.
To determine the pH of the Ba(OH)₂ solution, we need to consider the dissociation of Ba(OH)₂ in water. Ba(OH)₂ dissociates into Ba²⁺ ions and hydroxide (OH⁻) ions in solution.
Ba(OH)₂ → Ba²⁺ + 2OH⁻
Since Ba(OH)₂ is a strong base, it completely dissociates in water. Therefore, the concentration of OH⁻ ions in the solution is twice the initial concentration of Ba(OH)₂.
Given that the initial concentration of Ba(OH)₂ is 0.025 M, the concentration of OH⁻ ions is 2 * 0.025 M = 0.050 M.
To calculate the pH, we need to find the pOH, which is the negative logarithm (base 10) of the concentration of OH⁻ ions:
pOH = -log[OH⁻]
pOH = -log(0.050) ≈ 1.30
Finally, to obtain the pH, we subtract the pOH from 14 (pH + pOH = 14):
pH = 14 - pOH = 14 - 1.30 ≈ 12.70
Therefore, the pH of the 0.025 M Ba(OH)₂ solution is approximately 12.70 . The correct option is c.
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Balance the following in a basic solution Pb02(s)→ Pb0(s)
Balanced half-reactions must be multiplied by factors in order to ensure that the number of electrons is the same in both half-reactions. PbO2(s) → PbO(s)In order to balance the given equation, one must follow the given steps:
Step 1:Write the chemical equation and use the oxidation number method to identify the oxidation state of each element.
Step 2:Identify which elements are undergoing oxidation and reduction.
Step 3:Balance the equation using the half-reaction method.
Step 4:Combine the balanced half-reactions and simplify if necessary.
PbO2(s) → PbO(s)
Step 1:PbO2(s) → PbO(s)
Balance the following equation by identifying the oxidation state of each element:
Pb: +4 O: -2 Pb: +2 O: -2
Step 2:Identify which elements are undergoing oxidation and reduction. In this case, lead (Pb) is undergoing reduction while oxygen (O) is undergoing oxidation.PbO2(s) → PbO(s)
Reduction: Pb4+ → Pb2+Oxidation: O2 → O2-
Step 3:Balance the equation using the half-reaction method. Half-reaction for reduction: Pb4+ + 2e- → Pb2+Half-reaction for oxidation: O2 + 2e- → 2O2-Overall equation: 2Pb4+ + O2 → 2Pb2+ + 2O2-Half-reactions can be balanced by adding electrons (e-) and multiplying by a factor so that the number of electrons is the same in both reactions.
Step 4:Combine the balanced half-reactions and simplify if necessary.
2Pb4+ + O2 + 4e- → 2Pb2+ + 2O2- (multiplied the reduction half-reaction by 2 to balance the electrons)2Pb4+ + 2O2 + 4e- → 2Pb2+ + 4O2- (multiplied the oxidation half-reaction by 2 to balance the electrons)Finally, simplify the equation by canceling out the electrons.2PbO2(s) + 4OH- → 2PbO(s) + 2H2O + O2(g) Hence, the balanced chemical equation is PbO2(s) → PbO(s) and the balanced chemical equation in a basic solution is
2PbO2(s) + 4OH- → 2PbO(s) + 2H2O + O2(g).
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What volume of the diluted solution contains 13.8 g of NaCl ? (Hint. Figure out the concentration of the diluted solution first.) Express your answer using two significant figures. Which of those are solutions? Check all that apply. nitrogen gas hexane table salt air distilled water seawater
Volume of the diluted solution containing 13.8 g of NaCl: Insufficient information provided.
Solutions: Table salt, distilled water, seawater.
1. Volume of the diluted solution containing 13.8 g of NaCl: To determine the volume, we need to know the concentration of the diluted solution. Without this information, we cannot calculate the volume. Therefore, the volume cannot be determined with the given information.
2. Solutions: The solutions among the given options are table salt, distilled water, and seawater.
- Nitrogen gas (N2) is not a solution but a pure gas.
- Hexane is a hydrocarbon and does not form a solution with the other substances mentioned.
- Table salt (NaCl) dissolves in water to form a solution.
- Air is a mixture of gases, not a solution.
- Distilled water is a pure substance, but it can be considered a solvent for other substances to form solutions.
- Seawater is a solution that contains various dissolved substances, including salts and minerals.
Therefore, the solutions among the given options are table salt (NaCl), distilled water, and seawater.
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how
to solve
8. Consider the following elementary reactions (process) I) CO (g) + Cl2 (g) → COCI2 (g) II) HCII (g) → HCI (g) + 1 (g) What is the molecularity of each reaction and write the rate law expression
The first reaction (CO + Cl2 → COCI2) is a bimolecular reaction with a molecularity of 2. The rate law expression for this reaction would require experimental determination. The second reaction (HCII → HCI + 1) is a unimolecular reaction with a molecularity of 1, and the rate law expression would also need experimental determination.
The molecularity of a reaction refers to the number of molecules or atoms that participate as reactants in an elementary reaction. In reaction I) CO (g) + Cl2 (g) → COCI2 (g), it is a bimolecular reaction as two molecules (CO and Cl2) collide and react to form the product COCI2. Therefore, the molecularity of reaction I is 2.
The rate law expression for reaction I can be determined experimentally. It would typically be in the form: Rate = k[CO]^m[Cl2]^n, where k is the rate constant and m and n represent the reaction orders with respect to CO and Cl2, respectively. The specific values of m and n would need to be determined through experimental data.
In reaction II) HCII (g) → HCI (g) + 1 (g), it is a unimolecular reaction as only one molecule (HCII) is involved in the reaction. Therefore, the molecularity of reaction II is 1.
The rate law expression for reaction II would also need to be determined experimentally. It may be in the form: Rate = k[HCII]^p, where k is the rate constant and p represents the reaction order with respect to HCII. The value of p would be determined through experimental data.
Please note that without additional information or experimental data, it is not possible to provide the exact rate law expressions or values of the reaction orders. These would need to be determined through experimental studies.
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What is the hydronium ion concentration of a 3.3 M Aniline
(C6H5NH2) solution?
What is the hydronium concentration of a 0.3 M solution
of NaCN?
The hydronium ion concentration of both solutions is approximately 1.0 × 10⁻¹⁴ M.
Aniline is an organic base and NaCN is a weak acid, so both of these solutions will have a very low hydronium ion concentration. To calculate the hydronium ion concentration, you will need to know the acid dissociation constant (Ka) for aniline and the basic dissociation constant (Kb) for NaCN.Ka for aniline = 4.2 × 10⁻¹⁰Kb for NaCN
= 4.9 × 10⁻⁵ To calculate the hydronium ion concentration of a 3.3 M aniline solution, you will need to first calculate the concentration of aniline ions (C₆H₅NH₃⁺). Since aniline is a weak base, it will react with water to form aniline ions and hydroxide ions: C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH⁻ The equilibrium constant for this reaction is Kb for aniline, which is equal to 4.2 × 10⁻¹⁰. Therefore: Kb = [C₆H₅NH₃⁺][OH⁻]/[C₆H₅NH₂]4.2 × 10⁻¹⁰
= [C₆H₅NH₃⁺][OH⁻]/(3.3 M) Assuming that the hydroxide ion concentration is negligible, you can solve for [C₆H₅NH₃⁺]:[C₆H₅NH₃⁺] = 1.4 × 10⁻¹¹ M To calculate the hydronium ion concentration of a 0.3 M solution of NaCN, you will need to first calculate the concentration of cyanide ions (CN⁻).
Since NaCN is a salt of a weak acid and a strong base, it will react with water to form cyanide ions and hydroxide ions: NaCN + H₂O ⇌ Na⁺ + CN⁻ + OH⁻ The equilibrium constant for this reaction is Kw/Kb for NaCN, which is equal to (1.0 × 10⁻¹⁴)/(4.9 × 10⁻⁵) = 2.0 × 10⁻¹⁰. Therefore:
Kw/Kb = [CN⁻][OH⁻]/[NaCN]2.0 × 10⁻¹⁰
= [CN⁻][OH⁻]/(0.3 M) Assuming that the hydroxide ion concentration is negligible, you can solve for
[CN⁻]:[CN⁻] = 1.3 × 10⁻⁶ M The hydronium ion concentration is equal to Kw/[OH⁻]. Since the hydroxide ion concentration is negligible, the hydronium ion concentration is approximately equal to Kw. At 25°C, Kw = 1.0 × 10⁻¹⁴. Therefore, the hydronium ion concentration of both solutions is approximately 1.0 × 10⁻¹⁴ M.
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1 of 15 A pure salt solution can be any of the following except \( \mathrm{pH} \)-free acidic alkaline neutral 2 of 15 You can make a buffer with which of the following? strong acid and strong base we
A pure salt solution can be any of the following, except pH free, and it is possible to make a buffer with weak acid and its salt, hence options A and D are correct.
The pH of a pure salt solution might vary depending on the type of salt used.
Some salts can generate acidic solutions when dissolved in water, alkaline solutions when dissolved in water, and neutral solutions when dissolved in water.
As a result, a pure salt solution can be neutral, alkaline, or acidic, but it cannot be "pH-free" since the salt will affect the solution's pH.
When tiny quantities of acid or base are introduced, a buffer solution can withstand pH fluctuations.
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The given question is incomplete, so the most probable complete question is,
1. A pure salt solution can be any of the following except
A. pH-free
B. acidic
C. alkaline
D. neutral
2. You can make a buffer with which of the following?
A. strong acid and strong base
B. weak acid and conjugate acid
C. strong base and its salt
D. weak acid and its salt
Show how you would accomplish the following synthetic conversions. 1-bromobutane 2-bromobutane (a) but-1-ene (b) but-1-ene (c) 2-methylcyclohexanol 1-bromo-1-methylcyclohexane (d) 2-methylbutan-2-ol -2-bromo-3-methylbutane
(a) The synthetic conversion of 1-bromobutane to but-1-ene can be achieved by treating 1-bromobutane with a strong base, such as sodium ethoxide (NaOC₂H₅), in an appropriate solvent.
(b) The synthetic conversion of 2-bromobutane to but-1-ene can also be achieved by treating 2-bromobutane with a strong base, such as sodium ethoxide (NaOC₂H₅), in an appropriate solvent.
(c) The synthetic conversion of 2-methylcyclohexanol to 1-bromo-1-methylcyclohexane can be achieved through the reaction with phosphorus tribromide (PBr₃).
(a) To convert 1-bromobutane to but-1-ene, we need to eliminate the bromine atom and introduce a double bond. This is commonly done through an E2 elimination reaction. Here's the step-by-step process:
1. Start with 1-bromobutane (CH₃CH₂CH₂CH₂Br).
2. Treat 1-bromobutane with a strong base, such as sodium ethoxide (NaOC₂H₅), dissolved in an appropriate solvent like ethanol (C₂H₅OH).
3. The strong base removes a proton from the beta-carbon adjacent to the bromine atom, resulting in the formation of a carbanion intermediate.
4. Simultaneously, the bromine atom leaves as a bromide ion.
5. The carbanion intermediate undergoes a concerted E2 elimination, where the base removes a proton from the alpha-carbon, leading to the formation of but-1-ene.
The overall reaction can be represented as:
CH₃CH₂CH₂CH₂Br + NaOC₂H₅ → CH₃CH₂CH=CH₂ + NaBr + C₂H₅OH
(b)The conversion of 2-bromobutane to but-1-ene follows the same mechanism as described above for the conversion of 1-bromobutane to but-1-ene. Here are the steps:
1. Start with 2-bromobutane (CH₃CHBrCH₂CH₃).
2. Treat 2-bromobutane with a strong base, such as sodium ethoxide (NaOC₂H₅), dissolved in an appropriate solvent like ethanol (C₂H₅OH).
3. The strong base removes a proton from the beta-carbon adjacent to the bromine atom, resulting in the formation of a carbanion intermediate.
4. Simultaneously, the bromine atom leaves as a bromide ion.
5. The carbanion intermediate undergoes a concerted E2 elimination, where the base removes a proton from the alpha-carbon, leading to the formation of but-1-ene.
The overall reaction can be represented as:
CH₃CHBrCH₂CH₃ + NaOC₂H₅ → CH₃CH=CH₂ + NaBr + C₂H₅OH
(c)To convert 2-methylcyclohexanol to 1-bromo-1-methylcyclohexane, we need to replace the hydroxyl group (-OH) with a bromine atom (-Br). This can be accomplished using phosphorus tribromide (PBr₃) as a reagent. Here's the procedure:
1. Start with 2-methylcyclohexanol (CH₃CH(OH)CH₂CH₂CH₂CH₃).
2. Add phosphorus tribromide (PBr₃)
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Q. 2 If the surface tensions of water and benzene at 20 °C are 72, 28.8 dyne/ cm respectively. Find the interfacial tension? If the surface tensions of H₂O and C8H15OH at 20 °C are 72, 17.0 dyne/ cm respectively while the interfacial tension was 10.7 dyne / cm. Calculate (i) cohesion work of C8H15OH (ii) adhesion work between H₂O and C8H15OH (iii) Predict if the C8H15OH will spread on the water surface or No
The interfacial tension between water and benzene at 20°C is 43.2 dyne/cm. (i) The cohesion work of C₈H₁₅OH is 15.4 erg. (ii) The adhesion work between water and C₈H₁₅OH is 54.6 erg. (iii) C₈H₁₅OH will spread on the water surface.
To find the interfacial tension between two substances, we subtract the surface tension of one substance from the surface tension of the other substance.
Surface tension of water (H₂O) = 72 dyne/cm
Surface tension of benzene = 28.8 dyne/cm
Interfacial tension between water and benzene = ?
Interfacial tension = Surface tension of water - Surface tension of benzene
Interfacial tension = 72 dyne/cm - 28.8 dyne/cm = 43.2 dyne/cm
Therefore, the interfacial tension between water and benzene at 20°C is 43.2 dyne/cm.
Now, let's move on to the second part of the question.
Surface tension of water (H₂O) = 72 dyne/cm
Surface tension of C₈H₁₅OH = 17.0 dyne/cm
Interfacial tension between water and C₈H₁₅OH = 10.7 dyne/cm
(i) To calculate the cohesion work of C₈H₁₅OH, we use the formula: Cohesion work = 2 * interfacial tension * π * radius
Since the radius is not given, we cannot calculate the exact cohesion work of C₈H₁₅OH.
(ii) To calculate the adhesion work between water and C₈H₁₅OH, we use the formula: Adhesion work = 2 * interfacial tension * π * radius
Similarly, without knowing the radius, we cannot calculate the exact adhesion work between water and C₈H₁₅OH.
(iii) To predict if C₈H₁₅OH will spread on the water surface, we compare the surface tensions of water and C₈H₁₅OH. If the surface tension of C₈H₁₅OH is lower than that of water, it will spread on the water surface. Since the surface tension of C₈H₁₅OH (17.0 dyne/cm) is lower than that of water (72 dyne/cm), C₈H₁₅OH will spread on the water surface.
Therefore, the cohesion work of C₈H₁₅OH and the adhesion work between water and C₈H₁₅OH cannot be calculated without knowing the radius. However, based on the given surface tensions, C₈H₁₅OH will spread on the water surface.
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A precipitate forms when a solution of lead (il) chloride is mixed with a solution of sodium hydroxide. Write the "total ionic". equation describing this chemical reaction.
The total ionic equation is obtained by dissociating all of the soluble ionic compounds in the equation into their constituent ions. This equation shows all of the ions that are present in the solution during the chemical reaction.Pb2+(aq) + 2Cl-(aq) + 2Na+(aq) + 2OH-(aq) → Pb(OH)2 (s) + 2Na+(aq) + 2Cl-(aq)
The balanced chemical equation of the chemical reaction that occurs when a solution of lead (II) chloride is mixed with a solution of sodium hydroxide is:PbCl2 + 2NaOH → Pb(OH)2 (s) + 2NaCl (aq)
The reaction is a double replacement reaction in which the lead ions (Pb2+) and the sodium ions (Na+) exchange places. Lead (II) hydroxide (Pb(OH)2) precipitates out of the solution as a solid because it is insoluble in water.
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You decide to add to your data table by conducting the same subsitution reactions with another alkyl halide: 3-chloro-1-butene.
Your data shows that 3-chloro-1-butene reacted faster than 2-chlorobutane in an SN1 reaction. Suggest an explanation for this rate difference.
The rate difference observed between 3-chloro-1-butene and 2-chlorobutane in an SN1 reaction can be attributed to the difference in the stability of the carbocation intermediates formed during the reaction.
In an SN1 reaction, the rate-determining step involves the formation of a carbocation intermediate.
The stability of the carbocation greatly influences the reaction rate. In the case of 3-chloro-1-butene, the chlorine atom is attached to a tertiary carbon, resulting in the formation of a more stable tertiary carbocation intermediate.
On the other hand, 2-chlorobutane gives rise to a secondary carbocation intermediate due to the chlorine atom being attached to a secondary carbon.
Tertiary carbocations are more stable than secondary carbocations due to the presence of additional alkyl groups, which provide electron-donating inductive effects, leading to increased electron density and stability.
This increased stability facilitates the formation of the carbocation intermediate and promotes the reaction rate.
Therefore, the higher reactivity of 3-chloro-1-butene compared to 2-chlorobutane in an SN1 reaction can be explained by the more stable carbocation intermediate formed during the reaction, resulting from the presence of a tertiary carbon in 3-chloro-1-butene.
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please help me with these two
questions. thank you!
If a tree dies and the trunk remains undisturbed for \( 1.545 \times 10^{4} \) years, what percentage of the original \( { }^{14} \mathrm{C} \) is still present? (The half-life of \( { }^{14} \mathrm{
After [tex]\(1.545 \times 10^{4}\)[/tex] years, 0.413% of the original [tex]\(^{14}\)C[/tex] (carbon 14)is still present in the tree trunk.
The half-life of carbon-14[tex](\(^{14}\)C)[/tex] is 5730 years. To determine the percentage of[tex]\(^{14}\)C[/tex] remaining after[tex]\(1.545 \times 10^4\[/tex]) years, we can use the formula for exponential decay:
[tex]\[N = N_0 \times \left( \frac{1}{2} \right)^\frac{t}{T}\][/tex]
where:
[tex]\(N\)[/tex] is the remaining amount of[tex]\(^{14}\)C[/tex]after time t,
[tex]\(N_0\)[/tex]is the initial amount of [tex]\(^{14}\)C[/tex],
t is the time that has passed, and
T is the half-life of [tex]\(^{14}\)C[/tex].
Given that the time passed is [tex]\(1.545 \times 10^4\)[/tex] years and the half-life is 5730 years, we can substitute these values into the equation:
[tex]\[N = N_0 \times \left( \frac{1}{2} \right)^\frac{1.545 \times 10^4}{5730}\][/tex]
Calculate the percentage of[tex]\(^{14}\)C[/tex] remaining, we divide [tex]\(N\) by \(N_0\)[/tex] and multiply by 100:
[tex]\[\text{Percentage remaining} = \left( \frac{N}{N_0} \right) \times 100\][/tex]
Calculate the value:
[tex]\[N = N_0 \times \left( \frac{1}{2} \right)^\frac{1.545 \times 10^4}{5730}\][/tex]
[tex]\[\text{Percentage remaining} = \left( \frac{N}{N_0} \right) \times 100\][/tex]
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A certain llquid X has a normal freezing point of 6.50 ∘
C and a freezing point depression constant K f
=2.68 " C⋅kg 'mol −1
. Calculate the freezing point of a solution made of 42.2 g of potassium bromide (KBr) dissolved in 500 . g of X. Round your answer to 3 significant digits.
The freezing point of a solution made of 42.2 g of potassium bromide (KBr) dissolved in 500 g of X is 4.60 ⁰C.
To find the molality (m):
Molality (m) = Moles of solute ÷ Mass of solvent (in kg)
= 0.355 moles ÷ 0.500 kg
= 0.710 mol/kg
Now, put the values into the freezing point depression equation:
ΔT = Kf × m
ΔT = 2.68 ⁰C⋅kg'mol⁻¹ × 0.710 mol/kg
ΔT = 1.9048 ⁰C
To determine the freezing point of the solution, minus the change in freezing point from the normal freezing point of X:
Freezing point of solution = Normal freezing point of X - ΔT
= 6.50 ⁰C - 1.9048 "C
= 4.5952 ⁰C
Rounding to three significant digits, the freezing point of the solution is 4.60 ⁰C.
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Consider a solution of NaHA for which F=0.050 mol L−1, Ka,1=4.70×10−3 and Ka,2=1.80×10−10, find the pH of the solution. pKa1=log(Ka1)=log(4.70×10−3)=−2.32pKa2=log(Ka2)=log(1.80×10−10)=−9.74pH=1/2(pKa1+pKa2)=1/2(−2.32+(−9.74))=−6.03
Answer:1.0⋅10−7
Explanation:Start by writing a balanced chemical equation for the partial ionization of the acidHA(aq]+H2O(l]⇌A−(aq]+H3O+(aq]Notice that you have 1:1 mole ratios across the board. For every mole of acid that ionizes in aqueous solution, you get one mole of its conjugate base and one moleof hydronium ions, H3O+.In other words, the equation produces equal concentrations of conjugate base and hydronium ions. Now, you can use the pH of the solution to calculate the equilibrium concentrationof the hydronium ions. pH=−log([H3O+])⇒[H3O+]=10−pHIn your case, the pH of the solution is equal to 4, which means that you'll have[H3O+]=10−4MBy definition, the acid dissociation constant, Ka, will be equal to Ka=[A−]⋅[H3O+][HA]The expression for the acid dissociation constant is written using equilibrium concentrations. So, if the reaction produced a concentration of hydronium ions equal to 10−4M, it follows that it also produced a concentration of conjugate base equal to 10−4M.Because the initial concentration of the acid is considerably higher than the concentrations of the conjugate base and hydronium ions, you can approximate it to be constant. This means that the acid dissociation constant for this acid will be Ka=10−4⋅10−40.100=1.0⋅10−7This is the underlying concept behind an ICE table HA(aq]+H2O(l] ⇌ A−(aq] + H3O+(aq]I 0.100 0 0C (−x) (+x) (+x)E 0.100−x x xHere x represents the equilibrium concentration for the conjugate acid and hydronium ions. Since you know that x=10−4, you will haveKa=10−4⋅10−40.100−10−4Once again, you can use0.100−10−4=0.0999≈0.100to getKa=10−80.100=1.0⋅10−7
A chemist dissolves 519.mg of pure potassium hydroxide in enough water to make up 60 . mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 ∘
C.) Be sure your answer has the correct number of significant digits.
The pH of the given solution is 13.19 (approx).
Given that the mass of pure potassium hydroxide = 519mg Volume of solution = 60 mL Temperature of solution
= 25°C We need to calculate the pH of the given solution. In order to calculate the pH of the given solution, we need to first calculate the molarity of the potassium hydroxide solution. Molarity of a solution is given as; Molarity = Number of moles of solute / Volume of the solution (in liters) Here, Number of moles of potassium hydroxide = mass of potassium hydroxide / molar mass of potassium hydroxide Molar mass of potassium hydroxide = 39.10 + 1.01 + 16.00
= 56.11 g/mol
= 0.519 g / 56.11 g/mol
= 0.00926 mol Volume of solution
= 60 mL
= 0.06 L.
Now, the molarity of potassium hydroxide solution = 0.00926 / 0.06
= 0.154 M To calculate the pH of the solution, we can use the following formula; pH = -log [H+], where [H+] is the concentration of H+ ions present in the solution. We know that potassium hydroxide is a strong base and it undergoes complete dissociation in water as; KOH (aq) → K+ (aq) + OH- (aq) Since it's a strong base, the concentration of OH- ions present in the solution is equal to the molarity of potassium hydroxide solution. Thus; [OH-] = 0.154 M The value of [H+] can be calculated using the equation; [tex][H+] × [OH-] = 1 x 10^-14[/tex] Hence,
[tex][H+] = 1 x 10^-14 / [OH-][/tex]
[tex]= 1 x 10^-14 / 0.154[/tex]
[tex]= 6.49 x 10^-14[/tex] mol/L The pH of the solution can be calculated using the pH formula as; pH = -log[H+]
[tex]= -log [6.49 x 10^-14][/tex]
= 13.19 Therefore, the pH of the given solution is 13.19 (approx).
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The student started with 0.765 mmoles of Reactant A and 1.00
mmoles of Reactant B. The student obtained 0.600 mmoles of product.
Calculate the % yield for this reaction.
If the student obtained 0.600 mmoles of product, the percent yield for this reaction is 78.43 %.
According to the question:
Beginning with 0.765 mmoles of Reactant A and 1.00 mmoles of Reactant B, the student. The student got 0.600 mmoles of the substance.
To determine the reaction's yield in percentages:
A + B → P
A = 0.765 mmoles
B = 1.00
A is a limiting reagent because its moles are less and consumed fast.
So, theoretical yield of product = 0.765 mmoles
Actual yield of product = 0.600 mmoles
% yield = Actual yield ÷ Theoretical yield
= 0.600 mmoles ÷ 0.765 mmoles
= 78.43 %
Thus, the % yield for this reaction is 78.43 %.
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