1.The equilibrium constant for the oxidation of elemental iron metal, Fe, by oxygen, O2 to form hematite, according to the reaction 2Fe + (3/2)O2 = Fe2O3 is known to have the value K=exp(+68.8) at 1000C. The equilibrium constant for the formation of CO2 from carbon and O2, C + O2 = CO2 at the same temperature is known to be K = exp(+48.90). Is it possible to prepare elemental iron in high yield from hermatite by using carbon (from coke) at 1000C? Justify your answer.
2.Glycolic acid is often used in “facial peels” and has a Ka of 1.5x10^-4. If a movie star is to have a solution of pH 2.1 applied to her face at night, what concentration of glycolic acid solution should be made up (assuming that it is a monoprotic acid, and that it dissolves freely in water)
3.The solubility product for silver (I) chloride is 1.6x10^-10 while that for sliver (I) iodide is 8.5x10^-17. Suppose 5g of AgCl is put in a liter of 0.1M sodium iodide solution. If a solid is eventually isolated, what is it, and how did it form?
4.The Kb of ammonia, NH3, is 1.8x10^-5. What is the pH of a 2.000 millimolar solution of ammonia , to 3 significant digits.
5.A weak monoprotic acid was calculated, using the usual formula for Ka, to give a pH of 7.4 at a certain concentration. The student who did the calculation was puzzled. What would you advice him/her to do?
6.Why is it that biochemists often remember the pKa of weak acids, rather than Ka?
7.If we suppose that Kw = 1.0x10^-14 at 25C, would it be larger or smaller at 100C? What is the value of K for the reaction 8H + 8OH = 8H2O at 25C? Would this value be smaller, or larger, at 100C?
I looked only at #1. Seven prolems per post is six too many. Do you have any idea how long it takes to work and type in the answers? Too long. Most volunteers will not tackle one that takes that long to answer.
If you will note, equation 1 reversed + equation 2 times 3/2 will give you a balanced equation for Fe2O3 + C ==>>??
So K' for equation 1 will be 1/K1
K" for equation 2 will be (K2)3/2
Adding the reverse of equation 1 to equation 2 (times 3/2) will give a K for the reaction of K'*K". A positive K value indicates the spontaneity of the reaction under the conditions shown.
Thank You DrBob222! This answer helped alot!
1. In order to determine if elemental iron can be prepared in high yield from hematite using carbon at 1000C, we can compare the equilibrium constant for the oxidation of iron and the equilibrium constant for the formation of CO2.
The equilibrium constant, K, is a measure of the degree of completion of a reaction. A higher value of K indicates that the reaction goes to completion in the forward direction. If the forward reaction has a higher value of K compared to the reverse reaction, then the reaction is more likely to proceed in the forward direction.
Given that the equilibrium constant for the oxidation of iron is K = exp(+68.8) and the equilibrium constant for the formation of CO2 is K = exp(+48.90), we can see that the oxidation of iron has a much higher value of K compared to the formation of CO2.
Therefore, it is unlikely that elemental iron can be prepared in high yield from hematite using carbon at 1000C. The equilibrium favors the formation of hematite rather than the oxidation of iron.
2. To determine the concentration of glycolic acid solution needed to achieve a pH of 2.1, we can use the expression for the acidity constant, Ka.
Given that Ka = 1.5x10^-4, we can use the equation Ka = [H+][A-]/[HA], where [H+] represents the concentration of hydrogen ions, [A-] represents the concentration of the conjugate base, and [HA] represents the concentration of the acid.
Assuming that the concentration of glycolic acid is equal to the concentration of [HA], we can solve for [H+].
[H+] = sqrt(Ka * [HA])
Given that pH = -log[H+], we can substitute pH = 2.1 and solve for [HA].
[HA] = 10^(-pH) * [H+]
Substituting the values, we find [HA] = 10^(-2.1) * sqrt(Ka * [HA])
Simplifying the equation gives [HA]^3 = Ka^2 * (10^(-2.1))^2
Solving for [HA] gives [HA] = (Ka^2 * (10^(-2.1))^2)^(1/3)
Substituting the values, we can find the concentration of glycolic acid solution needed to achieve a pH of 2.1.
3. To determine the solid that is eventually isolated when 5g of AgCl is put in a liter of 0.1M sodium iodide solution, we can compare the solubility products of silver chloride and silver iodide.
The solubility product, Ksp, is a measure of the solubility of a compound in a solution. A smaller value of Ksp indicates lower solubility, while a larger value indicates higher solubility.
Given that the Ksp for AgCl is 1.6x10^-10 and the Ksp for AgI is 8.5x10^-17, we can compare the values.
The Ksp for AgCl is larger than the Ksp for AgI, indicating that AgCl is more soluble in solution compared to AgI.
However, when AgCl is put in a sodium iodide solution, a double displacement reaction occurs, resulting in the formation of AgI and NaCl. Since AgI has lower solubility compared to AgCl, it will precipitate out as a solid.
Therefore, the solid that is eventually isolated would be AgI, which forms due to the reaction between AgCl and NaI.
4. To find the pH of a 2.000 millimolar solution of ammonia (NH3), we need to calculate the concentration of hydroxide ions (OH-) using the base dissociation constant, Kb.
Kb represents the equilibrium constant for the reaction NH3 + H2O = NH4+ + OH-, where NH4+ is the conjugate acid of ammonia.
Given that Kb = 1.8x10^-5, we can use the equation Kb = [NH4+][OH-]/[NH3] to solve for [OH-].
Since the concentration of ammonia is 2.000 millimolar, we have [NH3] = 0.002 M.
Substituting the values, we have Kb = [NH4+][OH-]/0.002
Solving for [OH-] gives [OH-] = (Kb * 0.002)/[NH4+]
Assuming that the concentration of [NH4+] is negligible compared to 0.002, we can approximate [OH-] to be equal to (Kb * 0.002)/0.002
Substituting the values, we can solve for [OH-] and convert it to pH.
5. If a weak monoprotic acid gives a pH of 7.4 at a certain concentration, it implies that the acid is not completely dissociating in water. This can happen when the acid is weak and only partially ionizes.
To determine the exact behavior of the acid, further experimental measurements are needed. It is possible that the acid is partially ionizing at the given concentration, resulting in a lower concentration of hydrogen ions and a higher pH than expected.
To better understand the behavior of the acid, the student should perform additional experiments to measure the concentration of hydrogen ions at different concentrations of the acid and compare the results to the expected behavior based on the acid's dissociation constant, Ka.
6. Biochemists often remember the pKa of weak acids rather than Ka because the pKa value represents the negative logarithm of the acid's dissociation constant, making it easier to compare the strengths of different weak acids.
The pKa is a measure of the acidity of an acid, with lower pKa values indicating a stronger acid. By remembering the pKa values of weak acids, biochemists can easily compare and predict the behavior of different acids in biochemical reactions.
Additionally, the pKa value is directly related to the pH of a solution containing the acid through the Henderson-Hasselbalch equation. The pKa value can be used to calculate the pH of a solution at a given acid concentration, making it a useful tool in biological and biochemical studies.
7. The value of Kw, the ion product of water, is 1.0x10^-14 at 25C. At higher temperatures, such as 100C, the dissociation of water molecules into hydrogen ions (H+) and hydroxide ions (OH-) is more favorable due to increased thermal energy.
Therefore, Kw would be larger at 100C than at 25C.
The value of K for the reaction 8H + 8OH = 8H2O at 25C can be determined using the equation:
K = ([H2O]^8) / ([H]^8 * [OH]^8)
Given that water is in excess, we can assume [H2O] remains constant. Additionally, since the reaction is in a neutral aqueous solution, [H] = [OH]. Therefore, the equation can be simplified to:
K = [H2O]^8 / ([H]^8)^2
Using the value of Kw = 1.0x10^-14, which is equal to [H][OH], we can substitute [OH] = [H] = sqrt(Kw).
Simplifying, we find K = Kw^4
Substituting the value of Kw, we find K = (1.0x10^-14)^4
Therefore, the value of K for the reaction 8H + 8OH = 8H2O at 25C is (1.0x10^-14)^4.
To determine if the value would be smaller or larger at 100C, we need to consider the temperature dependence of the equilibrium constant. Generally, increasing the temperature favors the forward reaction, resulting in a larger equilibrium constant.
Therefore, the value of K for the reaction would be larger at 100C compared to 25C.