General Chemistry Review
Classify each of these reactions. Ba(ClO3)2⟶BaCl2 + 3 O2 a. exchange (double replacement) b. decomposition c. displacement (single replacement) d. combination (synthesis) NaNO2 + HCl⟶ NaCl + HNO2 a. exchange (double replacement) b. decomposition c. displacement (single replacement) d. combination (synthesis) CaO + CO2 ⟶ CaCO3 a. exchange (double replacement) b. decomposition c. displacement (single replacement) d. combination (synthesis) ZnSO4 + Mg ⟶ Zn + MgSO4 a. exchange (double replacement) b. decomposition c. displacement (single replacement) d. combination (synthesis)
Ba(ClO3)2⟶BaCl2 + 3 O2 b. decomposition NaNO2 + HCl⟶ NaCl + HNO2 a. exchange (double replacement) CaO + CO2 ⟶ CaCO3 d. combination (synthesis) ZnSO4 + Mg ⟶ Zn + MgSO4 c. displacement (single replacement)
Which molecule is most soluble in water? CH3OH CH3(CH2)4COOH C2H6 CH3(CH2)5CH2OH
CH3OH Water is a polar substance that forms hydrogen bonds. Therefore, compounds that are polar and capable of hydrogen‑bonding are soluble in water. Consider how molecular weight affects solubility. Water is a polar substance that forms hydrogen bonds. Therefore, compounds that are polar and capable of hydrogen‑bonding are soluble in water. Smaller compounds (lower molecular weight) are more soluble in water than larger compounds. The small polar compound methanol, CH3OH, is the most soluble in water. Pentanoic acid, CH3(CH2)4COOH, and 1‑heptanol, CH3(CH2)5CH2OH, are both polar but have greater molecular weights than methanol, so they are less soluble in water. Ethane, C2H6, is nonpolar and is not soluble in water.
What is the oxidation state of each element in COH2? What is the oxidation state of each element in FeBr2?
COH2 C = 0 O = -2 H= +1 FeBr2 Fe = +2 Br = -1
Label the molecular shape around each of the central atoms in the amino acid glycine.
Count the number of bonded groups and lone pairs on each of the central atoms. Bonded groups Lone pairs Geometry 2 0 linear 2 1 bent 2 2 bent 3 0 trigonal planar 3 1 trigonal pyramidal 4 0 tetrahedral The nitrogen atom has one lone pair and three bonded groups, which form a trigonal pyramidal geometry. The left-most carbon atom has no lone pairs and four bonded groups, which form a tetrahedral geometry. The right-most carbon atom has no lone pairs and three bonded groups, which form a trigonal planar geometry. The central oxygen atom has two lone pairs and two bonded groups, which form a bent geometry.
For the reaction 2 NH3(g) ↽⇀ 3 H2(g) + N2(g) the equilibrium concentrations were found to be [NH3]=0.250 M [H2]=0.330 M [N2]=0.800 M What is the equilibrium constant for this reaction? 𝐾eq=
((0.8)(.33)^3)/((.25)^2) 0.459936 The 𝐾eq expression for this reaction is 𝐾eq= ([H2]^3 x [N2]) / ([NH3]^2) Now substitute the appropriate values into the 𝐾eq expression.
Indicate the direction of polarity of each of the covalent bonds by placing the appropriate delta notation next to each end of the bond. C-O O-Cl O-F C-N Cl-C S-H S-Cl
+ C-O - - O-Cl + + O-F - + C-N - - Cl-C + - S-H + + S-Cl - The more electronegative atom has a stronger attraction for the bonding electrons and, therefore, carries a partial negative charge, whereas the less electronegative atom carries a partial positive charge. The table shows the electronegativity values for each atom. F 4.0 O 3.5 Cl 3.0 N 3.0 C 2.5 S 2.5 H 2.1 Therefore, the bond polarity of each bond can be represented using the partial charge notation.
The reaction A+B⟶C+D rate=𝑘[A][B]^2 has an initial rate of 0.0780 M/s. What will the initial rate be if [A] is halved and [B] is tripled? What will the initial rate be if [A] is tripled and [B] is halved?
0.351 M/s 0.0585 M/s According to the rate law, the reaction is first‑order in A and second‑order in B. If the concentration of a first‑order reactant is halved, then the reaction rate changes by a factor of (1/2)^1=1/2. If the concentration of a second‑order reactant is tripled, then the reaction rate changes by a factor of (3)^2=9. The overall reaction rate changes by a factor of 1/2×9. (1/2)×9×0.0780 M/s=0.351 M/s If the concentration of a first‑order reactant is tripled, then the reaction rate changes by a factor of (3)^1=3. If the concentration of a second‑order reactant is halved, then the reaction rate changes by a factor of (1/2)^2=1/4. The overall reaction rate changes by a factor of 3×(1/4). 3×(1/4)×0.0780 M/s=0.0585 M/s
If a buffer solution is 0.540 M in a weak acid (𝐾a=2.0×10^−6) and 0.140 M in its conjugate base, what is the pH? pH=
5.11 There are two equally valid ways to solve this problem. Method 1: Substitute the concentrations of the acid and conjugate base along with the pKa into the Henderson-Hasselbalch equation. Recall that pKa = −logKa. pH=p𝐾a + log([base]/[acid]) pH=5.70 + log([0.140]/[0.540]) = 5.11 Method 2: Substitute the concentrations of the acid and conjugate base along with the Ka into the Ka expression and solve for [H3O+]. Ka=([H3O+][A−])/([HA]) 2.0×10^−6 = ([H3O+][0.140])/([0.540]) [H3O+] = ((2.0×10−6)[0.540])/([0.140]) = 7.7×10^−6 M Convert [H3O+][H3O+] to pH. pH = −log(7.7×10−6 M) = 5.11 In both methods, you can treat the initial concentrations of HA and A− as approximately equal to their equilibrium concentrations because the common ion effect reduces the extent of ionization.
Identify the Lewis acid and Lewis base in each of the reactions. Cl− + AlCl3 ⟶ AlCl−4 BF3 + F− ⟶ BF−4 NH3 + H+ ⟶ NH+4
A Lewis acid is an electron acceptor. A Lewis base is an electron donor. Cl− + AlCl3 ⟶ AlCl−4 Cl− is an electron donor. AlCl3 is an electron acceptor. BF3 + F− ⟶ BF−4 BF3 is an electron acceptor. F− is an electron donor. NH3+H+⟶NH+4 NH3 is an electron donor. H+ is an electron acceptor.
Identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction. Cu(s)+2AgNO3(aq)⟶2Ag(s)+Cu(NO3)2(aq) Which substance gets oxidized? Which substance gets reduced? What is the oxidizing agent? What is the reducing agent?
Cu gets oxidized Ag+ gets reduced Ag+ is the oxidizing agent Cu is the reducing agent The net ionic equation is Cu(s) + 2Ag+(aq) ⟶ 2Ag(s) + Cu2+(aq) Oxidation is the loss of one or more electrons. Reduction is the gain of one or more electrons. Therefore, an atom or ion is reduced if it gains electrons. The substance that causes the oxidation is the oxidizing agent, and the substance that causes the reduction is the reducing agent.
Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released). Strongest bond A-B A-A B-B C-C B-C A-C Weakest bond
Endothermic: AB + C --> AC + B because A-B bonds are stronger than A-C bonds. A2 + C2 --> 2 AC is endothermic because A-A and C-C bonds are both stronger than A-C bonds. B2 + C2 --> 2 BC is endothermic because B-B and C-C bonds are both stronger than B-C bonds. Exothermic: A + BC --> AB + C is exothermic because B-C bonds are weaker than A-B bonds. A2 + B2 --> 2 AB is exothermic because A-A and B-B bonds are both weaker than A-B bonds. Endothermic reaction is the energy required to break the reactant bonds is more than the energy released during the formation of product bonds. In other words, an endothermic reaction has stronger reactant bonds (large energy requirement) and weaker product bonds (small energy release). Exothermic reaction is the energy required to break the reactant bonds is less than the energy released during the formation of product bonds. In other words, an exothermic reaction has weaker reactant bonds (small energy requirement) and stronger product bonds (large energy release).
Consider the redox reaction Fe(s) + Cu2+(aq) ⟶ Fe2+(aq) + Cu(s) Which substance gets oxidized? What is the oxidizing agent? Which substance gets reduced? What is the reducing agent?
Fe gets oxidized Cu2+ is the oxidizing agent Cu2+ gets reduced Fe is the reducing agent
Identify the conjugate acid for each base. conjugate acid of HCO3^-: conjugate acid of CO3^2−: conjugate acid of NH3:
H2CO3 HCO3- NH4+
Complete the abundance diagram by identifying the species present in an aqueous solution of 1.00 M acetic acid (CH3COOH). The 𝐾a of acetic acid is 1.8×10^−5. What is the approximate pH of a solution of 1.00 M CH3COOH? a. between 4.74 and 7.00 b. over 7.00 c. less than 4.74 Identify the major species at a pH of 3. a. CH3COO- b. H3O+ c. CH3COOH d. OH- Identify the minor species at a pH of 3. a. H3O+ b. OH- c. CH3COO- d. CH3COOH
H3O+ CH3COOH CH3COO- OH- c. b. & c. b. & c. The chemical equations to consider are CH3COOH(aq) + H2O(l) ↽⇀CH3COO−(aq) + H3O+(aq) H2O(l) + H2O(l) ↽⇀ H3O+(aq) + OH−(aq) The 𝐾a of acetic acid is 1.8×10^−5, and therefore the p𝐾a is p𝐾a = −log(𝐾a) = −log(1.8×10−5) = 4.74 At low pH values, the protonated form of acetic acid, CH3COOH, will be the predominant form of acetic acid in solution. As the pH increases, so does the amount of the acetate ion, CH3COO−, in solution. When the pH of the solution is equal to the pKa of acetic acid, there are equal amounts of acetic acid and the acetic ion in solution. Similarly for the autoionization of water, the hydronium ion is the predominant form at low pH values. When the pH is equal to 7 there are equal amounts of the hydronium ion and hydroxide ion in solution. To find the pH of a solution of 1.00 M acetic acid set up a concentration table to determine expressions for the equilibrium concentrations. Insert the expressions into the Ka expression and solve for x. 𝐾a=([H3O+][CH3COO−])/([CH3COOH]) = 𝑥^2/ (1.00−𝑥) 1.8×10^−5 = (𝑥^2)/(1.00−𝑥) 𝑥2 + (1.8×10^−5)𝑥 − (1.8×10^−5) = 0 Using the quadratic equation gives 𝑥 = [H3O+] = 4.23×10^−3 M pH = −log[H3O+]= −log(4.23×10^−3) = 2.37 The major form of acetic present below pH 4.74 is CH3COOH, and the major form present above the pKa is CH3COO−. These forms are present in roughly equal amounts at pH 4.74. Therefore at pH 3, CH3COOH is a major species and CH3COO− is a minor species. Similarly, water has a pKa of 7. At low pH values, H3O+ is present in greater amounts, whereas OH− is present at high pH values. Therefore at pH 3, H3O+ is a major species and OH− is a minor species.
Identify the conjugate base for each acid. conjugate base of H2SO4: conjugate base of HCO3^−: conjugate base of NH4^+:
HSO4^- CO3^-2 NH3
For the general reaction A+BC⟶ABC how would each event affect the rate of the reaction? increasing the temperature decreasing the concentration of BC increasing the concentration of A adding a catalyst
Increase reaction rate - increasing the temperature - increasing the concentration of A - adding a catalyst Decrease reaction rate - decreasing the concentration of BC Increasing the concentrations of the reactants increases the rate at which they collide. In other words, having more reactants results in more collisions and increases the rate of the reaction. Increasing the temperature increases the rate of the reaction by causing the reactants to move faster, increasing the rate at which reactants collide. Catalysts allow the reaction to take place through an alternative route with a lower activation energy. Therefore, the reactant molecules need less energy to react with each other, which means that less forceful collisions will result in a reaction. Increasing the concentration of a reactant, increasing the temperature, or adding a catalyst will increase the rate of a reaction.
Which type of bond exists in each compound? KCl BCl3 P4 Br2 CO SO2
KCl - ionic bond, The electronegativity of K is 0.82 and the electronegativity of Cl is 3.16. The difference is 2.34 units. Therefore, the bonding is ionic. BCl3 - polar covalent bond, The electronegativity of B is 2.04 and the electronegativity of Cl is 3.16. The difference is 1.12 units. Therefore, the bonding is polar covalent. The overall molecule is nonpolar due to its symmetric shape, but that will be covered in a later chapter. P4 - nonpolar covalent bond, There is no difference in the electronegativities of the P atoms. Therefore, the bonding is nonpolar covalent. Br2 - nonpolar covalent bond, There is no difference in the electronegativities of the Br atoms. Therefore, the bonding is nonpolar covalent. CO - polar covalent bond, The electronegativity of C is 2.55 and the electronegativity of O is 3.44. The difference is 0.89 units. Therefore, the bonding is polar covalent. SO2 - polar covalent bond, The electronegativity of S is 2.58 and the electronegativity of O is 3.44. The difference is 0.86 units. Therefore, the bonding is polar covalent.
Arrange the organic compounds from most soluble in water to least soluble in water.
Most soluble in water CH3OH CH3OCH3 CH4 Least soluble in water A hydrogen bond forms between the unshared electron pair of a strongly electronegative atom (O,N, or F)(O,N, or F) and a hydrogen atom bonded to a strongly electronegative O , N , or F atom. Alcohols (R−OH) are very soluble in water because they can form more than one hydrogen bond. The H atom in the OH group can form a hydrogen bond with the oxygen atom of water. The O atom of the OH group can form a hydrogen bond with a hydrogen atom from water. Ethers (R−O−R′) are somewhat soluble in water. The oxygen atoms have unshared pairs of electrons that can take part in hydrogen bonding; however, there are no hydrogen atoms that can take part in hydrogen bonding. Alkanes are not soluble in water. Alkanes contain only carbon and hydrogen atoms, and they are nonpolar. Therefore, these compounds arranged from most soluble to least soluble in water are CH3OH > CH3OCH3 > CH4
Complete the reaction between a Brønsted-Lowry acid and base. Phases are optional. HCl + NH3 ⟶
NH4^+ + Cl-
Identify the products formed in this Brønsted-Lowry reaction. HPO4^−2 + F− ↽⇀ acid + base
PO4^-3 base HF acid HPO4^2− is amphoteric, which means it may act as an acid or a base depending on the other reactant. F− cannot be an acid because it does not contain hydrogen. A negative charge is often indicative of a Brønsted‑Lowry base. As a base, F− will accept H+ to form the conjugate acid HF. Because F− is a base, HPO4^2− will behave as an acid. As an acid, HPO4^2− will donate H+ to form the conjugate base PO3^4−. HPO4^-2 + F− ↽⇀ HF + PO4^−3
In the context of small molecules with similar molar masses, arrange the intermolecular forces by strength. Hydrogen Bonding Dipole-Dipole Interactions London Dispersion Force
Strongest Hydrogen Bonding Dipole-dipole interactions London Dispersion Forces Weakest Dipole-dipole attraction exists between polar molecules. Dispersion forces, also known as London forces or van der Waals forces, are temporary dipoles that form between nonpolar molecules. Hydrogen bonding occurs when a hydrogen atom that is bonded to O,O, N,N, or FF is attracted to the O,O, N,N, or FF atom of another nearby molecule. Hydrogen bonding is stronger than dipole-dipole interactions because hydrogen bonding involves particularly polar bonds, those between HH and N,N, O,O, or F,F, and also a lone pair, to which the electron‑deficient hydrogen atom is particularly attracted. The temporary dipole that characterizes dispersion forces is typically weaker than the permanent dipole of dipole-dipole interactions. However, in large molecules, there is room for more areas of temporary polarity, which can have a compounding effect, resulting in very strong intermolecular attraction. Paraffin wax is an example of a nonpolar substance with a particularly high melting point.
A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) + H2O(l) ↽⇀ H3O+(aq) + A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.130 M H3O+]=2.00×10^−4 M, [A−]=2.00×10^−4 M Calculate the Ka value for the acid HA. 𝐾a=
The acid dissociation constant expression for the dissociation of the monoprotic weak acid HA is 𝐾a= ([H3O+][A−])/[HA] Insert the equilibrium concentrations and solve for Ka. 𝐾a=((2.00×10^−4)×(2.00×10^−4))/0.130 = 3.08×10^−7
At 25 °C, what is the hydroxide ion concentration, [OH−], in an aqueous solution with a hydrogen ion concentration of [H+]=2.6×10^−8 M? [OH−]=
The autoionization of water can be described by the equilibrium reaction H2O(l) ↽⇀ H+(aq) + OH−(aq) The equilibrium constant for this reaction is Kw, the ion product of water. The equilibrium constant expression for the autoionization of water is 𝐾w=[H+][OH−] where Kw is a constant equal to 1.0×10^−14. Substitute [H+][H+] and Kw into the equilibrium constant expression, and rearrange to solve for [OH−]. 1.0×10^−14 = (2.6×10^−8 M) [OH−] [OH−] = (1.0×10^−14)/ (2.6×10^−8) = 3.8×10^−7 M
Consider the reversible reaction A(g)↽⇀B(g) Which 𝐾 values would indicate that there is more B than A at equilibrium? a. 𝐾=4×10^8 b. 𝐾=9×10^-6 c. 𝐾=2000 d. 𝐾=0.1
The equilibrium constant, 𝐾, expresses the relationship between the concentrations of the species at equilibrium. In this case, 𝐾=[B][A] When [B] is greater than [A], 𝐾 is greater than 1. Of the given values, 4×10^8 and 2000 are greater than 1.
Calculate the [OH−] and the pH of a solution with an [H+]=6.4×10^−11 M at 25 °C. [OH−]= pH= Calculate the [H+] and the pH of a solution with an [OH−]=6.3×10^−9 M 25 °C. [H+]= pH= Calculate the [H+] and the [OH−] of a solution with a pH=1.39 at 25 °C. [H+]= [OH−]=
[OH−]= 1.55*10^-4 pH= 10.19 [H+]= 1.58*10^-6 pH= 5.8 [H+]= 4.07*10^-2 [OH−]= 2.45*10^-13
Chemical reactions occur when molecules or atoms collide in a way that the bonds between atoms break and new bonds form. Breaking the bonds of the reactants requires energy, whereas bond formation releases energy. Select the true statements regarding energy changes during a reaction. a. Increasing the concentration of reactants increases the number of collisions, and the reaction goes faster. b. The energy of a collision between atoms or molecules must be greater than or equal to the activation energy, 𝐸a, for bonds to be broken. c. If the heat of reaction, Δ𝐻, is positive, the energy of the products is lower than the energy of the reactants and the reaction is endothermic. d. The activation energy, 𝐸a, of the forward reaction is the difference between the energy of the products and the energy of the transition state. e. Decreasing the temperature decreases the kinetic energy of the reactants, and the reaction goes more slowly. f. When the activation energy is low, the reaction rate is slow. g. Reactants must collide with proper orientation and with energy greater than or equal to the activation energy for a reaction to occur.
a, b, e, g Two energy diagrams are shown. One shows an exothermic reaction with the energy of the products lower than the reactants; the other shows an endothermic reaction with the energy of the products higher than the energy of the reactants. The activation energy, 𝐸a, of a reaction is the energy required for the reaction to occur. In other words, it is the energy required for bonds to be broken. For the reaction to take place, reactants must collide with proper orientation and sufficient force that is greater than or equal to 𝐸a. The greater the 𝐸a, the slower the reaction; the lower the 𝐸a, the faster the reaction. Increasing the temperature increases the kinetic energy of the reactants. The reactants move faster, colliding more frequently and with more energy. Therefore, increasing the temperature increases the rate of the reaction, thus the reaction goes faster. Increasing the concentration of reactant increases the probability that reactants will collide because there are more molecules in the same amount of area. Therefore, increasing the amount of reactant increases the reaction rate. In an exothermic reaction, the energy of the products is lower than that of the reactants, and Δ𝐻 is negative. In an endothermic reaction, the energy of the products is higher than that of the reactants, and Δ𝐻 is positive.
Which molecules can form hydrogen bonds with water? a. CH3CH2CHO b. C2H6 c. CH3(CH2)3COOH d. CH3OH
a., c. & d. Compounds that are polar and form hydrogen bonds are soluble in water. Hydrogen bonds form between a strongly electronegative atom (O, N, or F) and a hydrogen atom bonded to a strongly electronegative atom. Therefore, the polar compounds CH3OH (methanol), CH3(CH2)3COOH (pentanoic acid), and CH3CH2CHO (propanal), are soluble in water. Ethane (C2H6) is nonpolar, and is not soluble in water. Examples of hydrogen bonds between an alcohol, carboxylic acid, aldehyde, and water are shown. Hydrogen bonds are represented as dashed lines.
What factors determine whether a collision between two reactant molecules will result in a reaction? a. The energies of the colliding particles must be equal to each other. b. The particles must collide with enough energy to meet the activation energy of the reaction. c. The particles must be charged. d. The particles must be in the correct orientation upon impact.
b & d Particles must collide in order for chemical reactions to take place, but they must collide with the correct orientation. Heat increases the rate at which particles collide and the force with which they collide, and so will usually increase the rate at which reactions occur. So, although heat can affect reaction rates, it is not the determining factor for their occurrence. Although pH can influence the reaction rates of many biological reactions, it is generally only relevant to acid-base reactions. The activation energy is the minimum required kinetic energy that particles must possess in order for a reaction to take place.
Select all of the true statements regarding chemical equilibrium. a. The concentrations of reactants and products are equal. b. The concentrations of reactants and products remain constant. c. Reactants are being converted to products and vice versa. d. The rates of the forward and reverse reactions are equal.
b, c, d. Equilibrium occurs when the rates of the forward and reverse reactions are equal. Reactants are being converted to products and vice versa. Even though both reactions are occurring, there is no change in concentration. In other words, each species is being produced and consumed at the same rate.
How does a buffer resist change in pH upon addition of a strong acid? a. The strong acid reacts with the weak acid in the buffer to form a weak base, which produces few H+ ions in solution and therefore only a little change in pH. b. The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH. c. The strong acid reacts with the strong base in the buffer to form a salt, which produces few H+ ions in solution and therefore only a little change in pH.
b. A buffer consists of a weak acid and its conjugate base (a weak base) in solution. For example, acetic acid and sodium acetate form a buffer (HOAc/OAc buffer). If you have a HOAc/OAc buffer and you add a stong acid (i.e., HCl or H2SO4), the strong acid reacts with some of the sodium acetate (weak base) to form acetic acid (weak acid). The result is still a buffer that that has slightly more acetic acid and slightly less acetate. The change in pH is small.
Which statement describes the action of a buffer composed of formic acid (HCOOH) and sodium formate (NaHCOO)? a. Both components, formic acid and sodium formate, neutralize added acid. b. Formic acid neutralizes added base, and sodium formate neutralizes added acid. c. Sodium formate neutralizes added base, and formic acid neutralizes added acid. d. Both components, formic acid and sodium formate, neutralize added base.
b. Neutralization reactions occur between acids and bases. Buffers, which are composed of an acid and its conjugate base, lessen the change in pH by reacting with added acid (H+) or added base (OH−). The formate ion (HCOO−) is the conjugate base of formic acid (HCOOH). HCOOH (aq)↽⇀ HCOO− (aq) + H+(aq) The formic acid component of the given buffer reacts with added OH− ions according to the chemical equation HCOOH (aq) + OH−(aq) ⟶ HCOO− (aq) + H2O(l) The sodium formate (NaHCOO) component of the given buffer exists in the solution as dissociated formate ions and sodium ions. The formate ions from sodium formate react with added H+ ions according to the chemical equation HCOO− (aq) + H+(aq) ⟶ HCOOH (aq) The reaction of the formic acid component of the buffer with hydroxide ions produces formate ions, and the reaction of the formate ion component of the buffer with hydrogen ions produces formic acid. Thus, the reaction of each component of the buffer regenerates the other component.
Select all the true statements. a. When an atom gains an electron, it becomes a cation. b. The K+ ion is formed when a potassium atom loses one electron. c. The Fe2+ and Fe3+ ions have the same number of protons. d. The Cu+ and Cu2+ ions have the same number of electrons. e. Anions carry a positive charge. f. The Cl− and Br− ions have the same number of electrons.
b. & c. Explanation: In a neutral atom, the number of electrons is equal to the number of protons, so the sum of their charges is zero. If an atom has lost an electron, there are more protons than electrons, and a positively charged ion, called a cation, is formed. When an atom gains an electron, it becomes a negatively charged ion, called an anion.
The energy of an electrostatic interaction between two charged atoms is dependent on the charges on the atoms, the distance between them, and the dielectric constant of the solvent. For example, the strength of a weak acid (Ka, acid dissociation constant) depends on the strength of the electrostatic interaction between a negatively charged carboxylic acid group and a proton. The solvent dielectric constant has a large influence on the pKa for weak acids. Select the statements that correctly explain the influence of two solvents, water, and hexane, on the pKa of acetic acid. a. In water, the association of CH3COO- and H+ will be favored because charge-charge interactions are stronger in water. b. The pKa will be higher in hexane. c. Nonpolar solvents like hexane will weaken charge-charge interactions resulting in more dissociation of H+. d. More H+ will dissociate in water because the dielectric constant is higher. e. Hexane cannot affect the pKa of acetic acid since hexane cannot donate or accept H+.
b. & d. The dielectric constant of a solvent is a measure of a solvent's ability to diminish the electrostatic force between two charged atoms (as compared to in a vacuum). Hexane has a very low dielectric constant (2), compared to water (80). This means that hexane will not attenuate the charge-charge interaction between CH3COO− and H+, and the pKa will be higher (i.e. the protonated form, CH3COOH, will be strongly favored). Nonpolar solvents like hexane will strengthen charge‑charge interactions, resulting in less H+ dissociation. Although hexane cannot ionize and is not involved in the protonation/deprotonation equilibrium, hexane still effects the dissociation of acetic acid by dictating the environment in which the charged species would have to form. Nonpolar environments prohibit the formation of charged ions since there are no polar interactions to stabilize charged species. The high dielectric constant of water means that charge-charge interactions will be weakened in this solvent. Therefore, acetic acid dissociation will be favored and more H+ will be formed. The dielectric constant is commonly used as an indicator of solvent polarity. A polar solvent will have a higher dielectric constant than a nonpolar solvent (e.g. hexane). The polar nature of water favors weak acid dissociation (lower pKa) because the separated ions will be stabilized by polar interactions with water.
Predict the approximate shape of a water molecule. The shape is:
bent Water has a bent shape. The four electron groups on the central atom (two lone pairs and two bonded atoms) form an approximate tetrahedron such that the H−O−H bond angle is approx. 109.5°. Repulsion between the lone pairs and the bonds reduces this bond angle somewhat (the actual bond angle is 104.5°).
Write the Henderson-Hasselbalch equation for a propanoic acid solution (CH3CH2CO2H, pKa = 4.874). ____ = _____ + log (___/____) Using the equation to calculate the quotient [A−]/[HA] at three different pH values. pH = 4.396 [A−]/[HA]= pH = 4.874 [A−]/[HA]= pH = 5.060 [A−]/[HA]=
pH = pKa + log ([A-]/[HA]) 0.333 1 1.535
What is the molecular geometry of the left carbon atom (circled) in acetic acid? What is the molecular geometry of the right carbon atom (circled) in acetic acid?
tetrahedral, trigonal planar In acetic acid, the left carbon atom has four bonded groups and no lone electron pairs, which form a tetrahedral geometry. The right carbon atom has three bonded groups and no lone electron pairs, which form a trigonal planar geometry. Trigonal planar is sometimes called triangular.
The curve 𝑦=1/ ((x^2)+1) is called the witch of Agnesi after the Italian mathematician Maria Agnesi (1718-1799) who wrote one of the first books on calculus. This strange name is the result of a mistranslation of the Italian word la versiera meaning "that which turns." Find equations of the tangent lines to the curve at 𝑥=±9. (Use symbolic notation and fractions where needed.) At 𝑥=9,x=9, the tangent line is 𝑦= __? At 𝑥=−9,x=−9, the tangent line is 𝑦= __?
x= 9 y= - (9/3362)x + (61/1681) x=-9 y=(9/3362)x + (61/1681) First, use equation of the tangent line to 𝑓 at 𝑥=𝑎. 𝑦=𝑓′(𝑎)(𝑥−𝑎)+𝑓(𝑎) Calculate the derivative of𝑓. 𝑓′(𝑥)=((0)((𝑥^2)+1)−1(2𝑥))/(𝑥^2+1)^2 = − 2𝑥/(𝑥^2+1)^2 https://www.symbolab.com/solver/tangent-line-calculator
Consider the rate law. rate=𝑘[A]^𝑥 Determine the value of 𝑥 if the rate doubles when [A] is doubled. 𝑥= Determine the value of 𝑥 if no change in rate occurs when [A] is doubled. 𝑥=
x=1 x=0 In this rate law, 𝑥 represents the order of the reaction with respect to A. A zero‑order relationship means that the rate is independent of concentration. A first‑order relationship means that the rate is directly proportional to concentration. A second‑order relationship means that the rate is directly proportional to the square of the concentration. If the rate doubles when [A] is doubled rate2/rate1 = 2=2^𝑥 𝑥=1 If no change in rate occurs when [A] is doubled rate2/rate1= 1=2^𝑥 𝑥=0
