(R),(R)-1,3-cyclohexadiol undergoes a ring flip. What happens to the stereochemistry of the compound as it undergoes the ring-flip?
Cyclohexane is a 6-membered ring with all substituents in either the axial or equatorial positions. When a ring-flip occurs, the axial substituents are moved to the equatorial position and the equatorial substituents are moved to the axial position. This then changes the stereochemistry of 1,3-cyclohexadiol. Both chiral centers will be flipped to their opposite orientation (R -> S). Thus, (R),(R)-1,3-cyclohexadiol becomes (S),(S)-1,3-cyclohexadiol.
The energy required to ionize oxygen is less than the energy required to ionize nitrogen. Why do we see this trend in ionization energies?
First, the ionization energy of an atom is the energy required to remove an electron from the atom. Next, we must look at the orbitals of the two atoms and how the electrons are organized within these orbitals. Both oxygen and nitrogen are composed of the following orbitals: 1s, 2s, and 2p. Both atoms contain two electrons in the 1s and 2s orbitals. However, oxygen has 4 electrons in the 2p orbitals while nitrogen only has 3 electrons. Since we know there are three 2p orbitals (px, py, pz), electrons will fill these three orbitals with one electron before adding another electron to the orbitals (making electron pairs). This is the energetically favorable process, since two electrons create repulsive energies and requires more energy. Therefore, oxygen will have one p-orbital with two electrons (higher energy) and two p-orbitals with one electron. If we ionize oxygen, the p-orbital with an electron pair will be reduced to just one electron (less energy). Since less energy means more stable, this ionization is favorable. However, nitrogen has three p-orbitals with one electron in each. If nitrogen is ionized, an empty p-orbital is formed. This does not create a more stable atom and thus requires more energy to ionize than oxygen does. The same trend can be seen throughout the periodic table.
In a beaker, 200 mL of a 2 M acetic acid (pKa = 4.75) solution was titrated using a 1 M sodium hydroxide solution. What is the pH of the solution at the equivalence point?
First, we need to determine the mols of acetic acid that are present in the beaker. We know the concentration of the stock solution is 2 M, meaning 2 mols per liter. We will multiply the concentration by the volume we used, making sure to use the correct units: 200 mL = 0.2 L -> 0.2L * 2 M = 0.4 mols acetic acid. At the equivalence point, equal mols of titrant (sodium hydroxide) and titrand (acetic acid) are in the beaker. Therefore, 0.4 mols of sodium hydroxide was added to reach the equivalence point. We need to calculate the total volume of the beaker now, for future use with concentrations. 0.4 mols sodium hydroxide / 1 M sodium hydroxide = 0.4 L of sodium hydroxide -> 0.2 L (original volume) + 0.4 L (sodium hydroxide added) = 0.8 L total volume. When acetic acid reacts with sodium hydroxide, acetate is generated. Acetic acid only loses one H+ and sodium hydroxide only has one -OH. This means the two will react 1:1. Therefore, at the equivalence point, all of the acetic acid and sodium hydroxide reacted and only 0.4 mols of acetate remains. We know that acetic acid is a weak acid, thus acetate must be a weak base. This means the 0.4 mols of generated acetate will react with water to form hydroxide. A RICE table must be set up in order to determine how much hydroxide is generated. Remember to divide 0.4 mols acetate by total volume (0.8 L) to get molarity (0.5 M). (After a RICE Table) Kb = [-OH][AcOH] / [AcO-] -> Solve for Kb: 14 - 4.75 = 9.25 = pKb -> 10^-9.25 = 5.62x10^-10 -> 5.62x10^10 = [x][x] / [0.5-x] -> we can ignore the x in the denominator using the shortcut rule -> (5.62x10^10 * 0.5) = x^2 -> x = 1.67x10^-5 = [-OH]. Finding pOH = -log(x) = 4.78 -> 14 - 4.78 = pH = 9.22.