Physics CH 9 - Acids/Bases

[26 in book]All of these compounds are considered to be acids by chemists. Which is not a Brønsted acid? a. BF3 b. H3PO4 c. HCl d. All of these are Brønsted acids

Answer: a. A Brønsted acid is an H+ donor. Since BF3 has no hydrogens, it is not a Brønsted acid.

[24 in book]Do you expect the equilibrium constant for binding carbon monoxide to hemoglobin (Hb) to be larger or smaller than the equilibrium constant for binding oxygen to hemo- globin? Explain. a. Larger b. Smaller c. Nearly zero d. Nearly infinite Hb + 4 CO = Hb(CO)4 Hb + 4 O2 = Hb(O2)4

Answer: a. A larger equilibrium constant indicates the reaction favors formation of the products to a greater extent. From your training, you know that carbon monoxide binds to hemoglobin more tightly than oxygen. Therefore, the equilibrium constant for the reaction of carbon monoxide is larger.

[28 in book]To be an effective buffer, the acid must A.Have a pKa nearly equal to the desired pH B.Be a strong acid C.Have a strong conjugate base D.All of these

Answer: a. The capacity of a buffer is highest when the pH is near the pKa of the acid.

[35 in book]What is the pH of a buffer that contains equal amounts of HCO3- and CO32-? a. 10.30 b. 7.00 c. 3.70

Answer: a. The pH of a buffer in which the concentration of the weak acid (HCO3-) is equal to the concentration of the weak base (CO32−) will be equal to the pKa of the acid.

[30 in book]Which conjugate acid-base pair is most important in buffering blood? a. H3PO4 and H2PO4− b. H2PO4- and HPO42− c. HPO42- and PO43−

Answer: b. Effective buffers have acids with pKa close to the desired pH. H2pO4− has pKa = 7.2. So, this acid and its conjugate base (HpO42() act as phosphate buffers in blood.

[33 in book] What is the conjugate base of bicarbonate, HCO3−? a. H2CO3- b. CO32- c. OH-

Answer: b. The conjugate base has lost an H+ ion, so there is one fewer hydrogen in the formula, and the charge has decreased by one.

[31 in book] What is the pH of a 0.001 M solution of HCl? a. 1 b. 2 c. 3 d. 7

Answer: c. HCl is a strong acid, so [H+] is the same as the formal concentration of HCl. pH = (-log(0.001)) = 3.

[32 in book] What is the hydrogen ion concentration in a solution whose pOH is 12.0? A.2 M B.1 × 10-12 M C.1×10-2M D.More information is needed.

Answer: c. If the pOH is 12, then the pH = 14 − 12 = 2. [H+] = 10−pH = 10−2 = 0.01 M.

[25 in book]One treatment for carbon monoxide poisoning is administration of oxygen. The high concentration of oxygen binds to free hemoglobin, including the Hb from the equilib- rium between Hb and CO. This is an application of: A.Henry's law B.Raoult's law C.Le Châtelier's principle D.Trying really hard not to kill the patient

Answer: c. In the presence of excess oxygen, any free hemoglobin in the equilibrium with carbon monoxide will be converted to oxyhemoglobin, reducing the amount of free hemo- globin. The equilibrium with carbon monoxide responds by releasing more free hemoglobin.

[27 in book]What is the conjugate acid of nitric acid (HNO3)? a. H+ b. H3O+ c. HNO2 d. H2NO3+

Answer: d. A conjugate acid has an additional H+ ion, so adding one more H+ increases the charge by one.

[29 in book]Given the pKa values for phosphoric acid H3PO4 = H+ + H2PO4− pK1 = 2.2 H2PO4− = H+ + HPO4−2 pK2 = 7.2 HPO4−2 = H+ + PO4−3 pK3 = 12.3 Which of these is the strongest base? a. H3PO4 b. H2PO4− c. HPO42− d. PO43−

Answer: d. The stronger the acid, the weaker the conjugate base, and vice versa. Therefore, the strongest base comes from the weakest acid. The weakest acid has the largest pKa.

[36 in book] You have 1 L of water. You add 1.0 × 10−8 mol of HCl, so now the molarity of the HCl is 1.0 × 10−8 M. What is the pH of the solution? Be careful! a. 8.00 b. 6.98 c. 7.00

[PAGE 517 FOR DETAILED EXPLANATION] b. This one is tricky! If you add even a tiny amount of HCl to pure water, the pH will not increase from 7 to 8. The solution is that there is another source of [H+]: the self-ionization of water. So this is an equilibrium problem. We start with the equilibrium reaction and the equilibrium constant: H2O = H+ + OH− Kw = [H+ ][OH− ]

[34 in book]If the Ka of bicarbonate is 4.8 × 10−11, what is the pKa of bicarbonate? a. 10.30 b. 3.70 c. 7.00 d. 14.00

a. pKa = −log(Ka) = log(4.8 × 10−11) = 10.32.