Acids and Bases because you are TERRRIBLE AT THEM

Question 1

What is the pH of a 0.010 M sodium hydroxide solution at 25°C?

Answer 1

12, You should immediately notice that choice D is the only option that is basic. To answer this question using math, however, first note that sodium hydroxide is a strong base that completely dissociates in aqueous solution. Therefore, the hydroxide ion concentration of this solution is also 0.010 M, or (in scientific notation) 1 x 10-2 M. Taking the negative logarithm of the hydroxide concentration gives a pOH of 2. Since pH + pOH = 14 (at 25ºC), we can calculate the pH by subtracting the pOH from 14, which yields pH = 14 - 2 = 12.

Question 2

Equation for pH

Answer 2

pH = -log [H+]

Question 3

Equation for pOH

Answer 3

pOH = -log [OH-]

Question 4

What will the pH of a solution with an H+ concentration of 10^-4 be?

Answer 4

A solution with an H+ concentration of 10-4 M will have a pH of 4

Question 5

What will the pOH of a solution with an OH concetration of 10^-9 be?

Answer 5

A solution with an OH- concentration of 10-9 M will have a pOH of 9.

Question 6

What equation relates pH and pOH?

Answer 6

pH + pOH = 14 at 25C

Question 7

Amphiprotic

Answer 7

Amphoteric (or amphiprotic) compounds can act as either an acid or a base, depending on the other reactants present. Additionally, some compounds may have multiple H atoms that can participate in acid-base chemistry. Such substances are known as polyprotic acids.

Question 8

Which of the following in amphiprotic? Acetic acid, HC2H3O2 Sodium acetate, NaC2H3O2 Sodium bicarbonate, NaHCO3 Sodium carbonate Na2Co3

Answer 8

Sodium bicarbonate, NaHCO3 The prefix “amphi-“ means “both.” Therefore, an amphiprotic species is one that can act as both an acid or a base. Sodium bicarbonate (choice C) dissolves in aqueous solution to produce sodium ions and bicarbonate ions. The former ion is neither acidic nor basic, but the bicarbonate ion, HCO3-, can act as a Bronsted-Lowry acid by loss of a hydrogen ion and can act as a B-L base by accepting a hydrogen ion to form carbonic acid, H2CO3.

Question 9

A scientist wished to prepare a buffer for an experiment to be conducted at pH 9.7. Which of the following organic acids would be the best choice for this experiment? Acetic acid (pKa = 4.76) Carbonic acid (pKa = 6.35) Tricine (pKa = 8.05) Taurine (pKa = 9.06)

Answer 9

Taurine (pKa = 9.06) To construct the best possible buffer, we should choose the organic acid with the pKa closest to the pH at which the experiment will take place (9.7). This gives us answer choice D. Note that an ideal buffer should have a pKa within 1 pH unit of the expected experimental conditions.

Question 10

What is the normality of a 0.015 M solution of phosphoric acid?

Answer 10

H3PO4, there are 3 protons. SO multiply the molarity by 3 to get 0.045 N.

Question 11

What is the pH of a 0.010 M perchloric acid solution?

Answer 11

Perchloric acid is a strong acid that completely dissociates in aqueous solution, so the hydrogen ion concentration is 1.0 x 10-2 M. pH = -log[H+] = -log[10-2] = 2.

Question 12

If a buffer solution of 0.6 M H2CO3 (aq) and 1.2 M HCO3- (aq) is in equilibrium at a pH of 6.4, what is the pK­a of H2CO3?

Answer 12

The most straightforward way to solve this problem is to use the Henderson-Hasselbalch equation. With the pH and the values of the acid and conjugate base given in the question stem, we can use the Henderson-Hasselbalch equation to solve for the answer. We are given that [A-] = 1.2 M and [HA] = 0.6 M. pH = pKa + log([A-]/[HA]) 6.4 = pKa + log(1.2/0.6) 6.4 = pKa + log(2) 6.4 = pKa + 0.3 pKa= 6.1. To estimate log(2), remember that log(1) = 0 and log(10) = 1. So, log(2) will be between 0 and 1, and certainly closer to 0.

Question 13

What is the Henderson Hasselbach equation?

Answer 13

pH = pKa + log([A-]/[HA])

Question 14

Which of the following acids is expected to generate the strongest hydrogen bonding? HI HF HCl HBr

Answer 14

HF Hydrogen bonding is a unique type of intermolecular force that occurs when H is bonded to very electronegative elements like F, O or N. Fluorine is the most electronegative element on the periodic table. Of the answer choices, only hydrogen fluoride is capable of hydrogen bonding.

Question 15

The equivalence point of the titration of tolbutamide with NaOH was reached by adding 50 mL of NaOH. How much NaOH would have achieved its half equivalence point?

Answer 15

25mL

Question 16

The equivalence point of the titration of tolbutamide (pKa= 5.3) with NaOH was reached by adding 50 mL of NaOH. What happens to the pH of this solution once 25 mL are added in the titration?

Answer 16

If 50 mL of NaOH represents the volume of titrant required to reach the equivalence point, then 25 mL is the volume added at the half equivalence point. At this point, one half of the original tolbutamide present will have been converted to its conjugate base, and their concentrations will be equal. According to the Henderson-Hasselbach equation, when the values of protonated acid and conjugate base are equal, pH = pKa + log 1 = pKa + 0 = pKa. At the half-equivalence point, pH of solution equals the pKa of the analyte. According to paragraph 3, tolbutamide’s pKa is 5.3.

Question 17

At the half equivalence point in a titration, what is the relationship between pH and pKa?

Answer 17

They are equal according to the henderson hasselbach equation

Question 18

The rates of diffusion of four drugs were tested: acetazolamide (pKa = 7.2), sulfadiazine (pKa = 6.5), warfarin (pKa = 5.0), and cephalexin (pKa = 3.6). Which drug will have the strongest conjugate base?

Answer 18

Acetazolamide

Question 19

Ten moles of the monoprotic, weakly acidic medication aspirin were added to water to make one liter of solution. If the pH of the resulting solution was 5.9, what is the approximate Kb for the non-diffusible form of aspirin?

Answer 19

0.1 Since we are given pH in the question stem, we will not be able to find Kb immediately. Instead, we need to calculate Ka and solve for Kb from that value. The Ka for the dissociation of a generic acid HA can be written as Ka = [H+][A-]/[HA], where all concentrations are measured at equilibrium. In the solution of aspirin described, the initial concentration of drug is 10 M. Since only a small amount of this weak acid will dissociate, this value is a good approximation for our final equilibrium [HA]. Next, we must find the proton concentration. Remember, [H+] = 10-pH. Here, the pH of the solution is 5.9, so [H+] = 10-5.9 M ~ 10-6 M. Since each HA molecule dissociates into equal parts [H+] and [A-], our value for [A-] must be 10-6 M as well. Returning to the Ka expression, Ka = [(10-6 M )(10-6 M)] / (10 - 10-6 M). Remember, we can estimate that [HA] = 10 M, even though its true final value is 10 M - 10-6 M! [(10-6 M )( 10-6 M)] / (10 M) = 10-12 M / 10 M = 10-13 M In water at 25°C, Ka•Kb = 10-14. Given this, Kb = 10-14/Ka = 10 -14/10-13 = 10-1 = 0.1.

Question 20

True or False: [H+]=10^-pH

Answer 20

true

Question 21

Mathy math: [(10^-6 M )( 10^-6 M)] / (10 M) =

Answer 21

[(10-6 M )( 10-6 M)] / (10 M) = 10-12 M / 10 M = 10-13 M

Question 22

The gastric juices in the stomach have a pH of approximately 2. What is the hydroxide ion concentration in this solution?

Answer 22

10-12 M

Question 23

Based on the titration shown in Figure 1, what is the pKa of oxalic acid?

Answer 23

3.5

The pKa occurs at the half-equivalence point. As NaOH is added during the titration, the concentration of [A-] will increase and the concentration of [HA] will decrease. When pH = pKa, the concentration of [A-] = [HA], and this point is the half-equivalence point. On a titration curve, this is approximately at the midpoint of the initial horizontal part of the curve. In Figure 1, this is approximately pH 3.5.

Question 24

Nitrous acid and its conjugate base are combined in a 1 to 5 ratio to form a 100 mL buffer solution. If the Ka of nitrous acid is approximately 4.0×10-4, what is the pH of the buffer solution?

Answer 24

4.1

Question 25

True or false: If Rafael has Acid X (pKa = 3.2) and Acid Y (pKa = 4.7), the conjugate base of Acid Y must be stronger than Acid X’s conjugate base at standard conditions.

Answer 25

true

Question 26

Arrhenius acid

Answer 26

Acid forms H+

Question 27

Arrhenius base

Answer 27

Bases form OH-

Question 28

Bronsted Lowry acid and base

Answer 28

Acid is an H+ donor, Base is H+ acceptor

Question 29

Lewis acid and base

Answer 29

Acid is electron acceptor, base is an electron donor

Question 30

Strong or Weak:

HI

Answer 30

Strong acid

Question 31

Strong or weak:

HBr

Answer 31

Strong acid

Question 32

Strong or Weak:

HCl

Answer 32

Strong acid

Question 33

Strong or Weak:

HNO3

Answer 33

Strong acid

Question 34

Strong or Weak:

HNO2

Answer 34

Weak Acid

Question 35

Strong or Weak:

H2SO4

Answer 35

Strong Acid

Question 36

Strong or Weak:

HClO3

Answer 36

Strong acid

Question 37

Strong or Weak:

HClO4

Answer 37

Strong acid

Question 38

Strong or Weak:

HF

Answer 38

Weak acid

Question 39

Strong or Weak

H3PO4

Answer 39

weak acid

Question 40

Strong or Weak

H2S

Answer 40

Weak acid

Question 41

Strong or Weak

NaOH

Answer 41

Strong base

Question 42

Strong or Weak:

H-

Answer 42

Strong base

Question 43

Strong or Weak:

(CH3)CO-

Answer 43

Strong base

Question 44

pKa+pKb=

Answer 44

14

Question 45

pH+pOH

Answer 45

14

Question 46

How do you calculate pka?

Answer 46

-log(ka)

Question 47

What is the relationship between pH and pKa of a good buffer?

Answer 47

A buffer should have a pka that is +- 1 of the pH of the solution.