BICH 409 FINAL EXAM REVIEW QUESTIONS WITH
VERIFIED ANSWERS
Kw equation - Answers - Kw = [H+][OH-]
Keq - Answers - [H+][OH-]/H2O
concentration of H+ and OH- in pure water - Answers - H+ = OH- = 10^-7M
pH = yo - Answers - -log[H+]
[H+] = 10^-pH
pOH = - Answers - -log[OH-]
[OH-] = 10^-pOH
as pOH increases, hydroxide ion concentration __________. - Answers - decreases
pKw = - Answers - -logKw = pH + pOH
pH + pOH = ? - Answers - 14
strong electrolytes - Answers - NaCl, KCl, K2SO4, HCl, HNO3, H2SO4, NaOH, KOH
what is the difference between weak and strong electrolytes? - Answers - strong
electrolytes completely dissociate in H2O, weak electrolytes partially ionize in water
which of the strong electrolytes are salts? - Answers - NaCl, KCl, K2SO4
which of the strong electrolytes are acids? - Answers - HCl, HNO3, H2SO4
which of the strong electrolytes are bases? - Answers - NaOH, KOH
Keq (equilibrium constant) = - Answers - products over reactants
Ka (acid dissociation constant) = - Answers - Keq * [H2O]
([55.5M])
what is the pH and pOH of a solution containing 0.014M NaOH? - Answers - pOH = -
log[OH-] = 1.85
pH = 14 - pOH = 12.15
how do you convert 1mM to M? - Answers - 1mM = 0.001M
,how do you convert 1uM to M? - Answers - 1uM = 10^-6M
The pOH of lye is 0.40, calculate the hydrogen ion and hydroxide ion concentrations. -
Answers - pOH = -log[OH-]
[OH-] = 10^-pOH
[OH-] = 0.3979M
Kw = [H+][OH-]
10^-14M^2 = [H+][0.3979M]
[H+] = 2.51 x 10^-14M
Henderson-Hasselbach Equation - Answers - pH = pKa + log[CB/CA]
lactic acid has a pKa of 3.86. If the concentration of lactate is 2.42M and the
concentration of lactic acid is 0.15M what is the pH of the solution? - Answers - pH =
pKa + log[CB/CA]
pH = 5.07
calculate the ratio of the concentration of acetate and acetic acid required in a buffer
system of pH 5.1. The pKa of acetic acid is 4.76. - Answers - pH = pKa + log[CB/CA]
pH - pKa = log[CB/CA]
5.1 - 4.76 = log[CB/CA]
0.34 = log[CB/CA]
[CB/CA] = 10^0.34
ratio = 2.19/1
a solution is buffered at a pH 6.5 by a weak acid and its conjugate base. the
concentration of a weak acid is 0.332mM and the concentration of the conjugate base
equals 1.667mM. what is the pKa of this weak acid? - Answers - pH = pKa + log[CB/CA]
pKa = pH - log[CB/CA]
pKa = 6.5 - log[0.01667M/0.00332M]
pKa = 5.8
which form is ionized in weak acids? - Answers - the protonated form is nonionized and
the deprotonated form is ionized
which form is ionized in weak bases? - Answers - the protonated form is ionized and the
deprotonated form is nonionized
acetic acid has a pKa of 4.76. at what pH is the acetic acid 3/4 ionized? - Answers - pH
= pKa + log[CB/CA]
pH = 4.76 + log[.75/.25]
pH = 5.24
,the pKa of the ammonium ion is 9.25. at what pH is ammonia 75% ionized? - Answers -
pH = pKa + log[CB/CA]
pH = 9.25 + log[.25/.75]
pH = 8.77
calculate the pKa of a propionic acid solution, given that when the concentration of
propionic acid is 0.10M and the propionate ion concentration is 0.43M, the pH is 5.50. -
Answers - pKa = pH - log[CB/CA]
pKa = 5.5 - log[0.43/0.10]
pKa = 4.87
calculate the pKa of formic acid, given that when the concentration of formic acid is
0.1M and the formate ion concentration is 0.018M, the pH is 3.01. - Answers - pKa = pH
- log[CB/CA]
pKa = 3.01 - log[0.018/0.01]
pKa = 3.75
water derives all its special properties from its:
a. small degree of ionization
b. high dielectric constant
c. high boiling and melting points
d. cohesiveness and adhesiveness
e. polarity and hydrogen-bonding capacity - Answers - e. polarity and hydrogen bonding
capacity
which of the following statements about hydrogen bonds is true?
a. hydrogen bonds occur between nonpolar molecules
b. hydrogen bonds account for the high boiling point of water
c. a hydrogen bond can be as strong as a covalent bond
d. hydrogen bonds form between positively and negatively charged ions
e. water is a weak hydrogen bond donor, but a strong hydrogen bond acceptor -
Answers - b. hydrogen bonds account for the high boiling point of water
you want to study an enzyme reaction at pH 4.0. which of the following weak acids
would make the best buffer solution for this pH based on the given Ka values?
a. phosphoric acid, Ka = 7.3 x 10^-3
b. bicarbonate, Ka = 6.3 x 10^-11
c. H2PO4, Ka = 6.3 x 10^-8
d. acetic acid, Ka = 1.7 x 10^-5
e. lactic acid, Ka = 1.4 x 10^-4 - Answers - e. lactic acid, Ka = 1.4 x 10^-4
what is the pH of 10^-4M HCl?
a. 11
b. 4
c. -4
d. 0.0001
, e. none of these - Answers - b. 4
pH = -log(10^-4)
when the pH equals the pKa of a weak acid, then:
a. the weak acid becomes a strong acid
b. the concentrations of the weak acid and its conjugate base are equal
c. the weak acid is completely ionized
d. the weak acid is completely protonated
e. the net charge of the weak acid is zero - Answers - b. the concentrations of the weak
acid and its conjugate base are equal
in a typical eukaryotic cell the pH is usually around 7.40. what is the hydrogen ion
concentration in a typical eukaryotic cell? - Answers - [H+] = 10^-pH
[H+] = 10^-7.40
[H+] = 4 x 10^-8M
the extensive hydrogen-bonding network of liquid water molecules around a nonpolar
molecule is called:
a. the rate at which nonpolar solutes intercalate into solution
b. a clathrate
c. an amphiphile
d. an amphiphilic molecule
e. a micelle - Answers - b. a clathrate
the large dipole moment of water is due to:
a. water's bent geometry
b. the electronegativity of the oxygen atom of water
c. the polarity of the O-H bond
d. all are correct - Answers - d. all are correct
what are the values for temperature, pressure, concentrations of the reactants and
products, and the [H+] in the biochemist's standard state? - Answers - T = 25C = 298K
P = 1 atm
concentration of all reactants and products = 1M
water = 55.5M
[H+] = 1 x 10^-7M (pH 7.0)
what are the equations for Gibbs' free energy? - Answers - deltaG = deltaH - (T)(delta
S)
what does the magnitude of the standard free energy change tell you? - Answers - -
how far away from equilibrium you are under standard conditions
what does the magnitude of the free energy change tell you? - Answers - - how far away
from equilibrium you are under the given conditions
VERIFIED ANSWERS
Kw equation - Answers - Kw = [H+][OH-]
Keq - Answers - [H+][OH-]/H2O
concentration of H+ and OH- in pure water - Answers - H+ = OH- = 10^-7M
pH = yo - Answers - -log[H+]
[H+] = 10^-pH
pOH = - Answers - -log[OH-]
[OH-] = 10^-pOH
as pOH increases, hydroxide ion concentration __________. - Answers - decreases
pKw = - Answers - -logKw = pH + pOH
pH + pOH = ? - Answers - 14
strong electrolytes - Answers - NaCl, KCl, K2SO4, HCl, HNO3, H2SO4, NaOH, KOH
what is the difference between weak and strong electrolytes? - Answers - strong
electrolytes completely dissociate in H2O, weak electrolytes partially ionize in water
which of the strong electrolytes are salts? - Answers - NaCl, KCl, K2SO4
which of the strong electrolytes are acids? - Answers - HCl, HNO3, H2SO4
which of the strong electrolytes are bases? - Answers - NaOH, KOH
Keq (equilibrium constant) = - Answers - products over reactants
Ka (acid dissociation constant) = - Answers - Keq * [H2O]
([55.5M])
what is the pH and pOH of a solution containing 0.014M NaOH? - Answers - pOH = -
log[OH-] = 1.85
pH = 14 - pOH = 12.15
how do you convert 1mM to M? - Answers - 1mM = 0.001M
,how do you convert 1uM to M? - Answers - 1uM = 10^-6M
The pOH of lye is 0.40, calculate the hydrogen ion and hydroxide ion concentrations. -
Answers - pOH = -log[OH-]
[OH-] = 10^-pOH
[OH-] = 0.3979M
Kw = [H+][OH-]
10^-14M^2 = [H+][0.3979M]
[H+] = 2.51 x 10^-14M
Henderson-Hasselbach Equation - Answers - pH = pKa + log[CB/CA]
lactic acid has a pKa of 3.86. If the concentration of lactate is 2.42M and the
concentration of lactic acid is 0.15M what is the pH of the solution? - Answers - pH =
pKa + log[CB/CA]
pH = 5.07
calculate the ratio of the concentration of acetate and acetic acid required in a buffer
system of pH 5.1. The pKa of acetic acid is 4.76. - Answers - pH = pKa + log[CB/CA]
pH - pKa = log[CB/CA]
5.1 - 4.76 = log[CB/CA]
0.34 = log[CB/CA]
[CB/CA] = 10^0.34
ratio = 2.19/1
a solution is buffered at a pH 6.5 by a weak acid and its conjugate base. the
concentration of a weak acid is 0.332mM and the concentration of the conjugate base
equals 1.667mM. what is the pKa of this weak acid? - Answers - pH = pKa + log[CB/CA]
pKa = pH - log[CB/CA]
pKa = 6.5 - log[0.01667M/0.00332M]
pKa = 5.8
which form is ionized in weak acids? - Answers - the protonated form is nonionized and
the deprotonated form is ionized
which form is ionized in weak bases? - Answers - the protonated form is ionized and the
deprotonated form is nonionized
acetic acid has a pKa of 4.76. at what pH is the acetic acid 3/4 ionized? - Answers - pH
= pKa + log[CB/CA]
pH = 4.76 + log[.75/.25]
pH = 5.24
,the pKa of the ammonium ion is 9.25. at what pH is ammonia 75% ionized? - Answers -
pH = pKa + log[CB/CA]
pH = 9.25 + log[.25/.75]
pH = 8.77
calculate the pKa of a propionic acid solution, given that when the concentration of
propionic acid is 0.10M and the propionate ion concentration is 0.43M, the pH is 5.50. -
Answers - pKa = pH - log[CB/CA]
pKa = 5.5 - log[0.43/0.10]
pKa = 4.87
calculate the pKa of formic acid, given that when the concentration of formic acid is
0.1M and the formate ion concentration is 0.018M, the pH is 3.01. - Answers - pKa = pH
- log[CB/CA]
pKa = 3.01 - log[0.018/0.01]
pKa = 3.75
water derives all its special properties from its:
a. small degree of ionization
b. high dielectric constant
c. high boiling and melting points
d. cohesiveness and adhesiveness
e. polarity and hydrogen-bonding capacity - Answers - e. polarity and hydrogen bonding
capacity
which of the following statements about hydrogen bonds is true?
a. hydrogen bonds occur between nonpolar molecules
b. hydrogen bonds account for the high boiling point of water
c. a hydrogen bond can be as strong as a covalent bond
d. hydrogen bonds form between positively and negatively charged ions
e. water is a weak hydrogen bond donor, but a strong hydrogen bond acceptor -
Answers - b. hydrogen bonds account for the high boiling point of water
you want to study an enzyme reaction at pH 4.0. which of the following weak acids
would make the best buffer solution for this pH based on the given Ka values?
a. phosphoric acid, Ka = 7.3 x 10^-3
b. bicarbonate, Ka = 6.3 x 10^-11
c. H2PO4, Ka = 6.3 x 10^-8
d. acetic acid, Ka = 1.7 x 10^-5
e. lactic acid, Ka = 1.4 x 10^-4 - Answers - e. lactic acid, Ka = 1.4 x 10^-4
what is the pH of 10^-4M HCl?
a. 11
b. 4
c. -4
d. 0.0001
, e. none of these - Answers - b. 4
pH = -log(10^-4)
when the pH equals the pKa of a weak acid, then:
a. the weak acid becomes a strong acid
b. the concentrations of the weak acid and its conjugate base are equal
c. the weak acid is completely ionized
d. the weak acid is completely protonated
e. the net charge of the weak acid is zero - Answers - b. the concentrations of the weak
acid and its conjugate base are equal
in a typical eukaryotic cell the pH is usually around 7.40. what is the hydrogen ion
concentration in a typical eukaryotic cell? - Answers - [H+] = 10^-pH
[H+] = 10^-7.40
[H+] = 4 x 10^-8M
the extensive hydrogen-bonding network of liquid water molecules around a nonpolar
molecule is called:
a. the rate at which nonpolar solutes intercalate into solution
b. a clathrate
c. an amphiphile
d. an amphiphilic molecule
e. a micelle - Answers - b. a clathrate
the large dipole moment of water is due to:
a. water's bent geometry
b. the electronegativity of the oxygen atom of water
c. the polarity of the O-H bond
d. all are correct - Answers - d. all are correct
what are the values for temperature, pressure, concentrations of the reactants and
products, and the [H+] in the biochemist's standard state? - Answers - T = 25C = 298K
P = 1 atm
concentration of all reactants and products = 1M
water = 55.5M
[H+] = 1 x 10^-7M (pH 7.0)
what are the equations for Gibbs' free energy? - Answers - deltaG = deltaH - (T)(delta
S)
what does the magnitude of the standard free energy change tell you? - Answers - -
how far away from equilibrium you are under standard conditions
what does the magnitude of the free energy change tell you? - Answers - - how far away
from equilibrium you are under the given conditions
