Biochem Chapter 2 Learning Objectives
How to find pKa
-logka
Identifying the Conjugate Base Which is the conjugate
(A) second (b) first (c) first (d) second
What is the H+ concentration of a solution with pH of (a) 3.82; (b) 6.52; (c) 11.11?
(a) 1.51 x 10^-4 (b) 3.20 x 10^-7 (c) 7.76 x 10^-12
A buffer contains 0.010 mol of lactic acid (pKa = 3.86) and 0.050 mol of sodium lactate per liter. (a) Calculate the pH of the buffer. (b) Calculate the change in pH when 5 mL of 0.5 M HCl is added to 1 L of the buffer. (c) What pH change would you expect if you added the same quantity of HCl to 1 L of pure water?
(a) 4.56 (b) ) Strong acids ionize completely, so 0.005 L 0.5 mol/L 0.002 mol of H is added. The added acid will convert some of the salt form to the acid form. Thus, the final pH is pH 3.86 log [(0.050 0.0025)/(0.010 0.0025)] 3.86 0.58 4.44 The change in pH 4.56 4.44 0.12, which rounds to 0.1 pH unit (c) HCl completely dissociates. So, when 5 mL of 0.5 M HCl is added to 1 L of water, [H] (0.002 mol)/(1 L) 0.002 mol/L 0.002 M pH log 0.002 2.7 The pH of pure water is 7.0, so the change in pH 7.0 - 2.7 4.3, which rounds to 4 pH units.
What is the pH of a solution that has an H+ concentration of (a) 1.75 × 10-5 mol/L; (b) 6.50 × 10-10 mol/L; (c) 1.0 × 10-4 mol/L; (d) 1.50 × 10-5 mol/L?
(a) 4.8 (b) 9.2 (c) 4 (d) 4.82 A pH to one decimal place (like 5.2) corresponds to a concentration known to one significant figure
Calculate the pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pKa = 4.76) of (a) 2:1; (b) 1:3; (c) 5:1; (d) 1:1; (e) 1:10.
(a) 5.06 (b) 4.28 (c) 5.46 (d) 4.76 (e) 3.76
The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in the protonated form (—NH+3) or as the free base (—NH2), because of the reversible equilibrium R—NH+3⇌ R—NH2 +H+ (a) In what pH range can glycine be used as an effective buffer due to its amino group? (b) In a 0.1 M solution of glycine at pH 9.0, what fraction of glycine has its amino group in the —NH+3 form? (c) How much 5 M KOH must be added to 1.0 L of 0.1 M glycine at pH 9.0 to bring its pH to exactly 10.0? (d) When 99% of the glycine is in its —NH+3 form, what is the numerical relation between the pH of the solution and the pKa of the amino group?
(a) 8.6, 10.6 (b) 9 = 9.6 + log[base]/[acid] -0.6 = log [NH2]/[NH3+] 0.25 = [NH2]/[NH3+] or, 0.25/[1 + 0.25] x 100 = 20% of total amino group is in the form of NH2 (c) we know that the amino group is about 1/5, or 20%, deprotonated at pH 9.0. Thus, in moving from pH 9.0 to pH 9.6 (at which, by definition, the amino group is 50% deprotonated), 30%, or 0.3, of the glycine is titrated. We can now calculate from the Henderson-Hasselbalch equation the percentage protonation at pH 10.0: This ratio indicates that glycine is 5/7, or 71%, deprotonated at pH 10.0, an additional 21%, or 0.21, deprotonation above that (50%, or 0.5) at the pKa. Thus, the total fractional deprotonation in moving from pH 9.0 to 10.0 is 0.30 0.21 0.51, which corresponds to 0.51 0.1 mol 0.05 mol of KOH Thus, the volume of 5 M KOH solution required is (0.5 mol)/(5 mol/L) 0.01 L, or 10 mL (d) [NH2] = 1-0.99 = 0.01 [HA] = 0.99 Feeding the value in the above equation we get, pH = 9.6 + log [0.01]/[0.99] = 7.6 Thus the pH = 7.6
(a) Under such conditions, pCO2 in the air space of the lungs rises above normal. Qualitatively explain the effect of these procedures on the blood pH. (b) A common practice of competitive short-distance runners is to breathe rapidly and deeply (hyperventilate) for about half a minute to remove CO2 from their lungs just before the race begins. Blood pH may rise to 7.60. Explain why the blood pH increases. (c) During a short-distance run, the muscles produce a large amount of lactic acid (CH2CH(OH)COOH; K2 = 1.38 × 10-4M) from their glucose stores. Why might hyperventilation before a dash be useful?
(a) The air in the bag becomes enriched in CO2, and inhaling this air increases the CO2 concentration in the body and blood and decreases blood pH. Blood pH is controlled by the carbon dioxide-bicarbonate buffer system, as shown by the net equation CO2 + H2O = H+ HCO3 During hypoventilation, the concentration of CO2 in the lungs and arterial blood increases, driving the equilibrium to the right and raising the [H]; that is, the pH is lowered. (b) During hyperventilation, the concentration of CO2 in the lungs and arterial blood falls. This drives the equilibrium to the left, which requires the consumption of hydrogen ions, reducing [H] and increasing pH. Hyperventilation, the rapid breathing sometimes elicited by stress or anxiety, tips the normal balance of O2 breathed in and CO2 breathed out in favor of too much CO2 breathed out, raising the blood pH to 7.45 or higher. (c) ) Lactate is a moderately strong acid (pKa 3.86) that completely dissociates under physiological conditions. This lowers the pH of the blood and muscle tissue. Hyperventilation is useful because it removes hydrogen ions, raising the pH of the blood and tissues in anticipation of the acid buildup.
(a) Does a strong acid have a greater or lesser tendency to lose its proton than a weak acid? (b) Does the strong acid have a higher or lower Ka than the weak acid? (c) Does the strong acid have a higher or lower pKa than the weak acid?
(a) greater tendency (b) higher ka (c) lower pka
The strongly polar, hydrogen-bonding properties of water make it an excellent solvent for ionic (charged) species. By contrast, nonionized, nonpolar organic molecules, such as benzene, are relatively insoluble in water. In principle, the aqueous solubility of any organic acid or base can be increased by converting the molecules to charged species. For example, the solubility of benzoic acid in water is low. The addition of sodium bicarbonate to a mixture of water and benzoic acid raises the pH and deprotonates the benzoic acid to form benzoate ion, which is quite soluble in water.
(a) hc1 (b) naoh (c) naoh
Alanine is a diprotic acid that can undergo two dissociation reactions (see Table 3-1 for pKa values). (a) Given the structure of the partially protonated form (or zwitterion; see Fig. 3-9) below, draw the chemical structures of the other two forms of alanine that predominate in aqueous solution: the fully protonated form and the fully deprotonated form. Of the three possible forms of alanine, which would be present at the highest concentration in solutions of the following pH: (b) 1.0; (c) 6.2; (d) 8.02; (e) 11.9. Explain your answers in terms of pH relative to the two pKa values.
(b) At pH 1.0, 1.3 pH units below the pKa of the carboxyl group, more than 90% of the carboxyl groups are protonated, and protonated amino groups predominate by a factor of more than 107. (c) At pH 6.2 the zwitterion predominates. This is 4 pH units above the pKa of the carboxyl group, so the vast majority of carboxyl groups are deprotonated. It is 3.5 pH units below the pKa of the amino group, so the vast majority of amino groups are protonated. (d) At pH 8.02 the zwitterion still predominates. The carboxyl groups are deprotonated and, with the pH still 1.6 units below the pKa of the amino group, the vast majority of amino groups are protonated. (e) At pH 11.9, 2.2 pH units above the pKa of the amino group, the vast majority of amino groups are deprotonated; and the carboxyl groups, at 9.6 pH units above their pKa, remain deprotonated
hydrophobic interactions
-The forces that hold the nonpolar regions of the molecules together - results from the system's achieving the greatest thermodynamic stability by minimizing the number of ordered water molecules required to surround hydrophobic portions of the solute molecules -By clustering together in micelles, the fatty acid molecules expose the smallest possible hydrophobic surface area to the water, and fewer water molecules are required in the shell of ordered water
Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 M HCl to a final volume of 100 mL with H2O.
-log(.075 = 1.12
What is the pH of a solution that contains 0.20 M sodium acetate and 0.60 M acetic acid (pKa = 4.76)?
4.28
Define: hypertonic
A solution with higher osmolarity than that of the cytosol; the cell shrinks as water moves out.
Define: hypotonic
A solution with lower osmolarity than the cytosol cell swells as water enters.
Which of these compounds would be the best buffer at pH 5.0: formic acid (pKa = 3.8), acetic acid (pKa = 4.76), or ethylamine (pKa = 9.0)? Briefly justify your answer.
ACETIC ACID, CLOSEST TO 5
If the ATP-binding site of an enzyme is buried in the interior of the enzyme, in a hydrophobic environment, is the ionic interaction between enzyme and substrate stronger or weaker than that same interaction would be on the surface of the enzyme, exposed to water? Why?
ATP site on the inside of the enzyme is stronger because the ionic attractive force is perpendicular to the inverse of the dielectric constant. And that the hydrophobic "solvent" has a lower dielectric constant than a polar solvent. Substances with high dielectric constants break down more easily when subjected to intense electric fields LOOK AT BOOK ANSWER
What is a buffer?
Buffers are aqueous systems that tend to resist changes in pH when small amounts of acid or base are added.
Define: osmosis
Bulk flow of water through a semipermeable membrane into another aqueous compartment containing solute at a higher concentration
What are the three reversible equilibria in the sodium-bicarbonate buffer system?
CO₂ (d) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻ When H+ is added to blood as it passes through the tissues, reaction 1 in Figure 2-21 proceeds toward a new equilibrium, in which [H2CO3] is increased. This in turn increases [CO2(d)] in the blood (reaction 2) and thus increases the partial pressure of CO2(g) in the air space of the lungs (reaction 3); the extra CO2 is exhaled. When H+ is lost from the blood, the opposite events occur: more H2CO3 dissociates into H+ and HCO−3 and thus more CO2(g) from the lungs dissolves in blood plasma.
Define amphipathic
Compounds that contain regions that are polar and regions that are nonpolar
Effect of Holding One's Breath on Blood pH The pH of the extracellular fluid is buffered by the bicarbonate/carbonic acid system. Holding your breath can increase the concentration of CO2(g) in the blood. What effect might this have on the pH of the extracellular fluid? Explain by showing the relevant equilibrium equation(s) for this buffer system.
Dissolving more CO2 in the blood increases [H] in blood and extracellular fluids, lowering pH: CO2(d) + H2O ⇌ H2CO3 ⇌ H+ HCO3
Why does untreated diabetes cause acidosis
Do to no being able to use glucose as fuel, tissues use stored fatty acids as their primary fuel. This dependence on fatty acids results in the accumulation of high concentrations of two carboxylic acids causing the body to go into acidosis In individuals with untreated diabetes mellitus disrupts the uptake of glucose from blood into the tissues and forces the tissues to use stored fatty acids as their primary fuel. dependence on fatty acids results in the accumulation of high concentrations of two carboxylic acids, β-hydroxybutyric acid and acetoacetic acid (a combined blood plasma level of 90 mg/100 mL, compared with <3 mg/100 mL in control (healthy) individuals; urinary excretion of 5 g/24 hr, compared with <125 mg/24 hr in controls). Dissociation of these acids lowers the pH of blood plasma to less than 7.35, causing acidosis
Explain why ethanol (CH3CH2OH) is more soluble in water than is ethane (CH3CH3).
Ethanol contains an alcohol group which is polar while ethan just contains an alkyl group which is nonpolar
Calculate the cocentrations of acetic acid (pKa = 4.76) and sodium acetate necessary to prepare a 0.2 M buffer solution at pH 5.0.
First, calculate the required ratio of conjugate base to acid. pH pKa log ([acetate]/[acetic acid]) log ([acetate]/[acetic acid]) pH pKa 5.0 4.76 0.24 [acetate]/[acetic acid] 100.24 1.7 [acetate]/[acetate + acetic acid] 1.7/2.7 0.63 (two significant figures) Thus, 63% of the 0.2 M buffer is acetate and 27% is acetic acid. So at pH 5.0 the buffer has 0.13 M acetate and 0.07 M acetic acid.
In a typical analysis, 15 mL of an aqueous solution containing an unknown amount of acetylcholine had a pH of 7.65. When incubated with acetylcholinesterase, the pH of the solution decreased to 6.87. Assuming there was no buffer in the assay mixture, determine the number of moles of acetylcholine in the 15 mL sample.
Given that pH = -log[H+], we can calculate [H+] at the beginning and at the end of the reaction. At pH 7.65, log[H+] = -7.65, so [H+] = 10-7.65 = 2.24x10-8 M At pH 6.87, log[H+] = -6.87, so [H+] = 10-6.87 = 1.35x10-7 M The difference in [H+] is 1.35x10-7 M - 2.24x10-8 M = 1.13x10-7 M. Since the reaction is quantitative, the same amount of acetylcholine is initially present. (0.015L)(1.13x10-7 M) = 1.7x10-9 mol of acetycholine
Which of the following aqueous solutions has the lowest pH: 0.1 M HCl; 0.1 M acetic acid (pKa = 4.86); 0.1 M formic acid (pKa = 3.75)?
HC1 has the lowest ph
The interactions between biomolecules are often stabilized by weak interactions such as hydrogen bonds. How might this be an advantage to the organism?
Hydrogen bonds keep both the strands of a double helix together. Sometimes, the hydrogen bonds between the strands need to be broken to access the genetic code in the dna by replication enzymes. A replication bubble is formed. As the hydrogen bonds are weak, it is easier to break them without heat and much energy. If the bonds are not hydrogen bonds, then it would be impossible to separate the strands of DNA Bio-molecular interactions generally need to be reversible; weak interactions allow reversibility
Why does ice float in water?
In ice, each water molecule forms four hydrogen bonds forming a crystal lattice as compared to liquid water where each water molecule forms 3.4 hydrogen bonds each. This crystal lattice structure makes ice less dense than liquid water
draw structure of ice; why less dense than liquid h20
In ice, each water molecule forms four hydrogen bonds, the maximum possible for a water molecule, creating a regular crystal lattice. By contrast, in liquid water at room temperature and atmospheric pressure, each water molecule hydrogen-bonds with an average of 3.4 other water molecules. This crystal lattice structure makes ice less dense than liquid water, and thus ice floats on liquid water draw figure 2-2
List bond strength from strongest to weakest (intra/intermolecular)
Intramolecular: covalent bonding (network) ionic bonding covalent bonding (molecule) intermolecular: hydrogen bonding dipole-dipole ion-dipole
What is the pH of a solution containing 0.12 mol/L of NH4Cl and 0.03 mol/L of NaOH (pKa of pKa of NH + 4 / NH 3 is 9.25)?
NH4+Cl- + NaOH ---> NH3 + H2O + NaCl 0.12 0.03 0 initial 0.09 0 0.03 final it forms bufferr. pH= pKa + log {[NH3]/ [NH4Cl]} pH = 9 .25 + log (0.03/0.09) =9.25 + (-0.48) = 8.77 Answer: 8.77 the 0.12 M is equal to the [NH3] and [NH4]. the 0.03 NaOH is a base and will be a base in the solution, so you have to subtract it from the acid concentration. pH = 9.25 + log [0.03]/[0.09] = 8.77
Define: isotonic
Solutions that are similar in osmolarity to that of the cytosol
You have been observing an insect that defends itself from enemies by secreting a caustic liquid. Analysis of the liquid shows it to have a total concentration of formate plus formic acid (Ka = 1.8 × 10-4) of 1.45 M; the concentration of formate ion is 0.015 M. What is the pH of the secretion?
Solve the Henderson-Hasselbalch equation for pH. pH pKa log ([conjugate base]/[acid]) Given the Ka of formic acid (Ka 1.8 104 ), you can calculate pKa as log Ka 3.7. If the concentration of formate formic acid 1.45 M and the concentration of formate is 0.015 M, then the concentration of formic acid is 1.45 M 0.015 M 1.435 M. log ([formate]/[formic acid]) log (0.015/1.435) 2.0 pH 3.7 2.0 1.7 (two significant figures)
Define: micelle
Stable structures of amphipathic compounds in water
In a hospital laboratory, a 10.0 mL sample of gastric juice, obtained several hours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of NaOH was required. The patient's stomach contained no ingested food or drink; thus assume that no buffers were present. What was the pH of the gastric juice?
The equation to use here is M1V1 =M2V2. (0.010 L)x = (0.0072)(0.1) x = 0.072 M The pH of this concentration can be found by using pH =-log[H] where [H] is the concentration of the acid. pH = -log(0.072) = 1.14 GASTRIC = ACID HCl + NaOH -------> NaCl + H2O Volume of gastric juice (HCl) = 10.0 mL = 0.01 L letthe molarity of HCl = M molarity of NaOH = 0.1M volume of NaOH = 7.2 mL ∴ Number of moles of NaOH = 0.1 M * 0.0072L =0.00072 mol In the balancedequation 1 mol of HCl is neutralized by 1 mol of NaOH . ∴ Number of molesof HCl neutralized = 0.00072mol ∴ M * 0.01L = 0.00072 mol Molarity of HCl = M = 0.072 M [H+] = 0.072 M pH = - log[H+] = - log 0.072 M = 1.142
given 0.10 M solutions of acetic acid (pKa = 4.76) and sodium acetate, describe how you would go about preparing 1.0 L of 0.10 M acetate buffer of pH 4.00.
Use the Henderson-Hasselbalch equation to calculate the ratio [Ac]/[HAc] in the final buffer. pH pKa log ([Ac]/[HAc]) log ([Ac]/[HAc]) pH pKa 4.00 4.76 0.76 [Ac]/[HAc] 100.76 The fraction of the solution that is Ac [Ac]/[HAc Ac] 100.76/(1 100.76) 0.148, which must be rounded to 0.15 (two significant figures). Therefore, to make 1.0 L of acetate buffer, use 150 mL of sodium acetate and 850 mL of acetic acid.
Calculation of Blood pH from CO2 and Bicarbonate Levels Calculate the pH of a blood plasma sample with a total CO2 concentration of 26.9 mM and bicarbonate concentration of 25.6 mM. Recall from page 67 that the relevant pK2 of carbonic acid is 6.1.
Use the Henderson-Hasselbalch equation: pH pKa log ([bicarbonate]/[carbonic acid]) If total [CO2] 26.9 M and [bicarbonate] 25.6 M, then the concentration of carbonic acid is 26.9 M 25.6 M 1.3 M. pH 6.1 log (25.6/1.3) 7.4 (two significant figures) In blood, the total CO2 concentration measured is equal to the sum of the concentrations of the bicarbonate ion and the carbonic acid. Thus, in this problem, [H2CO3] + [HCO2-] = [CO2] [H2CO3] = [CO2] - [HCO2-] = 1.3mM
A compound has a pKa of 7.4. To 100 mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0 M hydrochloric acid. What is the pH of the resulting solution?
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. (0.03 L) (1 mol/L) = 0.03 moles HCl are added, so A- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
The components of poison ivy and poison oak that produce the characteristic itchy rash are catechols substituted with long-chain alkyl groups
Wash the area with soap, water, and baking soda (sodium bicarbonate). In order to neutralize its effect it would be better to treat or wash the hands with basic solutions which can be soap water or sodium bicarbonate(baking soda) because as soap water and sodium bicarbonate is basic in nature so it can neutralize the alcoholic group(acidic in nature)in catechols to form salt and water and the effect of poison would be reduced.
What are the unusual properties of water?
Water has a higher melting point, boiling point and heat of vaporization than most other common solvents because it has great internal cohesion from attractions between adjacent water molecules These orbitals describe a rough tetrahedron, with a hydrogen atom at each of two corners and unshared electron pairs at the other two corners (Fig. 2-1a). The H—O—H bond angle is 104.5°, slightly less than the 109.5° of a perfect tetrahedron because of crowding by the nonbonding orbitals of the oxygen atom. each hydrogen atom bears a partial positive charge (δ+), and the oxygen atom bears a partial negative charge equal in magnitude to the sum of the two partial positives (2δ−). As a result, there is an electrostatic attraction between the oxygen atom of one water molecule and the hydrogen of another (Fig. 2-1b), called a hydrogen bond. The sum of all the hydrogen bonds between H2O molecules confers great internal cohesion on liquid water.
Describe the ionization of water and the extent to which it occurs
Water ionizes into H+ and OH- to a small extent on it's own --Hydrogen bonding between water molecules makes the hydration of dissociating protons virtually instantaneous. The degree of ionization of water at equilibrium (Eqn 2-1) is small; at 25 °C only about two of every 109 molecules in pure water are ionized at any instant.
Define: Van der Waals forces
Weak attractions that occur between two dipoles of atoms causing the two nuclei to come closer together
Define: hydrophobic effect
When amphipathic compounds cluster together to present the smallest hydrophobic area to the aqueous solvent and the polar regions are arranged to maximize their interaction with the solvent
how does a buffer system work
Whenever H+ or OH− is added to a buffer, the result is a small change in the ratio of the relative concentrations of the weak acid and its anion and thus a small change in pH. The decrease in concentration of one component of the system is balanced exactly by an increase in the other. The sum of the buffer components does not change, only their ratio change
equation for ka
[h+][A-]/[HA] 1.74 × 10 − 5 M
Define: hydrogen bond
a type of weak chemical bond formed between the slightly positively charged hydrogen atoms of one molecule and the slightly negatively charged atoms of another.
(a) Write out the acid dissociation reaction for hydrochloric acid. (b) Calculate the pH of a solution of 5.0 × 10-4M HCl. (c) Write out the acid dissociation reaction for sodium hydroxide. (d) Calculate the pH of a solution of 7.0 × 10-5M NaOH.
a) HCl = H+ + Cl- [H+] = [HCl] = 5.0 x 10-4 M pH = -log[H+] = -log(5.0 x 10-4) = 3.3 b) NaOH = Na+ + OH- [OH-] = [NaOH] = 7.0 x 10-5 M pOH = -log[OH-] = -log(7.0 x 10-5) = 4.2 pH = 14 - pOH = 14 - 4.2 = 9.8
Aspirin is a weak acid with a pKa of 3.5 (the ionizable H is shown in red): It is absorbed into the blood through the cells lining the stomach and the small intestine. Absorption requires passage through the plasma membrane, the rate of which is determined by the polarity of the molecule: charged and highly polar molecules pass slowly, whereas neutral hydrophobic ones pass rapidly. The pH of the stomach contents is about 1.5, and the pH of the contents of the small intestine is about 6. Is more aspirin absorbed into the bloodstream from the stomach or from the small intestine? Clearly justify your choice.
at higher ph, aspirin becomes more deprotonated. Aspirin is better absorbed in the stomach because it offers a more acidic environment
when is buffer most effective
buffer is most effective in the range of 1 pH unit above and below its pKa (!!!!!!!)
Define: clathrates
crystalline compounds of nonpolar solutes and water
What is Kw and what is its value? (@ 25 C)
designates the product (55.5 M)(Keq), the ion product of water at 25 °C. kw= [h+][oh-]= (Keq)(55.5M) kw= [H+][oh-]= (55.5 M)(1.8 x 10^-16 M)= 1.0 x 10^-14 m^2
Define: bond dissociation energy
energy required to break a bond
The glass electrode used in commercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. To convert these responses to a pH reading, the electrode must be calibrated against standard solutions of known H+ concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4 · H2O; FW 138) and disodium hydrogen phosphate (Na2HPO4; FW 142) needed to prepare 1 L of a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M (see Fig. 2-16). See Problem 24 for the pKa values of phosphoric acid.
https://memberfiles.freewebs.com/71/70/84717071/documents/pob5e_solutions_ch02.pdf
One way to make vinegar (not the preferred way) is to prepare a solution of acetic acid, the sole acid component of vinegar, at the proper pH (see Fig. 2-15) and add appropriate flavoring agents. Acetic acid (Mr 60) is a liquid at 25 °C, with a density of 1.049 g/mL. Calculate the volume that must be added to distilled water to make 1 L of simulated vinegar
https://www.chegg.com/homework-help/questions-and-answers/one-way-make-vinegar-gourmet-choice-prepare-solution-acetic-acid-acid-component-vinegar-pr-q4125267
Define hydrophilic and hydrophobic
hydrophilic- high affinity to water (polar molecules that dissolve easily in water) (polar solvent: water) hydrophobic- repelling of water (nonpolar) (lipids/waxes) (nonpolar solvent: benzene/chloroform)
What interactions are involved when ionic, polar, nonpolar or amphipathic solutes are mixed with water
ionic: dissolve in water due to water stabilizing the + and - charges of ions. Water dissolves salts such as NaCl by hydrating and stabilizing the Na+ and Cl− ions, weakening the electrostatic interactions between them and thus counteracting their tendency to associate in a crystalline lattice (Fig. 2-6). Water also readily dissolves charged biomolecules, including compounds with functional groups such as ionized carboxylic acids (—COO−), protonated amines, and phosphate esters or anhydrides. Water replaces the solute-solute hydrogen bonds linking these biomolecules to each other with solute-water hydrogen bonds, thus screening the electrostatic interactions between solute molecules. polar: dissolve easily in water due to their hydrophilicity; They readily form between an electronegative atom (the hydrogen acceptor, usually oxygen or nitrogen) and a hydrogen atom covalently bonded to another electronegative atom (the hydrogen donor) in the same or another molecule nonpolar: do not dissolve well in water; The nonpolar nature of these gases and the decrease in entropy when they enter solution combine to make them very poorly soluble in water. addition to water may therefore result in a small gain of enthalpy; the breaking of hydrogen bonds between water molecules takes up energy from the system, requiring the input of energy from the surroundings. In addition to requiring this input of energy, dissolving hydrophobic compounds in water produces a measurable decrease in entropy. amphipathic: form micelles in water due to hydrophobic interactions. When an amphipathic compound is mixed with water, the polar, hydrophilic region interacts favorably with the water and tends to dissolve, but the nonpolar, hydrophobic region tends to avoid contact with the water (Fig. 2-7a). The nonpolar regions of the molecules cluster together to present the smallest hydrophobic area to the aqueous solvent, and the polar regions are arranged to maximize their interaction with the solvent (Fig. 2-7b), a phenomenon called the hydrophobic effect.
For a weak acid with a pKa of 6.0, calculate the ratio of conjugate base to acid at a pH of 5.0.
log(conjugate base/acid) = 5.0-6.0 = -1.0 (conjugate base/acid) = 10^-1.0 = 0.1
Constancy of pH is achieved primarily by biological buffers:
mixtures of weak acids and their conjugate bases.
What molar ratio of HPO 2 − 4 to H 3 PO − 4 in solution would produce a pH of 7.0? Phosphoric acid (H3PO4), a triprotic acid, has three pKa values: 2.14, 6.86, and 12.4. Hint: Only one of the pKa values is relevant here
pH pKa log ([conjugate base]/[acid]) log ([HPO4 2]/[H2PO4 ]) pH pKa 7.0 6.86 0.14 [HPO42]/[H2PO4 ] 100.14 1.38 1.4 (two significant figures)
What is the Henderson-Hasselbalch equation?
pH = pKa + log [A-]/[HA]
how do i find ph!
ph= -log (h+)
find h from ph
put ph here 10^-ph
Define: osmolarity
the concentration of a solution expressed as the total number of solute particles per liter. (ic)
isotonic
two different solutions that have the same concentration of solute in them Surrounded by an isotonic solution, a cell neither gains nor loses water
calculate pH or pOH at 25C given h+ or oh- concentration of a solution
use kw! kw= 1 x 10^-14 m^2
Define: colligative properties
vapor pressure, boiling point, melting point (freezing point) and osmotic pressure colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. the effect of solutes on all four properties has the same basis: the concentration of water is lower in solutions than in pure water.
