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Homework 3
- What is the [OH–] in a 0.32 M HCl solution? What is the pH? HCl - > H+(aq) + Cl-(aq) [H+] = 0.32 M pH = - log [H+] pH = - log [0.32] pH = - [-0.49] pH = 0. pH + pOH = 14 pOH = 14 – pH pOH = 14 – 0. pOH = 13. pOH = - log [OH-]
- pOH = log [OH-] 10 - pOH^ = [OH-] 10 - 13.5^ = [OH-] 3.2 x 10 -^14 = [OH-]
- You are a lab technician in a biochemistry lab and asked to prepare a buffered solution at pH 9.3? Which buffer would you choose from the list?
- Rank the water solubility of the following compounds: From most soluble (most polar) to least soluble (least polar): c, b, a
- For the reaction at 25°C, the value of the standard free-energy change, ΔG°, is – 70.4 kJ. 2 NO(g) + O2(g) à 2 NO2(g) ΔH° = – 114.1 kJ, ΔS° = – 146.5 J/ K i. Calculate the value of the equilibrium constant, Keq, for the reaction at 25°C. Converting the units first: ΔG° = – 70.4 kJ x 1000 J/ 1kJ = - 70,400 J T = 273 + 25 = 298 K Keq = e−ΔG°′/RT Keq = e−ΔG°′/RT = e−(−70,400 J · mol−1)/(8.314 J / K · mol)(298 K) = e28. = 2.19 x 10^12 ii. Indicate whether the value of ΔG°^ would become more negative, less negative, or remain unchanged as the temperature is increased. Justify your answer. Since ∆G° = ∆H° − T·∆S°, as the temperature increases, ∆G° would be less negative (or become positive) because ∆S° < 0.
- A solution is made by mixing 50 mL of 2.0 M K 2 HPO 4 and 25 mL of 2.0 M KH 2 PO 4. The solution is diluted to a final volume of 200 mL. What is the pH of the final solution? The pK corresponding to the equilibrium between H 2 PO 4 −^ (HA) and HPO 4 2−^ (A−) is 6. (Table 2-4).
Let X be the volume (L) of 3 .0 M solution of boric acid is added. X x 3 .0 moles / L = X· 3 .0 moles / L Sodium borate = 10 mM x 1 M / 1000 mM x 800 mL x 1 L / 1 000 mL = 0.008 moles pH = pKa + log [sodium borate] / [boric acid] 9 .75 = 9.24 (Table 2-4) + log [0.008 moles] / [ X· 3 .0 moles / L] 10 0.51^ = [0.008] / [3·X] 3.24 x 3·X = 0. 9.71·X = 0. X = 0.0008 23 L x 1000 mL / 1L = 0.8 23 mL
- Calculate the pH of a 2.0 L solution containing 10 mL of 5 M acetic acid and 10 mL of 1 M sodium acetate. Acetic acid … Ka = 1.74 × 10−5^ , pKa = 4.76 (Table 2-4) Volume of Solution = 2.0L Acetic acid volume = 10mL x 1L / 1000 mL = 0.01 L Sodium Acetate volume = 10 mL x 1L / 1000 mL = 0.01 L For Acetic acid: C1V1 = C2V C 1 x 2 .0 L = 5 x 0.01L C1 = 0.05 L / 2.0 L C1 = 0.025M For Sodium Acetate salt: C1V1 = C2V C 1 x 2.0 L = 1 x 0.01 L C1 = 0.01 L / 2.0 L C1 = 0.005M Calculate pH using Henderson - Hasselbalch equation: pH = pKa + log [salt] / [acid] pH = 4.7 6 + log (0.005/0.025) pH = 4.7 6 + log 0. pH = 4.7 6 - 0. pH = 4.
- A solution is made by mixing 1.0 mL of 1.0 M acetic acid (pK = 4.56, Ka = 1.74 x 10–^5 ) with one 500 mL of pure water. Calculate the pH of the resulting solution (assume the total volume is 1.0 L). Since the total volume is assumed to be 1.0 L, we will be adding 499 mL of water in addition to the 500mL to prepare the solution. (total pure water volume = 999 mL) CH 3 COOH ↔ CH 3 COO-(aq) + H+(aq) Molarity of acetic acid = 1.0 M Number of moles = 1.0 mL x 1 L/ 1000 mL x 1.0 M = 0.001 moles Moles CH 3 COOH ↔ CH 3 COO-(aq) + H+(aq) Initial 0.001 moles 0 0 Change 0.001 – x + x + x Ka = [CH 3 COO-] [H+] / [CH 3 COOH] 1.74 x 10–^5 = [x][x]/ [0.001 – x] 1.74 x 10–^5 = x^2 / 0.001 – x 1.74 x 10–^8 – 1.74 x 10–^5 x = x^2 1.74 x 10–^8 – 1.74 x 10–^5 x – x^2 = 0 Using the quadratic equation: X = (-b ± √$^ 2 − 4 )*) / 2a X = (1.74 x 10–^5 + +(− 1. 74 x 10 – 5 )^ 2 − 4 (− 1 )( 1.74 x 10–^8 )) / 2 (-1) = - 0.00014 ( A pH value cannot be negative so we will take the other positive quadratic value) X = (1.74 x 10–^5 - (^) +(− 1. 74 x 10 – 5 )^ 2 − 4 (− 1 )( 1.74 x 10–^8 )) / 2 (-1) = 0. X = 0. pH = - log [H+] pH = - log [0.00012 M] pH = - (-3.9) pH = 3.
- Identify the sequence of the following polypeptide.
(a) How to determine the pI of a peptide: Step 1: First write out the pKa values of the amino acids from low to high so in our case the pK’s of the ionizable side chains (Table 4-1) of the peptide with the sequence ATLDAK are: 9.87 (N-terminal (NH 3 +) of Ala) 3.90 (side chain of Asp) 10.54 (side chain of Lys) 2.16 (C-terminal (COO-) of Lys) Writing them in sequence from low to high: 2.16, 3.90, 9.87, 10. Step 2: Drop the pH below each of the pKa values (greater than the pKa value below it for example pH 3.0 was selected for pKa value of 3.90 which was above pKa value of 2.16) of the amino acids and determine the net charge of the peptide (bolded) for each pH value (in blue) PKA VALUES OF AMINO ACIDS IN PEPTIDE
PH VALUES 2.0 3.0 7.0 10.0 12.
CHARGE OF N-TERMINAL +1 +1 +1 0 0
CHARGE OF C-TERMINAL 0 - 1 - 1 - 1 - 1
CHARGE OF ASP SIDECHAIN 0 0 - 1 - 1 - 1
CHARGE OF LYS SIDECHAIN +1 +1 +1 +1 0
NET CHARGE OF THE PEPTIDE +2 +1 0 - 1 - 2
Step 3 : Calculate the pI by averaging the two pKa values that are just before and just after the zero net charge ( so the pKa values 3.90 and 9.87)
pI = (3. 90 + 9.87) / 2 = 6. 89 (b) The net charge at pH 7.0 is 0 (as drawn above).
- The following molecule is a: a) Amino acid b) Dipeptide c) Tetrapeptide d) Tripeptide e) polypeptide N-term C-term
