For the equilibrium below, pKa₁ = 7.9 and pKa2 = 10.1. At what pH is [HA] equal to [A²-]?
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![For the equilibrium below, pKa₁ = 7.9 and pKa2 = 10.1. At what pH is [HA] equal to
[A²-]?
H₂A <=> H+ + HA <=> 2H+ + A²-
OpH = 7.9
pH < 7.9
OpH > 10.1
pH = 10.1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F691d833d-667d-47e7-89c1-c8c6926199ae%2F2a220091-ae8c-44ab-a9df-f7e1930a513b%2F7fxc1rb_processed.jpeg&w=3840&q=75)
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- A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) = H*(aq) + A¯(aq) The equilibrium concentrations of the reactants and products are [HA] = 0.290 M, H+| = 3.00 × 10¬4 M, and [A-] = 3.00 x 10-4 M. Calculate the value of pKa for the acid HA. pKaA patient presents to the emergency department in a critical condition. The patient’s blood was tested and the concentration of carbonic acid (H2CO3) was found to be 4.0×10-4 M and bicarbonate ion (HCO3-) 3.062×10-3 M. Ka = 4.3 × 10-7 H2CO3(aq) ⇌ HCO3-(aq) + H+(aq) Show all calculations including equation(s) used. Calculate the pH of the patients’ blood. Based on your answer would you say the patient is suffering from Acidosis or Alkalosis?A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) = H+ (aq) + A¯(aq) The equilibrium concentrations of the reactants and products are [HA] = 0.220 M, [H+] = 3.00 × 10−4 M, and [A¯] = 3.00 × 10−4 M. Calculate the value of pKa for the acid HA. pKa =
- Which of the equations below best represents how excess acid in the blood plasma can be removed in the lungs by breathing? H+ (aq) + 0. 2(g) OH + H₂ CO (aq) OH+ + O (aq) 2(g) → H, CO. H + O₂ (aq) 2(g) 3(aq) 3(aq) → H₂O(g) + CO2(g) + CO2(g) → H₂O H, CO3(aq) → HCO, OH t + HCO3(aq) → H₂ CO3(g) → H₂O(g) (aq) → H, CO3(aq) 3 (aq) + CO2(g) H₂O(g) + CO2(g)Draw dipeptide Arg-Thr at pH 6.0. Circle the six atoms that are restricted in one plane. Calculate its approximate pI assuming pKa for NH3+ is 10.0, pKa for COOH is 3.0, PKR for Arg is 12.0Carbonic acid, H₂CO₃, undergoes the following acid base equilibria: H₂CO₃,⇌H⁺(aq) + HCO₃⁻(aq) ⇌ H⁺(aq) + CO₃²⁻(aq). The pKa values for the first and second steps are 3.6 and 6.3, respectively. The pH of human blood is 7.37 and this is tightly regulated. Which of the following statements most accurately describes the concentrations of the different species in the above reaction. A) [H₂CO₃(aq)] is much larger than the other species B) [H₂CO₃(aq)] and [HCO₃⁻(aq)] are much larger than the other species C) [HCO₃⁻(aq)] and [CO₃²⁻(aq)] are much larger than the other species D) [CO₃²⁻(aq)] is much larger than the other species
- Quinine ( C20 H24 N2 O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, pK, = 5.1 and pK, = 9.7 ( pKp = – log Kp). Only 1 g quinine will dissolve in 1920.0 mL of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction | Q+ H2O= QH+ + OH- described by pK, where Q = quinine. pH =The acidity of the stomach is maintained by the H*/K* ATPase in parietal cells of the gastric mucosa. These cells have an internal pH = 7.4. The H*/K* ATPase transports H* across the cellular membrane into the stomach where pH = 0.8. (a) Given a membrane potential of 65 mV (inside negative), calculate AG in kJ/mol for the transport of H* into the stomach at 37 °C. Show your work. (b) The hydrolysis of ATP is directly coupled to H* transport and the stoichiometry of the transport reaction is such that 1 mol of ATP is hydrolyzed for every 1 mol of H* transported into the stomach. ATP is hydrolyzed by the ATPase on the inside of the cell. Under the conditions given in part (a) and assuming the steady-state concentrations of ATP, ADP and P₁ given below, is the transport of H* in to the stomach thermodynamically favorable at 37 °C? Show the calculation that supports your answer. Steady-state intracellular concentrations: ATP = 4.2 mM ADP = 360 μM Pi = 5.4 mM For ATP hydrolysis at 37 °C: ATP…The reaction quotient is Q=1.6×10-26 Part B What pH is needed to produce this value of Q if the concentration and pressure values are [Br2]=2.50×10−4M , [Br−]=11.65M, [SO42−]=9.50M, and PSO2=3.50×10−5atm ? Express your answer numerically to two decimal places.
- The skeletal structures of the two amino acids, glycine and lysine, are given below along with the values of the relevant acid dissociation constants (pKa). NH,* PK, - 10.79 - HạN*CH2CO,¯ S pK¸=2.35 (CH2)4 - H3N*CHCO,- pK,=9,78 lysine (Lys) glycine (Gly) pK, = 9.18 pK, - 2.16 For an aqueous solution of glycine alone, calculate the value of pH at which the ratio of the concentration of neutral glycine zwitterions to the concentration of protonated cation is 102. On your under each of the following conditions. In the blank, provide the total charge of the dipeptide. (Example: If the charge is two plus write the answer simply as 2, if the charge is negative two write the answer as -2). draw the skeletal structure of the dipeptide, Lys-Gly, when it is solvated in an aque us solution i. pH= 1 ii. pH= 12In a 0.1000 M acetic acid solution at 25 degrees celsius , the acid ionizes to the extent of about 1.34 %. Since each molecule of acetic acid which ionizes produces 1 H+ ion and 1 C2H3O2- ion, the concentration in the solution are: HC2H3O2 < -----------> H+ + C2H3O2-Example The reference range for blood pH is 7.35 – 7.42. What is this range expressed as [H+] in nmol l-1? Calculation: pH = - log [H+] 7.35 = - log [H+] [H+] = antilog -7.35 = 4.47 x 10-8 mol.l-1 = 44.7nmol.l-1 similarly: pH 7.42 = 38.0 nmol.l-1 Range [H+] = 38 – 44.7 nmol.l-1 If the blood pH decreases in an acidosis from 7.42 to 7.15, what is the change in [H+] in nmol.l-1? so this is example on the worksheet, but i still understand how to answer the question.
