Suppose a biochemist has 10 mL of a 1.0 M solution of a compound with two ionizable groups at a pH of 7.80. She adds 10.0 mL of 1.0 M HCl, which changes the pH to 4.20. The pKą value of one of the groups is pK₁ = 4.10 and it is known that pK₂ is between 7 and 10. What is the exact value of pK₂? s pK₂ =

Suppose a biochemist has 10 mL of a 1.0 M solution of a compound with two ionizable groups at a pH of 7.80. She adds 10.0 mL of 1.0 M HCl, which changes the pH to 4.20. The pKą value of one of the groups is pK₁ = 4.10 and it is known that pK₂ is between 7 and 10. What is the exact value of pK₂? s pK₂ =

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter15: Equilibria Of Other Reaction Classes

Section: Chapter Questions

Problem 82E: The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they...

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The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, -NH2, and a carboxylic acid group, -CO2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid. CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3. (a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH. (b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the -NH2 and -CO2- groups.)
Consider the titration of a weak acid that has a pKa = 4.00. Suppose a chemist was going to perform a titration on 50.0 mL of 0.050 0 M of the weak acid using 0.500 M NaOH. (A) What would be the pH of the solution after 0.00 mL of 0.500 M NaOH has been added?(B) What would be the pH halfway to the equivalence point of the titration?(C) What would be the pH at the equivalence point in the titration? (D) What would be the pH of the solution after 6.00 mL of 0.500 M NaOH has been added? Use the following reasoning when solving this problem. Because at the equivalence point the moles of strong base (0.500 M NaOH) added as a titrant equal the moles of weak acid, HA (50.0 ml of 0.050 M) in the solution, we can start by calculating the volume of 0.500 M NaOH needed to reach the equivalence point. To do this we simply equate the moles of HA to moles of NaOH by using the dilution formula. i.e. MaVa = MbVb where a represents HA and b represents NaOH. To calculate volume of NaOH needed to reach…
Cacodylic acid (dimethylarsinic acid, HAsO2(CH3)2) is a common buffer used in biology to prepare and fix biological samples. It has a pKa value of 6.3.Calculate the pH of a buffer solution prepared by mixing 10.0 mL of 0.100 M NaOH with 20.0 mL of 0.150 M cacodylic acid.Express your answer to two decimal places. pH
Two solutions of an unknown slightly soluble salt, A(OH)2, were allowed to equilibrate—one at 25 °C and the other at 80 °C. A 15.00 mL aliquot of each solution is titrated with 0.200 M HCl. 6.37 mL of the acid is required to reach the endpoint of the titration at 25 °C, while 62.60 mL are required for the 80 °C solution. Assuming that the change in enthalpy is negligible over this temperature range, calculate ΔH
Which of the following acids (listed with pKa values) and their conjugate base should be used to form a buffer with a pH of 8.10? HC7H5O2, pKa = 4.19 HF, pKa = 3.46 HClO, pKa = 7.54 H2SO3, pKa = 1.77 HClO2, pKa = 1.96
Strong base is dissolved in 665 mL of 0.600 M weak acid (K₁ = 4.09 × 10−5) to make a buffer with a pH of 3.96. Assume that the volume remains constant when the base is added. HA(aq) + OH(aq) H,O(l) + A-(aq) Calculate the pKa value of the acid, and determine the number of moles of acid initially present. pK₂ = initial amount: When the reaction is complete, what is the concentration ratio of conjugate base to acid? [A-] [HA] = How many moles of strong base were initially added? amount added: mol HA mol OH-
The titration of CH3COOH (Ka = 1.76x10-4; pKa = 3.754; Kb = 5.68x10-11) with NaOH is an example of a weak acid/strong base titration. A 40 mL aliquot of a 0.060 M solution of HCOOH is titrated by 0.060 M KOH. What is the pH of the solution at each of the following points: a) initially b) after 1 mL of base has been added c) after 20 mL of base has been added d) at the equivalence point e) after 45 mL of bases has been added
An enzyme-catalyzed reaction was carried out in 5.0 mL of 0.15 M phosphate buffer pH 7.2. As a result of the reaction, the pH of the solution increased to pH 7.4. How many moles of OH- were produced during the reaction? H2PO4- + H2O ⇆ HPO42- + H3O+, pKa = 7.2 The answer is 8.50 × 10-5 moles But how do you solve it?
What volume, in milliliters, of 0.180 M NaOH should be added to a 0.140 L solution of 0.029 M glycine hydrochloride (p Kal = 2.350, pK₁2 = 9.778) to adjust the pH to 2.50? NaOH volume = mL
You want to prepare 100.0 mL of a buffer solution with a pH of 4.95 by mixing appropriate volumes of 0.10 M CH3CO2H and 0.10 M NaCH3CO2. How many milliliters (mL) of the acetic acid and sodium acetate solution are needed? (CH3CO2H has Ka = 1.8 x 10–5) (A) 75.0 mL of 0.10 M CH3CO2H + 25.0 mL of 0.10 M NaCH3CO2; (B) 62.0 mL of 0.10 M CH3CO2H + 38.0 mL of 0.10 M NaCH3CO2; (C) 38.0 mL of 0.10 M CH3CO2H + 62.0 mL of 0.10 M NaCH3CO2; (D) 25.0 mL of 0.10 M CH3CO2H + 75.0 mL of 0.10 M NaCH3CO2;
What ratio of sodium oxalate (Na2C2O4) and sodium hydrogen oxalate (NaHC2O4) is required for a buffer with a pH of 4.5? The Ka of the acid is Ka2 of oxalic acid (H2C2O4) = 6.4 x 10–5.
A 15.0 mL sample of 0.20 M of an unknown weak acid is titrated with 0.15 M NaOH at 25 oC. At 25 oC, the pKa of the weak acid is 4.06. What is the pH when 20.0 mL of titrant has been added to the analyte? Enter the answer to two decimal places.