A very weak acid has à Ka= 4.37 xM? ()What is the [H+] of a solution of this acid that is 2.65

The acids are substances that can readily donate a proton. Bases are molecules that receive a…

Answered: A very weak acid has à Ka= 4.37 x M?…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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pH of solution 14.00 12.00 10.00 8.00 6.00 First 4.00 equivalence point 2.00 0 First midpoint Second equivalence point Third midpoint pH = pKa=12.32 Third equivalence point HPO4(aq) + OH(aq) PO4(aq) + H2O(1) Second midpoint pH = pKa = 7.21 H2PO4 (aq) + OH(aq) HPO42 (aq) + H2O(1) Using the Henderson- Hasselback equation, show how to create 2L of a 0.1 M KPhos pH 7.5 buffer using K2HPO4 and KH2PO4. The chart to the left should help you understand what pKa to start with. Show your work. pH = pKa = 2.16 H3PO4(aq) + OH(aq) H2PO4(aq) + H2O(l) 25.0 50.0 75.0 100.0 Volume of NaOH added (mL)
The pKa values of a compound with two ionizable groups are pK1 = 4.10 and pK2 between 7 and 10. A biochemist has 10 mL of a 1.0 M solution of this compound at a pH of 8.00. She adds 10.0 mL of 1.00 M HCl, which changes the pH to 3.20. What is pK2?
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.
Can you please calculate the following?

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A very weak acid has à Ka= 4.37 x M? (-) What is the [H+] of a solution of this acid that is 2.65