EE 34% hu 9:12 PM Q. E Chrome File Edit View History Bookmarks People Window Hel
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EE 34% hu 9:12 PM Q. E Chrome File Edit View History Bookmarks People Window Help Mastering Chemistry: CH 14 HW https:// session. masteringchemistry.com /myct/ 77486250 GEN CHEM 1120 Spring 2017 Signed in as Hussain Al sidran Mastering Che Out CH 14 HW t Relating pKa and pKb Resources previous 4 of 6 l next GEN CH My Courses l t Relating pKa and pKb Another way to express acid strength is by using pKa: 2e Course Home pKa log Ki Learning Goal: Another way to express base strength is by using pKb To understand the relation between the strength of Assignment pKb log Kb an acid or a base and its pKa and pKb values TITLE IONS The degree to which a weak acid dissociates in solution is given by its acid-ionization constant, Part A CH 14 HW View Ka. For the generic weak acid, HA. A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, HA(aq) A (aq) +H (aq) Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid and the acid-ionization constant is given by but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in A 1H preventing cholesterol buildup in arteries. To find the pKa ofX-281, you prepare a 0.076 Mtest solution of X-281 at 25.0 C. The pH of the Similarly, the degree to which a weak base reacts solution is determined to be 2.40. What is the pKa of X-281? with H20 in solution is given by its base-ionization constant, Kb. For the generic weak base, B Express your answer numerically. B(aq) +H20 BH (aq) OH (aq) and the base-ionization constant is given by LBH l OH pKi Submit Hints My Answers Give Up Review Part Part BExplanation / Answer
Lets write X-281 as HA
pH = -log [H+]
2.40 = -log [H+]
[H+] = 3.98*10^-3 M
HA <—> H+ + A-
0.076 0 0 (initial)
0.076-x x x (at equilibrium)
x= [H+] = 3.98*10^-3
Ka = [H+][A-]/[HA]
Ka = x*x/(0.076-x)
= (3.98*10^-3)*(3.98*10^-3)/(0.076 - 3.98*10^-3)
= 2.199*10^-4
pKa = -log Ka
= -log (2.199*10^-4)
= 3.658
Answer: 3.658
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