Part A A new potential heart medicine, code-named X-281, is being tested by a ph

Question

Part A A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, 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 preventing cholesterol buildup in arteries To find the pKa of X-281, you prepare a 0.078 MM test solution of X-281 at 25.0 C. The pH of the solution is determined to be 2.40. What is the pKa of X-281? Express your answer numerically. pKa 4.96 Submit Hints My Answers Give Up Review Part incorrect, Try Again; 5 attempts remaining Part B This question will be shown after you complete previous question(s) Part C At a certain temperature, the pH of a neutral solution is 7.38. What is the value of Kw at that temperature? Express your answer numerically using two significant figures. Submit Hints My Answers Give Up Review Part Side vious Next Zoom Move Text Select Annotate Part A What is the H concentration for an aqueous solution with poH 4.20 at 25 C? Express your answer to two significant figures and include the appropriate units. CH 3.24.10 11 M Submit Hints My Answers Give Up Review Part

Explanation / Answer

For a weak acid, HA

HA          +             H2O        =             H3O+      +             A-

C                                                          0                            0

C(1-)                                                 C                         C

K = [H3O+][ A-]/[ HA][ H2O]

Ka = K[H2O] = [H3O+][ A-]/[ HA]

Ka = (C)2/ C(1-) = 2C                  (as <<1)

= (Ka/C)

[H3O+] = C = C * (Ka/C) = (KaC)

pH = - log[H3O+] = - log (KaC)

= 0.5 ( pKa - logC)

So pKa = 2 pH + logC

= 2*2.4 + log0.078

= 3.69

H2O = H+ + OH

Kw = [H+ ][OH -]

= [H+]2   (as [H+ ] = [OH -])

pH = 7.38

So [H+ ] = 10 -7.38 = 4.17*10-8

Kw =[H+]2 = (4.17*10-8)2 = 1.74 * 10­-15

pOH = 4.2

pH = 14-4.2 = 9.8

[H+ ] = 10 -9.8 = 1.58*10-10

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