PRINCIPAL CHEMISTRY II TEST II 8. (a) Which conjugate acid-base pair is the best

Question

PRINCIPAL CHEMISTRY II TEST II 8. (a) Which conjugate acid-base pair is the best choice to make a buffer at pH 7-4 according to the Ka table on the right? Acid Ka CH.COOH 18 10 7.1 x 10 C.HisNO S 6.3 x 10 | CsHnN:04S | 3.3 × 10 C4HI2NO 8.7 × 1 CoH19NOS | 4.0×10-1 Best choice: acid formula and base formula (b) If a buffer contains equal volumes of 0.75 M C HI2NO CI and 0.45 M C.HINOs, what is the pH of the buffer? 9. A 25 mL 0.50 M NHs solution is added into a 20 mL. 0.30 M HCI solution. The Kb of NH, is 1.8 x10. What is the final pH of the mixture? 10. The ionization of adenine (CsHsNs), a nucleobase, is listed below Kai = 7.08 × 10-5 Kaz- 1.58 x 10-10 For a 0.50 M adenine hydrochloride aqueous solution (CsH&NsCl;, completely ionized to CoH-Ns and Ch in water), calculate [C.H&Ns; 1.ICT H1ICHNs). ICsHNs1, and [OH) Page.6 2017 Fall Savannah State University

Explanation / Answer

if pH = pka , the acidic buffer capacity is maxium.

so that, pH = 7.4

pka = -logka

7.4 = -logka

ka = 3.98*10^-8

suitable acid is

acid = C8H18N2O4S

base = C8H17N2O4S^-

B) pka = -logka

       = -log(8.7*10^-9)

       = 8.06

pH = pka + log(base/acid)

let us take , acid = 100 ml

no of mol of acid = 0.75*100 = 75 mmol

no of mol of base= 0.45*100 = 45 mmol

pH = 8.06+log(45/75)

     = 7.84

9) pkb of NH3 = -log(1.8*10^-5) = 4.74

   no of mol of NH3 = 25*0.5/1000 = 0.0125 mol

   NO of mol of HCl = 20*0.3/1000 = 0.006 mol

pH = 14 - (pkb + log(acid/base))

     = 14- (4.74+log(0.006/(0.0125-0.006))

     = 9.3

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