Question

26. -/0.1 points 0/4 Submissions Used My Note A buffer solution contains 0.71 mol of hydrogen peroxide (HOOH) and 0.41 mol of sodium hydrogen peroxide (NaOH) in 7.60 L. The Ka of hydrogen peroxide (HOOH) is ka = 2.4e-12. (a) What is the pH of this buffer? pH = (b) What is the pH of the buffer after the addition of 0.36 mol of NaOH? (assume no volume change) pH = (c) What is the pH of the original buffer after the addition of 0.28 mol of HI? (assume no volume change) pH =

Answer 1

anss Given Hydrogen peroxide (Hook) = 0.71mel 0.4 moi absodium hydrogen peroalde (Nato H) = 7.602 The ka of hydrogen peroxide (HooH) = 2.4612 Henderson - Hasselbalch Equation --foq 24850) + lyg ( = 1.619 + log (0:5774) = 1.6197 -0.2385 pH = 1:38 ]

6 Given Nao H = 0. 36 mol The reaction is Hoot + NaOH - > Naoo H +420 moles of Hou H = 0.471'-0.36 = 0.35 mol moles of Nash = 0.41 +0:36 = 0.77mol pH = pka + log ( [hash] [Hook] suliy da tha e - sayfa 4 anos") # log ( 04*] = 11.6197 +log (2-2) = 11.6197+0.3424 = 11.96

(pH = 11.96 Given HI = 0.28 mol The reaction is NaOH + HI - > HOOH + NOI moles of Hoo H = 0.71 +0.28 P = 0.99 mol moles of NaOH = 0.47-0.28 1 = 0.13 mod = 0.13 moll po = pka t dog [field] - day tot by =-69 (24x139) #dog [ grond = 11.619 -0.8817 = 10.73

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