Question

(a) Calculate the percent ionization of 0.00660 M acetic acid (Ka = 1.8e-05). % ionization = % (b) Calculate the percent ionization of 0.00660 M acetic acid in a solution containing 0.0220 M sodium acetate. % ionization = % + -10.1 points 0/4 Submissions Used A buffer solution contains 0.70 mol of hydrogen peroxide (HOOH) and 0.87 mol of sodium hydrogen peroxide (NaOH) in 5.80 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 0.38 mol of NaOH? (assume no volume change) pH = (c) What is the pH of the original buffer after the addition of 0.16 mol of HI? (assume no volume change) pH =

Answer 1

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© cone of acetic acid given = ka = 1.88105 0.00660M (Pane-17 al we know mat where and degree of dimaciahom. la conccetnation. Smce a in very very eus man I in care of > weak acid. PYYT Som ka= ca? 1-8810 1 0.0066 ga 0.02777 1 op a= 2.72:10

© Butter solution' we ure Handeirion eq"; [PAGE -27 pro paano [ACID? pne pka + log (SAL7] pha - log ka + log (SAL? [acan] PNE -109 (.4.16 ") + log (0705:00) Ton- 11.71 hoonQ = 0.87+0.38 of Naon lop © After addition of 0.38 mole hoon+ Naon - Hoona + H2O moles at noon left after min Rxn = 10.70- 0.38 Ni 0.32 males Now again weinig handerion can ph= pka + log fs 4.77 Pns - 809 (7.4V101) + 109 (1.25/.) Toma 12.2] 0.3215.80)

© 10 atter addition of 0.16 moles of ni [PAGE-3] Hoona + H1 noont na? .. moler at noong een after min Rxn = 0.87-0.16 = 0.71 moles ut noon honned after min Ryn = 0.70+0.10 = 0.86 1 PPT YTT pn= -log Kar Jag bety - log (2.481652) + log(0.2016 1.0.80/5.00 Ton= 11.53 11.53