Question

A buffer solution contains 0.69 mol of hydrogen peroxide (HOOH) and 0.39 mol of sodium hydrogen peroxide (NaOOH) in 6.60 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 0.61 mol of NaOH? (assume no volume change) pH =

(c) What is the pH of the original buffer after the addition of 0.13 mol of HI? (assume no volume change) pH =

Answer 1

5 NA OOH = 0.39 mo) HCOOH = 0.09 ka = 2.4 x 1012 pia = -log(2-4*10/2) =) pka = volume = 6.6 lit 11.61978 a) PH= pka + log caout 2 - 11.619 + ley og 001] pH = 11.37 Ames - 6) addition of 0.61 mol of Noon HCOOH + NAOH - NUOOH + H20 initial 0.69 0.61 0:39 final 0.08 pH = pka + log [Na00H.] ) 11-619 + ley 1 Y THOOHT CHOOH] pH = 12.71 Ane c) addition of 0.13 mol of HI Na OOH + HI - HOUH + NA I initial 0.39 0.13 0.69 fineel 0.26 - 0.82 PH= pka + leg (N000HJ . CHOOH) 0.26 3 ))-6 32 PH = 11.12