Question

A buffer solution contains 0.68 mol of hydrosulfuric acid (H2S) and 0.63 mol of sodium hydrogen sulfide (NaHS) in 8.30 L. The Ka of hydrosulfuric acid (H2S) is Ka = 9.5e-08.

A buffer solution contains 0.68 mol of hydrosulfuric acid (H2S) and 0.63 mol of sodium hydrogen sulfide (NaHS) in 8.30 L. The K, of hydrosulfuric acid (H2S) is Ka = 9.5e-08. (a) What is the pH of this buffer? pH = (b) What is the pH of the buffer after the addition of 0.17 mol of NaOH? (assume no volume change) pH = (c) What is the pH of the original buffer after the addition of 0.14 mol of HI? (assume no volume change) pH =

Answer 1

. A buffer solution contains 0.68 mole of Mys and 0-43 mole of Natts in 8:30L ... [hus] = 0.68 () Ka of H2S 29.52-08 (given) 8.30"? [Nalts] = 0.63 a les 1918x10-8 28-30). pka : 7.022 From the expression of Hendensm equation of acidic (@ ph apika + log [Nahts? >7-022 + log (0-63/8-30) buffer of His ; 170327 000 (0.68/8-20) 27.022 – 0.033 -6.989 (b) After addition of oltmol of Nast; it will beact with Onlfmol of H₂s to from o. If mol of Nalts, Naot + H2S - Nahls + H2O Thus, the numbers of moles of His left aftero Meaction = 0.68 – 0.17 = 0.5lmol. and, the numbers of moles of Natts will be -(0.63+0.17) - 0.8 mole.

Need [H₂S) = 0.51 (M) (Neglecting volume . 8.30 [nalts] = 0.8 to change after a 890 ( M a ddition of Naost) Now the pH of this beffers solution); PH > pka + log (es nakts] 076022+ log(0.8/8.30) "70.51/8.30) 7.0227 0.1955 - 702175 (as str (C) Like Naost, HI is also strong acid , so it will fully bademy dissociates to produce equimotáz amount of Ht and 1. Thus the dissociative equilibmum of H2S coile vepress. the solution) As it has an buffer action so Oolf mole of Ht will Combine with the Conjugate base (As) of Has to from more molecular His . .. Thus, the number of moles of H₂S in the medium become a Bottorve 0.68+0.14) 20.82 moles

and, nember of miles of Nat's o(0:63–017). 50.49 mol Now, Ches] 0.62 mol , 0:82(0) ts? 2049 (0) 8.30L 8.30 Noco, the pH of the origin buffers aftero odbition of 0.14 mol of Hi, ен» phot love 37.022+ 1090048:30) (0.848:30) 27.022 – 0:2236 = 6:7984