Question

A buffer solution contains 0.402 M NaHSO2 and 0.248 M K S03. Determine the pH change when 0.071 mol HI is added to 1.00 L of the buffer. pH change =

Answer 1

Butter solution The pH of a butter solution can be Calulated by using the Henderson-Hasselball equation, pH pka tlog [ Base] Tacid) Base] = concentration of pka base logra [auid = concentration of acid ka aid dissociation constant Sulfurous (Hason) is a diproto acid He dissociation can be written as, Ha fog H+ +H505 ka, = 1.389 x 10-2 Hoog P H+ + so2 first we kaz=6.5x108 K2 503 calulate the pH of Nall soz and buffer system before addition of HI. H505 - Sog congiugate base ka = 695x108 According to Henderson Hasselbake [Hs03-] = 0.402M [so] = 0,248m Acid equation, pH putatlog - - Esse] pH = -log (6.5x108) tlog PH & 7.187 – 0.2097 0. 248 (8.402) o Hi to buffer system pH = 6.977 pH after addition of molarity a moles volume (L) so, moles of Hsong in solution 0.402 mile 1 l - 0.402 mote. 0.402 m of 1.002 & 1.00L

moles of m 100L of 0.248 m solution = 0.248 mole 12 0.402 0.248 mole moles of HI added = 0.071 mol so, the dissociation equilibrium can be written as, IC Etable S032 H H 503 Titial: 0.248 o.of Change: -6.071 -0.071 to:071 Equilibrium: 0.177 o 0.473 in [s032] = 0.177 more I L =061 77M and [hoz) = 0.473 more 1 L = 0.473 According to Hende raison - Hasselbalch equation pH a pka tlog ? [ 5032-) Itsoſ pH= 87.18 7 + log 0.177) pH = 2473 7.187 -0.4268 - pH = 6.76 So, PH decreases on addition of - HI, so the pH change = pH after addition , pH betoon (6.76 - 6977 addition of HL note; - -0.2166 Absolute Value pH HI decreases on al by 02168 canke taken) addition unit. ore Sign in PH But tanen indicates the absoluteralue decrease can be

6 V wel = 9.2076 ECH PH a 9.2076+ log10:257) 00.43 6 thea] 0.437m first we calculate pH before addition of koh to butter system Henlear. wear aude HCM is a very It dissociates ACN t H+ + CN- Aud Conjug te base ka = 6.2x10-10 and pka= -log ka = -10g (SLXIO According to Henderson Hassel balew equation in pHapka trog tcm-] - [CM] = 0.257 - pH = 8.977 pH in after addition of koll saoles of CN in 1.00h solution of 0. 257M = 0.257 moles, 11 1.602 - d. 257 moles and moles of HCN in 1.002 of 0.437m = 0.437 mole LE 1. oL = 0.437 mole and moles of KOH = 0.093 mol. dissociation equilibrium can be dritten we HCN + oH T CN titzo Initial! 0.437 0.093 01257 No Change: -0.093 -0.093 +0.093 Mo o 0.35 no Equiübrium, 0.344 " [HCM] = 0.344 more - о, эчч and [Na] = 0.85 more = 0.35m solu ton Now dissociation

According to Henderson - Hasselbalch equation, pH aquating [EN] [HCM] = pH = 9.2076 trog (0.35 0.344 pH = 9.2115 so pH increases kolta kott. on addition of pH change (sh after addition of Kold PH before addition of kot (9.2115 - 8 .977 = 0.2381