Question

2. Calculate the pH of a 0.10 M aqueous solution of sodium acetate, NaC2H302. (2 pts)

Answer 1

Sodium Acetate is a salt of weak as and strong base. It will dissociate completely in solution as CHE COONa > Chicco + Nat - Sodium Acetate The CH₂coo ions so produced react with water to produce or ions which are responsible for baric nature of solution _Cacoo_+H 3 CHCOOH + OH Assuming concentration of water as consta the dissociation constant of above equilibrium is given by - --Ko= [checoon] [en] =0 fchycoo} ek is base dissociation constant of Checool Dissociation.constant is the ratio of concentration of products to that of Beactants Also it is known that Kack = 10-14 ning ka of acetic acid (ch, cooH) 1.8x10-5 get the value of K as 10-14 - 10-14 - Ka- 1:8.8.10-5

So K. S. ssxia's Now we can construct an ICE table for the equilibrium crecootto CH₃COOH +OH ( where I stands for initial concentration c for change in concentration and... E for equilibrium concentration). ICE table is: cycoo CH₂ cooh oh Im) cm) E (M) 0.10 x (0.10 ) The formula is къ - сцоонЈСон) Echzceo-] where [Chroor [on] and [ch, coo] are equilibrium concentrations of - CH₂cooth, ot and choo- respectively using values from ICE table, we get k () () -

using value of t we get s.ss x10-10 = x² - 0.10 x Since value of k is very small so value of x is assumed to be very small as compared to 0.10 this we x as compared to o.10 which - 0.10-re Qulo cand Thus we have -Siss x10.10 or 0:10 x² = 0.1oxs ssx 10-1 30.555 X 10-10

so x= rosss x10-10 se 0:245 x 10-5 Since [or] = x la tony = 0:745x10-5 We can calculate pot using the pot= -log[or] (0:745x10-5) -f=5.128) = 5:128. formula As se pt + pot=14 pt = 14-pot This pt = 14-5.128 = 8.872 -- This pt of 0.10m aqueous solution sodium acetate is 18.872 of