Question

Calculate the pH of the solution obtained by titrating 40 ml of 0.35 M HNO2(aq) with 0.4 M NaOH(aq) to the equivalence point. Take Ka= 4.6x10^M.

Answer 1

At the equivalence point, the number of equivalents of acid is equal to the number of equivalents of base.

(N1V1) acid = (N2V2)base

But the PH will not be equal to 7 at the equivalence point. This is because we are tirrating a weak acid HNO2 against a strong base NaOH. Hence the PH will be slightly greater than 7.

Volume of HNO₂ = 4ome. Molarity of HNG = 0.35M Molarity of thg & NOT Normality of HNO2 is same as molality as HNO2 has only one acidic proton HNO2 z Ht+ Noz Molarity of NaOH = 0.4M Naon → Nat + oH Loney one of ion) - Holae Normality of NaoH is same as Molarity at equivalence point (NV) HNO2 = (N2V2) NaOH V = 40x0.35 0.4 - U2 = 35 me no of moles of HNO2 = Mix, = 40 0.35 MX 40 ml 1000 m = 0.014 no of moles of NaOH = Miy = 0.4 x 35 - oooly 100D ANG + NaOH ♡ Nanoz t H₂O Initial 0 0 D0014 60014 0.014 moles 0-014 moles o final o Total volume of solution = Vol of HNO2 + vol of Naoh =40+ 35 = 75 me Molarity of Nano₂ = no of moles of NaNO2 = rosto vin dit = 0ooly x1000 Scanned with CamScanner = p. 186 M US75

H,0 - MaxSo, t 0.186 00186-1 ино, + 40 0 0 Initial final 40 sind base dissociation constant Rakbekw Ka= acid dissociation (endorid Kw = 10-14 Kb = base discoriolion constant. ka = 4.6410-14 Kb = 1014 = 2.17% 10!! - 4.6x10-4 0.160 = 5 kb = [HAJO₂] [na.on ] = x² [Nano, 2017x10 11 = ? assuming is very im 2 small t-186- 2 0.166 = 4.0362 X 10^12 x=200094 106 POH = -log [omi] = -log(2.0096106) = 5.698 PHIPOH = 14 PH = -301 Ic: Scanned with CamScanner