Question

Which of the following solutions would have the lowest buffer capacity? 0.10 M H2CO3, 0.15 M NaHCO3 2.0 MHC2H302, 2.0 M KC2H302 1.0 M HCOOH, 1.15 M NaHCOO 0.50 MH3PO4, 0.35 M KH2PO4

Answer 1

0.18 I acid O Buffer t 0.10M ho 0 .15M NaHCO3 + c = + + + 40,- . NaHCO₃ - Nat et HCO ₂ g. 15 pH = pka at log o 15 = 6.35 + Log 1.5 monte = 6.52 Adding 0.03 M NaOH to buffer - H₂ CO & OH - HO & HCO3 EH, 56 [H2CO3 ] -10.10 -0.039 M 10.04 M - [NaHCO₃ ] =(0.15 + 0.03) - 0.18M pH = pka at log & salt] = 6.35 & Log ol = 6.759 Butter 2 2.0M HG H₂O₂ 2.0M FC ₂ H 302 HC, H302 3 **+ H20,- TIS K C2H302 4t & Ca 4302" pH = pka & log (salt] =4.76 at log 2 = 4.76 O pacid] Adding 0.03M Noot to buffer - H C₂ H₂O2 + oH H₂O + H₂O2- [ H C H302] 's 2.0 -0.03 = 1.97 [ K2 C2 H₂O2] = 2.0+ 0.03 = 2.03 pH = pka + log [salt i Log Iacid = 4.76 + 18:03 4.773 ] to 71.97 Butter -3 1.0M HCOOH 1.15M NaHCoo H coolet A coor + H+ Matcoo - oor tamat pH = pka at log Salt 2 3.75 + log 1.15 - 3.8106 Adding 0.03 M Maolt to buffer a нсон + он - ну о «н соо- [HCOOH J = 1.0 -0.03 - 0.97 M. [ Nath coo] = 1.15 + 0.03 = 1.18 M. pH = pka & log and = 3.75 + log 1.18 1.0 (0.97 - 3.946

Butter 4 0.50 0.35 M KH₂PO4 som H₂POg H₂ POA H3 Pod - Ht & H₂Poon KH₂ Po r t & H₂PO4 itt pH = pka -2.005 [ sath] 0.35 - 2.16 + 02. 0.50 Adding 0.03 NaOH to buffer as [H₂ Pod] =(0.50-0.03) M= 0.47 km [KH₂ Pod ] (0.35 +0.03 M = 0 spka +.log Esald) - 2.16+ nog 0.53 – 2.212 I acid] 0.53 M + 0.03 M = Buffost sphe Buffer 1 6.759 - 6.52 = 0.239 Buffer 2 4.773 -4.76 20.013 Butter capacity 0.03 = 0.1255 0.239 0.03 0.013 . 2.30 Buffer 3 3.946-3.810 0.136 0.03 0.136 0.2205 2 Buffer 4 2.212 - 2.005 0.03 0.20 ] 20. 1449 Thus butter 1 (0.10 M H₂CO3 of 0.15M NaHCO₃) has lower buffere capacity: 1