Question

4) Show your work to get full credit. (a) How much in g) sodium acetate (CH3COONa) must be added to 100 mL 0.1 M CH3COOH solution to prepare a buffer solution with pH 4.90? (assume no change in volume)

Answer 1

Moles of HC2H302 Volume of buffer V We know, Moles = Molarity (M) x Volume (V in L) So Moles HC2H302 = 0.100 * 0.100 mol = 0.010 mol Let moles of NaC2H302 added = ? mol = 0.100 M = 100.00 ml = 0.100 L 4.90 Given Ka= 1.80E-05 We know pKa = -log (Ka) = -log [ 1.8E-05 ] = 4.745 ; pH of buffer solution = Both HC2H302 and NaC2H302 act as buffer system Henderson equation for buffer solution is pH=pKa + log [Base/Acid] here the Base is C2H302- and acid is HC2H302 pH = pka + log (C2H302-/ HC2H302] 4.90 = 4.745 + log (C2H302- / HC2H302] 4.90 - 4.745 = log (C2H302-/HC2H302] 0.1553 = log (C2H302-/HC2H302] [C2H302-/HC2H302] = 10.00 ^ 0.15527 [C2H302-/ HC2H302] = 1.4298 [C2H302-) = 1.4298 x [HC2H302] = 1.4298 X 0.0100 mol Moles of [C2H302-) = 0.0143 mol Mass of sodium acetate = Mole x Molar Mass = 0.0143 mol x = 1.17 g 82.030 g mol-1

H+ addition to buffer: When H+ is added to the buffer, it reacts with base C2H302- and produce HC2H302. So during the addition of H+ to the buffer; C2H302- Conc/moles will decrease and HC2H302 moles/conc will increase as shown in the below table. so Moles of H+ added = 0.0025 mol C2H302 - (aq) + H+ (aq) <-----> HC2H302 (aq) Initial 0.0143 0.0025 0.010 Change -0.0025 -0.0025 + 0.0025 Equilibrium 0.0118 0 0.0125 pH = pka = 4.745 = 4.745 = 4.745 = 4.72 + log (C2H3O2- / HC2H302] + log [ 0.0118 = 0.0125 ] + log [ 0.9440 ] + -0.025 OH-addition to buffer : When OH- is added to the buffer, it reacts with acid HC2H302 and produce C2H302-. So during the addition of OH-to the buffer; C2H302- Conc/moles will increase and HC2H302 moles / conc will decrease as shown in the below table. so Moles of OH- added = 0.0025 mol HC2H302 (aq) + OH- (aq) <-----> C2H302- (aq) + H2O (1) Initial 0.010 0.0025 0.0143 Change -0.0025 -0.0025 + 0.0025 Equilibrium 0.0075 0 0.0168 pH = pKa = 4.745 4.745 4.745 = 5.09 + log (C2H302- / HC2H302] + log [ 0.0168 0.0075 ] + log [ 2.2400 ] + 0.3502 = 4.742

6 CHFJ 20. IM [IF] = 0.14 pl = 3.45 HF/F i a Acidic suffer Apply Hendersone aquation pH pka + los [kr] CHI) 3.45 I tka + log (O.1 TOM) 3.45 = pka + loa (1) 3.45 pka +(0) pka - 3.45 -3.45 0.00035421 Kar 10 kar 3.5 4X10