Question

For the following problems, use the pK,' values given in Table 2-3 2-6 How many mL of 0.1 M sodium acetate should be added to 100 mL of 0.1M acetic acid to make a buffer of pH 5.1? What is the molarity of the resulting buffer with respect to acetate (acetate + acetic acid)?

Answer 1

Solution of Let volume of ^ sodium acetate volume of a socium aret added = V ml so that total volume of buffer = (100tv) me solution Initially Molauity of 0.1 mole Sodium acelale ((H3Coona) No. of moles = Molarity x volume 20.l molex vxOPK =vxco* mole and Molaruty of acetic and = 0,1 molel." - No of moles of <H3cool = 0.1 molelax 100nd 1000kmet 210? mole After mining um dim CHICOONa and goone 0-1 M CH3COOH, the molarity of both (Hz CooNa and CH3C001) can be calulaled as LetzcooNa] =Vx10*mole (100+V) x1036 (100 tu) M = (Too tum

and [CHE COOH) = - 10-2 mole (v+ 100)x1024 - M Osing Han Value of (100+) " Hendersm - Hasselbalch equation, the u can be calulated as pH = pka (ct13(00H1) + log [CH3COONa] [CH: (004) 0.1 V 5.1 = 4.76 + log (100+01 m. 108 3 L 100+4) M => 0.34 = s ſ o = 200-34) 7 100 2.187 = Tv = 218.7 ml

= 218.7 ml Thus, volume of sodium acetate must be added to make a buffer of pH 5.1 The resulting molarity of CH₃COONa & CH₂ COOH core [CH3COONa] = 0.1V (100 FV) M 0.1 X 218.7 (100+ 218.7) Auch 21.87 318. 7M 0.0686 M and [CH3COOH ] = foet 10 (100+v} M 10 318.7 M 0.0314 M