Question

Develop a recipe for 1.00 L of pH 4.50 buffer using only 17.4 M HC2H3O2 and 3.00 M NaC2H3O2 (and deionized water). Be sure that it canaccommodate additioin af at least 0.50 mol of strong acid or base with delta pH less than or equal to 1.

Answer 1

pH = pka + log [NaC2H3O2]/[HC2H3O2]      , pka of HC2H3O2 = 4.76

4.5 = 4.76 + log [NaC2H3O2]/[HC2H3O2]

[NaC2H3O2] = 1.82 [HC2H3O2]

NaC2H3O2 moles = 1.82 HC2H3O2 moles

Let   volume of sodium acetate be V then volume of HCH3O2 is 1-V

then 3 x V = 1.82 x17.4 (1- V)            ( Moles = M x V)

3V = 31.668 -31.668V

V = 0.913465 L = 913.5 ml

hence we need 913.5 ml of sodium acteate and 86.5 ml of acetic acid to get required buffer

acid moles = M x V = 17.4 x ( 86.5/1000) = 1.5

sodium acetate moles = M x V = 3 x 0.9135 = 2.74

lets suppose 0.5 mol strong acid added then   acetic acid moles = 1.5+0.5 = 2

sodium acetate moles = 2.74-0.5 = 2.24

pH = 4.76+log ( 2.24/2) = 4.8

if 0.5 mol base adde , then acetic acid moles = 1.5-0.5 = 1

sodium acetate moles = 2.74+0.5 = 3.24

pH = 4.76+log ( 3.24/1)

   = 5.27

hence our buffer will not change by 1 unit when 0.5 mol acid or base added