Question

A weak acid, HA, has Ka = 1.4 x 10-5. At 312 nm, HA has  = 1.54 x 104 M-1 cm-1, and A- does not absorb. What absorbance will be recorded at 312 nm for a 6.17 x 10-5 M solution of HA buffered at pH 5.43? Assume that Beer's Law applies and that the spectrophotometer is not subject to stray light.

Answer 1

HA (aq) <---> H+ (aq) + A-(aq)

Initially [HA] = 6.17 x 10^ -5 M , [A-] = 0

at equilibrium [HA] = 6.17 x 10^ -5 -X , [A-] = X ,

given pH = 5.43 , [H+] = 3.715 x 10^ -6 M

Ka = [H+] [A-] / [HA]

1.4 x 10^ -5 = ( 3.715 x 10^ -6) (X) / ( 6.17 x 10^ -5 -X)      

8.64 x 10^-10 - 1.4 x 10^-5 X = 3.715 x 10^-6 X

X = 4.88 x 10^ -5

[HA] = ( 6.17 x 10^-5) - ( 4.88 x 10^-5) = 1.293 x 10^-5

Absorbance = constant x Concetration

Absorbance = 1.54 x 10^4 x 1.293 x 10^ -5

         = 0.2