the selectivity coefficient (kli ,h ) for a li ion-selective electrode is 4.00 x 10-4. when this electrode is placed in 3.44 x 10-4 m li solution at ph 7.20, the potential is -0.333 v. what would be the potential (in mv) if the ph were lowered to 1.10?

The equilibrium constant for the reaction of displacement by one ligand of another ligand in a compound with the substrate is the selectivity coefficient. Binding selectivity is crucial in biology and chemical separation processes.

According to question,

E1 = constant + 0.05916 log( [Li+] + kLi[H+])

E1 = constant + 0.05916 log( [[tex]3.44 * 10^-^4[/tex]] + [tex]4..0 * 10^-^4 * 10^-^7^.^2[/tex])

We know,

E2 = constant + 0.05916 log( [[tex]3.44 * 10^-^4[/tex]] + [tex][4..0 * 10^-^4 * 10^-^1^.^1][/tex])

subtracting E1 from E2

E2 - E1 = 0.05916 log( [[tex]3.44 * 10^-^4[/tex]] + [tex][4..0 * 10^-^4 * 10^-^1^.^1][/tex] -( [[tex]3.44 * 10^-^4[/tex]] + [tex]4..0 * 10^-^4 * 10^-^7^.^2[/tex])

E2 - E1 = 0.05916 log ([tex]\frac{3.44*10^-^4 *4 *10^-^1^.^1}{3.44 * 10^-^4 }[/tex])

E2 - E1 = 0.0022 = 2.2 mv

E2 = E1 +2.2

E2 = -333 + 2.2 = -330.8mv

The potential is -330.8mv.

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