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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?

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The potential is -330.8mv.

What is a selectivity coefficient?

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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