The selectivity coefficient kkh for a k ion selective...

The selectivity coefficient, KK+,H+, for a K+ ion-selective electrode is 0.00539. When the electrode is placed into a 1.48 × 10-5 M K+ solution at pH 6.60, the potential is -0.154 V vs. SCE. If the pH of the solution is lowered to 1.21 and the ionic strength is kept constant, what is the potential of the electrode? Assume ẞ = 1 for the electrode.

Transcript

00:01 Hi now we will discuss about how to find the e2 the electrode potential value if the selectivity coefficient k for the lithium electrode is given that 4 into 10 power minus 4 and the concentration of lithium ions is 3 .44 into 10 power minus 4 molar at ph 7 .2 the cell is showing the electrode potential minus 0 .33 as it is potential holds and if the ph is lower to 1 .1 you if it is lowered from 7 .2 to decrease to 1 .1 quantity, then what is the potential we have to find out? so first let us consider e is equal to we can write the potential equation is equal to constant plus 0 .0591 log l .i plus kli and concentration of hb here at ph 7 .2 let us consider the potential as even even we have to calculate even is equal to constant plus 0 .0 591 log lithium concentration is given that 3 .44 into 10 power minus 4 plus k value for lithium ion selectivity coefficient is 4 into 10 minus 4 into h plus ion concentration.

01:50 Here we have not provided with the h plus ion concentration but we have given the ph ph is equal to 7 .2.

01:57 We know the relation between the ph and the h plus ion concentration ph is equal to negative logarithm of the h plus ion concentration as ph is given we can write the h plus ion concentration as 10 power minus ph that is 10 power minus 7 2 that is the concentration of the h plus ions substitute this value in this equation 10 power minus 7 .2 now we have to solve for the e1 e1 is equal to here here e1 is given minus 0 .33 holds and we have to find the e2 e2 is equal to constant plus 0 .0 591 log 3 .44 plus 4 into 10 power minus 4 into 10 power ph is lower to 1 .1 so 10 power minus 1 .1 this is a e2 now as we have to find the e2 minus e1 e1 is equal to 0 .0591 log 3 .44 into 10 power minus 4 plus 4 into 10 power minus 4 into 10 power minus 1 .1 minus 0 .0 .0591 log 3 .44 into 10 power minus 4.

03:56 By solving this we'll get 0 .0591 log we'll take it as common into 3 .44 into 10 power minus 4 plus 4 into 10 power 10 power.

04:09 Minus 5 .8 divided by 3 .44 into 10 power minus.

04:17 So by solving this we'll get e2 minus e1, e2 minus e1 as 0 .0022 volts.

04:27 And here we have to find the e2.

04:30 We know e1 value...

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