Question

The selectivity coefficient, KLi+, H+, for a Li+ ion selective electrode is 4 x 10^-4. When this electrode is placed in 3.43 x 10^-4 M Li+ solution at pH 7.2, the potential is -0.343 V versus S.C.E. What would be the potential if the pH were lowered to 1.2 and the ionic strength were kept constant? (Hint: At pH 7.2 the effect of H+ will be negligible because the concentration of H+ is much less than that of Li+) (Keep three sig. fig. in your final answer.)

          The selectivity coefficient, KLi+, H+, for a Li+ ion selective electrode is 4 x 10^-4. When this electrode is placed in 3.43 x 10^-4 M Li+ solution at pH 7.2, the potential is -0.343 V versus S.C.E. What would be the potential if the pH were lowered to 1.2 and the ionic strength were kept constant? (Hint: At pH 7.2 the effect of H+ will be negligible because the concentration of H+ is much less than that of Li+)
(Keep three sig. fig. in your final answer.)
        
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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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The selectivity coefficient, KLi+, H+, for a Li+ ion selective electrode is 4 x 10^-4. When this electrode is placed in 3.43 x 10^-4 M Li+ solution at pH 7.2, the potential is -0.343 V versus S.C.E. What would be the potential if the pH were lowered to 1.2 and the ionic strength were kept constant? (Hint: At pH 7.2 the effect of H+ will be negligible because the concentration of H+ is much less than that of Li+) (Keep three sig. fig. in your final answer.)
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Transcript

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00:01 Hello student, given this question we have to find out the cell potential at ph 1 .2.
00:06 The cell potential at ph 7 .2 and the concentration of solution and the selectivity coefficient, everything is given.
00:14 So how we will find that? here we can use the equation of ethel that is equal to ionic strength plus 0491 divided by n, love the concentration of solution plus k into 10 raise to higher as 14 plus ph.
00:27 From the first ph 7 .2 we can find out the i unix strength is that will be equal to e -cell value that is minus 3443 on converting into what it will be minus 0 .134 3 volt minus 0 .1091 divided by 1 .1 low of 3 .4 4 into 10 10 x2 minus 4 into 4 into 10 x 2 minus 4 into 10 3 into 10 3 minus 14 plus 7 .2 1 2 in this year we will get the ionic strength values minus 0 .1384 next we have to find out the easel value for the ph of 1 .2 so here the 8l value will be equal to we can use the same this equation the ionic strength value that is minus 0 .1 138 plus 0 .0591 divided by 1 club the concentration of solution that is 3 .4 4 into 10 x2 minus 4 the k values 4 into 10 x 2 minus 4 into 10 3 to minus 14 plus here the page is 1 .2 so calculating all this here we will get out e -cell value that is equal to minus 0 .138...
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