When an ion-selective electrode for X was immersed in 0.0520 M XCI, the measured potential was...

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Question

science chemistry

Answer 1

Answer #1

Given that,

E_HCL = 0..0380 V

E_X = 0.0530 V

[X⁺] = 0.0520 M

Let's consider the Nernst equation ,

E = E⁰ - (0.0591/n)*log Q

E = E⁰ - (0.0591/n)*log(1/[X⁺]) . --------(1)

0.0380 = E⁰ - (0.0591/1)*log(1/0.0520)

0.0380 = E⁰ - 0.0756

E⁰ = 0.11

Now, substitute the above value in eq(1)

E = E⁰ - (0.0591/n)*log(1/[X⁺])

0.0530 = 0.11 - (0.0591/1)*log(1/[X⁺])

0.964 = log(1/[X⁺])

1/[X⁺] = 10^0.964

1/[X⁺] = 9.20

[X⁺] = 0.108 M

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Answer 2

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