Question

A glass membrane electrode is constructed that has a selectivity for Na vs Li, K, and Ag with the following selectivity coefficents: K = 1/2800; KNax = 1/2750; and KwaAg = 1/3015. Answer the following questions: 1. What is the potential for the Na ISE when immersed in the following solutions: a) A 0.0100 M Nacl solution b) A 0.1100 M NaOH Solution c) A 0.5701 M NaNO Solution 2. Compute the potential change for each of the solutions in question 1 if they also contain the following: a) 0.1000 M KCl b) 1.2500 M Li2SO c) 2.500 M AgN0 3. Give the difference between the two potentials for each of the systems you worked out in question 2 above:

Answer 1

1. Potential of Na⁺ ISE immersed in solutions : (z =1)

E = E^o+ 0.0592* log (Na⁺ ) (at 25 C)

since ISE and no other data, we can take E- E^o = E'

E' = 0.0592* log (Na⁺ )

a. 0.0100 M NaCl :  E' = 0.0592* log (0.01 ) = -0.1184 V

b. 0.1100 M NaOH :  E' = 0.0592* log (0.11 ) = -0.05675 V

c. 0.5701 M NaNO₃ :  E' = 0.0592* log (0.5701 ) = -0.01445 V

2. In presence of other ions :

E' = 0.0592* log ((Na⁺ )+(K_{Na+/M *}(M⁺ ) )

K_Na+/M = selectivity coefficient, it is a numerical measure of how well the membrane can discriminate against the interfering ion.

A. 0.0100 M NaCl : E' = 0.0592* log ((Na⁺ )+(K_{Na+/M *}(M⁺ ) )

- 0.1000 M KCl  E' = 0.0592* log ((0.01)+(3.64*10^-4 _*(0.1 ) ) = -0.1183 V

- 1.25 M Li₂SO₄ E' = 0.0592* log ((0.01)+(3.57*10^-4 _*(2*1.25) ) = -0.1162 V

- 2.500 M AgNO₃l  E' = 0.0592* log ((0.01)+(3.32*10^-4 _*(2.5 ) ) = -0.1164 V

B 0.1100 M NaOH : E' = 0.0592* log ((Na⁺ )+(K_{Na+/M *}(M⁺ ) )

- 0.1000 M KCl  E' = 0.0592* log ((0.11)+(3.64*10^-4 _*(0.1 ) ) = -0.05674 V

- 1.25 M Li₂SO₄ E' = 0.0592* log ((0.11)+(3.57*10^-4 _*(2*1.25) ) = -0.05633V

- 2.500 M AgNO₃l  E' = 0.0592* log ((0.11)+(3.32*10^-4 _*(2.5 ) ) = -0.05656 V

C. 0.5701 M NaNO₃ : E' = 0.0592* log ((Na⁺ )+(K_{Na+/M *}(M⁺ ) )

- 0.1000 M KCl  E' = 0.0592* log ((0.5701)+(3.64*10^-4 _*(0.1 ) ) = -0.01445 V

- 1.25 M Li₂SO₄ E' = 0.0592* log ((0.5701)+(3.57*10^-4 _*(2*1.25) ) = -0.01441 V

- 2.500 M AgNO₃l  E' = 0.0592* log ((0.5701)+(3.32*10^-4 _*(2.5 ) ) = -0.01441 V

3. Differences in potential with and without interference :

dE = E' - E'i

A. 0.0100 M NaCl :  E' = -0.1184 V

  - 0.1000 M KCl = -0.0001  V

- 1.25 M Li₂SO₄    = -0.0022 V

- 2.500 M AgNO₃   = -0.0020V

B. 0.1100 M NaOH :  E' = -0.05675 V

- 0.1000 M KCl  = -0.00001 V

- 1.25 M Li₂SO₄ = -0.00042 V

- 2.500 M AgNO₃  = -0.00019 V

C. 0.5701 M NaNO₃ :  E' = -0.01445 V

- 0.1000 M KCl ~ 0.0000 V

- 1.25 M Li₂SO₄  ~ -0.0004 V

- 2.500 M AgNO₃l ~ -0.0004 V