You need to be able to get an answer of 2.14 x 10 -7 M for part c. A lead ion-se

Question

You need to be able to get an answer of 2.14 x 10-7 M for part c.

A lead ion-selective electrode measured a potential of -33.9 mV when immersed in 50.00 mL of water from a polluted stream. Upon the addition of 0.5 mL          of a 0.105 mM solution of Pb(NO3)2 the potential changed to -11.2 mV.

a)  Write an equation for the response of this electrode, ignoring the response of any other elements (i.e., we will assume selectivity coefficients for non-lead species are all zero) (4 points).

b) Assuming = 1 in the above equation, write expressions for the response of the electrode to the original sample and to the spiked sample (6 points).

c) Solve the equations you wrote in (b) to find [Pb2+] in the original sample (6 points).

Explanation / Answer

An ideal I.S.E. (Ion-selective electrode) allows only select –ion can be transported thereby creates a

potential difference.

For a charged l species diffusing from a region of activity, A1, to a region of activity, A2, there is a free energy change given by G=-RTln(A1/A2), which is converted to electrical work

-RTln(A1/A2)=-nFE (for positive species)

Since, E=RT/zF ln (A1/A2) =0.05916/z ln(A1/A2) at 298K

Z=charge of species

a) E=0.05916V/2 ln (A1/A2)

A1=activity of Pb2+ in water

A2=activity of Pb2+ in cell electrolyte

For low concentration, activity of Pb2+=[Pb2+]

E= 0.05916V/2 ln{ A (Pb2+) (water)/(Pb2+)(cell electrolyte)}

-33.9 *10^-3 V= 0.05916V/2 ln{ [Pb2+] (water)/[Pb2+](cell electrolyte)}…………………(1)

b)for original sample , when mean ionic activity =1,activity=concentration

-33.9 *10^-3 V= 0.05916V/2 ln{ [Pb2+] (water)/[Pb2+](cell electrolyte)}…………………(1)

For spiked sample,

E= 0.05916V/2 ln{ [Pb2+] (spiked)/[Pb2+](cell electrolyte)}

[Pb2+](spiked)=[Pb2+](water)+ [Pb2+](added)

Moles of Pb2+ (added)=0.105mM*0.5ml=0.105*10^-3 M*0.5/1000L=0.0525 *10^-6 moles

[Pb2+](added)=0.0525 *10^-6 moles

Consequently,

E= 0.05916V/2 ln{ ([Pb2+]water + 1.040 M)/[Pb2+](cell electrolyte)}

-11.2 *10^-3 V = 0.05916V/2 ln{ ([Pb2+]water + 1.040 M)/[Pb2+](cell electrolyte)}………………..(2)

c) equation 1-equation2,

-33.9 *10^-3 V-(-11.2 *10^-3 V) = 0.05916V/2 ln [Pb2+] (water)/ ([Pb2+]water + 0.0525 *10^-6 moles)

-22.7*10^-3 V=0.02958 V ln [Pb2+] (water)/ ([Pb2+]water + 0.0525 *10^-6 moles)

0.879* = ln [Pb2+] (water)/ ([Pb2+]water + 0.0525 *10^-6 moles)

Or,e^(-0.879)= [Pb2+] (water)/ ([Pb2+]water + 0.0525 *10^-6 moles)

Or,0.415=[Pb2+] (water)/ ([Pb2+]water + 0.0525 *10^-6 moles)

Or,2.41=1+(0.0525*10^-6)/[Pb2+]

Or,1.41=0.0525*10^-6/[Pb2+]

Or,[Pb2+]=0.0525*10^-6 moles/1.41=0.0372 *10^-6 moles

[Pb2+]=0.0372 *10^-6 moles/ volume=0.0372 *10^-6 moles/50 ml*1000ml/L=0.744 *10^-6=7.4*10^-7M

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