The following cell was constructed to find the difference in Ksp between two nat

Question

The following cell was constructed to find the difference in Ksp between two naturally occurring forms of CaCO3(s), called calcite and aragonite.

Pb(s) | PbCO3(s) | CaCO3 (s, calcite) | buffer (pH 7.00) || buffer (pH 7.00) | CaCO3 (s, aragonite) | PbCO3(s) | Pb(s)

Each compartment of the cell contains a mixture of solid PbCO3 (Ksp = 7.4 x 10-14) and either calcite or aragonite, both of which have Ksp ~ 5 x 10-9 . Each solution was buffered to pH 7.00 with an inert buffer, and the cell was completely isolated from atmospheric CO2. The measured cell voltage was -1.8 mV. Find the ratio of solubility products, Ksp (for calcite) / Ksp (for aragonite).

Explanation / Answer

As we know the nernest equation given below we can find out the ratio of solubility products,

E=E0 - (2.303* R*T/nF) * log Q. ......1)

Where, R= universal gas const.= 8.314 J/K.mol

T= temperature in kelvin

F= Faraday const. = 96500 C/mol

n= no of mol of electron transferred in cell reaction/half reaction

   Q= reaction quotient

Let at 298 K the value of quantity (2.303* R*T/F) = 0.0592 V

= 59.2 mv

Let n=1, E0=0, Q=K [ksp ratio of calcite and aragonite]

& measured cell voltage ,E= -1.8 MV

Put all value in equation 1).

So, -1.8 = -59.2 * logK

logK = 0.0304

K= 1.0309

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