Question

Compare the calculated Ksp valueof 1.9x10^3 with the literature value of CuCO3(s) 2.4x10^-10

mistake, my calculated ksp is 1.92x10^-3

Compare your calculated Ksp values with the literature value (% error) 3 8. Determining a Solubility Product Constant Kup Ecell Use the half-equations to write the solubility equation for CuCOs spこ Show your calculation for the solubility product constant, Ksp, for Cuco, n c -3.2834 966ss 1o

Answer 1

The cell potential of an electrochemical reaction is calculated directly by the use of the Nernst's equation:

o RI E=E"--Ln(Q) (1)

Where:

E is the cell potential (V)

Eº is the standard cell potential (V)

R is the ideal gas constant (R = 8.314 J/mol*K)

T is the absolute temperature (K)

n is the number of moles of electrons transferred in the electrochemical reaction (mol e^-)

F is the Faraday constant (F = 96485 mol e^-)

Q is the reaction quotient

Given the reaction:

$CuCO_3_{(s)}\rightleftharpoons Cu^{+2}_{(aq)}+CO_3^{-2}_{(aq)}$

The quotient reaction (Q) includes only aqueous and gas species, and it is written as:

In equilibrium, no more reaction is being carried out because concentrations are constant in time. Cell potential is zero and quotient reaction is equal to the equilibrium constant, it is to say:

Substituting these conditions in equation (1)

o R1 R.F Ln(KSp)

Using the logarithmic property:

Ln(KSP) 2.303Log(KSP)

The previous equation is rewritten as:

2.303 Log(KSP 0-E"

Set a room temperature of T = 298.15K and substituting R and F values in this equation, we have:

$0 = E^{\circ}-\frac{2.303(8.314 \frac{J}{mol*K})(298.15K)}{n(96485\frac{coulomb}{mol\, e^-})}Log(K_{SP})$

0.0592 Log(Ksp)

Solving for K_SP:

$K_{SP}=10^{\frac{nE^{\circ}}{0.0592}}\;\;\;\;(2)$

The number of electrons transferred is observed in the half cell reaction:

Cu+2 + 2e-→ Cu

Where n = 2

With a given value of standard cell potential of Eº = -0.097V and substituting these values in equation (2), the solubility product constant of CuCO₃ is:

$K_{SP}=10^{\frac{-2*0.097}{0.0592}}=5.28*10^{-4}$

With a literature value of K_{SP(literature)} = 2.4*10^-10, the percent error is given by:

$\% \,error = \frac{\left |K_{SP(experimental)}-K_{SP(literature)} \right |}{K_{SP(literature)}}*100$

Substituiting known values, the percent error is:

$\% \,error = \frac{\left |5.28*10^{-4}-2.4*10^{-10} \right |}{2.4*10^{-10}}*100$

% error-219.999.900%