Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution. the solubility of cu(oh)2(s) is 1.72 x10–6 g/100. ml of solution at 25° c.(i) write the balanced chemical equation for the dissociation of cu(oh)2(s) in aqueous solution. cu(oh)2 < ===> cu 2+ + 2 oh –(ii) calculate the solubility (in mol/l) of cu(oh)2 at 25 °c.(1.72 x10–6 g/0.100 l)(1 mol/97.5 g) = 1.76 x10–7 mol/l(iii) calculate the value of the solubility-product constant, ksp, for cu(oh)2 at 25 °c. ksp = [cu 2+][oh –]2 = [1.76 x10–7][3.53 x10–7]2 = 2.20 x10–20

i) Cu(OH)2(aq) → Cu^2+(aq) + 2OH-(aq)

ii)  1.76 *10^-7 mol/L

iii) 2.2 *10^-20

Explanation:

Step 1: Data given

The solubility of Cu(OH)2(s) is 1.72 * 10^–6 g/100 mL of solution at 25° C.

(i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.

Cu(OH)2(aq) → Cu^2+(aq) + 2OH-(aq)

(ii) Calculate the solubility (in mol/L) of Cu(OH)2 at 25 °C.

1.72 * 10^–6 g /0.100L = 1.72 *10^-5 g/L

1.72*10^-5 g/L / 97.56 g/mol = 1.76 *10^-7 mol/L

(iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25 °C.

For 1 mol Cu(OH)2 we have 1 mol Cu^2+ and 2 moles OH-

This means if there reacts X of Cu(OH)2 there will be produced X if Cu^2+ and 2X of OH-

Ksp = [Cu^2+]*[OH-]²

Ksp = 1.76 *10^-7 * 2*(1.76 *10^-7)²

Ksp = 2.18 *10^-20 ≈ 2.2*10^-20

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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