Question

I need help with a virtual solubility lab.

I picked 3 different solutions:
PbCl₂, Sr(IO₃)₂, and PbCO₃.

I added .1 grams of each solution to 100.00mL of distilled water.

With the following data, how do I calculate the solubility (in units of g/100mL) for each solution and get the Ksp values?

Species (aq) H+ Molarity 2.08253e-7 OH 2.08253e-7 Cl 0.00719150 (Pb+2 0.00359575 Species (s) grams Temperature: 45.63°C 45.6 deg pH: 6.68


I also have these questions to answer:


If you can walk me through how to do the calculations stepwise, I'd really appreciate it!

Answer 1

(1) Ksp and solubility calculation for PbCl₂

Given, [Pb²⁺] = 0.00359575 M

[Cl^-] = 0.00719150 M

PbCl₂(s) $\rightleftharpoons$ Pb²⁺(aq) + 2Cl^-(aq)

Here, solubility of PbCl₂(s) is equal to the concentration of Pb²⁺.

solubility of PbCl₂ = [Pb²⁺] = 0.00359575 M = 0.00359575 mol/L

Molar mass of PbCl₂ = 278.1 g/mol

So,

solubility of PbCl₂ (in g/100mL) = solubility x molar mass = 0.00359575 mol/L x 278.1 g/mol = 0.999978075 g/L = 0.0999978075 g/100mL

Hence, solubility of PbCl₂ = 0.0999978075 g/100mL.

Now,

Ksp = [Pb²⁺][Cl^-]² = (0.00359575)(0.00719150)² = 1.86x10^-7

Hence, Ksp for PbCl₂ = 1.86x10^-7

(2) Ksp and solubility of Sr(IO₃)₂

Sr(IO₃)₂(s) $\rightleftharpoons$ Sr²⁺(aq) + 2IO₃^-(aq)

Solubility of Sr(IO₃)₂ (s) = [Sr²⁺] = 0.0000293222 mol/L

Molar mass of Sr(IO₃)₂ (s) = 437.4 g/mol

Solubility of Sr(IO₃)₂ in g/100 mL = 0.0000293222 mol/L x 437.4 g/mol = 0.0128255303 g/L = 0.00128255303 g/100mL

Hence, the solubility of Sr(IO₃)₂ is 0.00128255303 g/100mL.

Now,

Ksp = [Sr²⁺][IO₃]² = (0.0000293222)(0.0000586445)² = 1.008x10^-13

Hence, the solubility of Sr(IO₃)₂ is 1.008x10^-13

(3) Ksp and solubility of PbCO₃

PbCO₃(s) $\rightleftharpoons$ Pb²⁺(aq) + CO₃^2-(aq)

Solubility of PbCO₃ (s) = [Pb²⁺] = [CO₃^2-] = 1.95081x10^-9 mol/L

Molar mass of PbCO₃ =267.21 g/mol

Solubility of PbCO₃ in g/100mL = 1.95081x10^-9 mol/L x 267.21 g/mol =521.2759401x10^-9 g/L =  5.212759401x10^-8 g/100mL

Hence, the solubility of PbCO₃ is 5.212759401x10^-8 g/100mL

Now,

Ksp = [Pb²⁺][CO₃] = (1.95081x10^-9)(1.95081x10^-9) = 3.8056596561x10^-18

Hence, the solubility of PbCO₃ = 3.8056596561x10^-18 g/100 mL

Now Post Lab questions:

1. As we know that if the H⁺ reacts with the anion to for weak acid then only the pH would have the effect on solubility otherwise it is not.

(i) In PbCl₂, Cl^- to form HCl which is a strong acid. So, pH would have no effect.

(ii) In Sr(IO₃)₂, IO₃^- reacts with H⁺ to form HIO₃ which is also a strong acid. So, pH would have no effect.

(iii) In PbCO₃, CO₃^2- forms HCO₃^- with H⁺. So, in this case as we increase pH, CO₃^2- is removed by the below equilibrium to move the reaction in forward direction and PbCO₃ would be more soluble.

CO₃^2-(aq) + H⁺(aq) $\rightleftharpoons$ HCO₃^-(aq)

2. As we know that the decrease in temperature leads to decrease in solubility of insoluble compounds. So, Ksp would decrease as this experiment is conducted in ice-water bath.

Let me know if you have any queries regarding this.