Question

# Consider the equilibrium reaction for the saturated solution of a salt, A B 3 : A...

Consider the equilibrium reaction for the saturated solution of a salt, A B 3 : A B 3 (s)? A 3+ (aq)+3 B ? (aq) For each A B 3 formula unit that dissolves in water, the number of B ? ions is three times the number of A 3+ ions. In a saturated solution of the salt A B 3 , the number of formula units of the salt dissolved equals the number of dissolved A 3+ ions. If you know the concentration of the dissolved ions in a saturated solution, the solubility of the solute, S , can be calculated for the system. Solubility can be expressed as a molar concentration. The molar concentration or the molarity of the solution, M , is equal to the number of moles of solute per liter of the solution: molar concentration (M) = moles of solute volume of solution (L)

Question:

Part C Go to the Slightly Soluble Salts tab. Set the volume to 1.0× 10 ?16 L . Select the salt thallium sulfide from the dropdown menu. In the text box provided, set the total number of ions under the data for "cations" to 8 to make the solution saturated. Calculate the solubility in M of thallium (I) sulfide for a saturated solution. Express your answer numerically in moles per liter to three significant figures. I dont know which thalium sulfide it is so if you could do both i would appreciate it/

thallium (I) sulfide = TI2S   , Ksp = 6 x 10^-22

Tl2S ------------------------> 2 Tl+   + S-2

2S           S

Ksp = [Tl+]^2 [S-2]

6 x 10^-22 = (2S)^2(S)

S = 5.31 x 10^-8 mol / litre

solubility = 5.31 x 10^-8 mol / litre

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