The solubility product constant Ca(OH)2 is 5.02 *10-5. What is the solubility of this compound?
Correct answer: C. 0.023M
- A. 0.020M
- B. 0.012M
- C. 0.023M
- D. 0.037M
Explanation
Solubility is the ability of a solute to dissolve into a solvent. It is calculated by creating an equilibrium equation and solving for the concentration of dissolved ions.For this question, we will need to set up the equilibrium constant equation for calcium hydroxide.Ca(OH)2⇌Ca2++2OH−; Ksp=[Ca2+][OH−]Note that the solid compound is not included in the equilibrium expression. Now we can work on finding solubility. For each mole of calcium hydroxide dissolved, one mole of calcium ions and two moles of hydroxide ions are released. Mathematically, we can equate this ratio to the ion concentrations in the equilibrium calculation.Ca(OH2): Ksp=5.02∗10−5=(x)(2x)2In this calculation, x is the solubility. Given the solubility constant, we can solve for x.5.02∗10−5=4x3→xCa(OH)2=0.023M
Last updated
About Chemical Equilibrium
Reversible reactions reach dynamic equilibrium when forward and reverse rates become equal, and Le Chatelier's principle predicts the effect of changing concentration, pressure or temperature. The chapter also covers solubility product, the common ion effect, buffer action and the conditions used in Haber's process, with equilibrium shifts distinguished from changes in the equilibrium constant.
Practise Chemical Equilibrium
625 free Chemical Equilibrium MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1.8 x 10 -5 is the dissociation constant of:
1 mole of CH3COOH and 1 mole of C2H5OH react to produce 2/3 mole of CH3COOC2H5. The equilibrium constant is
1 mole of ethyl alcohol was treated with one mole of acetic acid at 25°C. 2/3 of the acid changes into ester at equilibrium. The equilibrium constant of the reaction will be:
2HF ⇌ H2 + F2 PCI5 ⇌ PCI3 + Cl2The statement that is false about the Ke of both the reactions is;
2O3 ⇌ 3O2 Ke = 1055 at 25°C, it explains