The equilibrium concentrations of and in a saturated solution of are and respectively.
Further explanation:
The equilibrium constant between the compound and its ion, when dissolved in solution, is known as solubility product constant. It is denoted by . The solubility product constant is used to calculate the product of the concentration of ions at equilibrium. Higher the solubility product constant more will be the solubility of the compound.
The general reaction is as follows:
The expression to calculate the solubility product for the general reaction is as follows:
Here,
is the solubility product constant.
is the concentration of ions.
is the concentration of ions.
The dissociation of occurs as follows:
Consider the molar solubility of to be S. Therefore, after dissociation, the concentration of and are S and 2S respectively.
The formula to calculate the molar solubility of is as follows:
…… (1)
Substitute S for , 2S for and for in equation (1).
Simplifying the above equation for S,
Solving for S,
Therefore the equilibrium concentration of is 0.002 M.
The equilibrium concentration of is calculated as follows:
Therefore the equilibrium concentration of is 0.004 M.
Learn more:
1. Sort the solubility of gas will increase or decrease: <u>brainly.com/question/2802008</u>.
2. What is the pressure of the gas?: <u>brainly.com/question/6340739</u>.
Answer details:
Grade: School School
Subject: Chemistry
Chapter: Chemical equilibrium
Keywords: Molar solubility, saturated solution, PbF2, Pb2+, F-, S, 2S, 0.002 M, 0.004 M, equilibrium concentration, Ksp, 3.2*10^-8.