If 0.0490g of AgIO3 DISOLVES per liter of solution, calculate the solubility product constant.
Why did the molecule go to school? Because it wanted to be a solubility "product"ive member of society!
To calculate the solubility product constant (Ksp), we need to use the equation:
Ksp = [Ag+][IO3-]
The solubility of AgIO3 is given as 0.0490g/L. We need to convert this into concentration, using the molar mass of AgIO3 (234.77 g/mol):
0.0490g/L / 234.77 g/mol = 0.0002087 mol/L
Now, since AgIO3 dissociates into Ag+ and IO3-, the concentration of Ag+ and IO3- are equal and can be represented as x. Therefore:
[Ag+] = [IO3-] = x
Substituting x into the equation for Ksp:
Ksp = (x)(x) = x^2
So, the solubility product constant (Ksp) is simply x^2.
To calculate the solubility product constant (Ksp), we first need to write the balanced chemical equation for the dissolution of silver iodate (AgIO3) in water:
AgIO3 (s) -> Ag+ (aq) + IO3- (aq)
From the equation, we can see that 1 mole of AgIO3 dissolves to form 1 mole of Ag+ ions and 1 mole of IO3- ions.
Next, we need to calculate the concentration of Ag+ and IO3- ions in the solution.
Given that 0.0490 grams of AgIO3 dissolves per liter of solution, we can convert this mass to moles using the molar mass of AgIO3.
Molar mass of AgIO3 = atomic mass of Ag (107.87 g/mol) + atomic mass of I (126.90 g/mol) + 3 x atomic mass of O (16 g/mol) = 282.87 g/mol
Number of moles of AgIO3 = mass / molar mass = 0.0490 g / 282.87 g/mol = 1.732 x 10^-4 mol
Since 1 mole of AgIO3 produces 1 mole of Ag+ ions and 1 mole of IO3- ions, the concentration of Ag+ and IO3- ions in the solution is also 1.732 x 10^-4 M.
The solubility product constant (Ksp) is the product of the concentrations of the dissociated ions raised to the power of their stoichiometric coefficients.
Ksp = [Ag+] x [IO3-] = (1.732 x 10^-4 M) x (1.732 x 10^-4 M) = 3.00 x 10^-8
Therefore, the solubility product constant (Ksp) for silver iodate (AgIO3) is 3.00 x 10^-8.
To calculate the solubility product constant (Ksp), we need to use the information given. The solubility product constant is the product of the concentrations of the cations and anions in a saturated solution, each raised to the power of their stoichiometric coefficient.
In this case, we have the compound AgIO3, which dissociates into Ag+ and IO3-. Let's assign the variable x to the concentration of Ag+ in moles per liter.
Now let's write the balanced chemical equation for the dissociation of AgIO3:
AgIO3 (s) ⇌ Ag+ (aq) + IO3- (aq)
The concentration of Ag+ is equal to x, and the concentration of IO3- is also equal to x, since the compound dissociates in a 1:1 ratio.
Since we are given that 0.0490 g of AgIO3 dissolves per liter of solution, we need to convert this mass to moles. The molar mass of AgIO3 can be calculated by adding the atomic masses of Ag (107.87 g/mol) and IO3 (127.96 g/mol). Thus, the molar mass of AgIO3 is 235.83 g/mol.
Converting the given mass to moles:
moles of AgIO3 = mass / molar mass = 0.0490 g / 235.83 g/mol = 0.0002077 mol
Since AgIO3 dissociates in a 1:1 ratio, the concentration of Ag+ and IO3- is also 0.0002077 M.
Now we can calculate the Ksp:
Ksp = [Ag+] * [IO3-] = (0.0002077 M) * (0.0002077 M) = 4.31 x 10^-8
Thus, the solubility product constant (Ksp) for AgIO3 is 4.31 x 10^-8.
AgIO3 ==> Ag^+ + IO3^-
Ksp = (Ag^+)(IO3^-)
Solubility
AgIO3 = grams/molar mass.
Plug that in for Ag^+ and for IO3^- and calculate Ksp.