The solubility product of copper(II) iodate, Cu(IO3)2, is 4 x 10-9 mol3 dm-9 at 25 °C.
What is the solubility of copper(II) iodate in g dm-3?
A 1.85 × 10-2
B 4.14 × 10-1
C 5.21 × 10-1
D 6.56 × 10
Also please i'd really appreciate if someone could explain me also how to do it.. like from step 1 to end ~
Thanks.. @@..
Cu(IO3)2 ==> Cu^2+ + 2IO3^-
Ksp = (Cu^2+)(IO3^-)^2
Make and ICE chart and substitute into Ksp expression.
(Cu^2+) = x
(IO3^-) = 2x
Solve for x, which will be in M, and multiply by molar mass to find grams.
To find the solubility of copper(II) iodate in g dm-3, we need to use the solubility product constant (Ksp) and the molar mass of copper(II) iodate.
The solubility product expression for copper(II) iodate, Cu(IO3)2, can be written as:
Ksp = [Cu2+][IO3-]2
Given that the solubility product constant (Ksp) is 4 x 10^-9 mol3 dm-9, we can assume that the concentration of Cu2+ and IO3- ions are equal since the formula for copper(II) iodate is Cu(IO3)2. Therefore, let's represent the concentration of Cu2+ and IO3- ions with the variable 'x'.
Ksp = (x)(x^2) = x^3
Now, let's solve for 'x':
x^3 = 4 x 10^-9
Taking the cube root of both sides:
x = (4 x 10^-9)^(1/3)
x ≈ 0.0159 mol dm-3
To convert the solubility from mol dm-3 to g dm-3, we need to multiply the solubility by the molar mass of copper(II) iodate, which is:
Cu(IO3)2:
Copper (Cu) - Atomic mass: 63.55 g/mol
Iodine (I) - Atomic mass: 126.90 g/mol
Oxygen (O) - Atomic mass: 16.00 g/mol
Molar mass of Cu(IO3)2 = (63.55 g/mol) + (2 * 126.90 g/mol) + (6 * 16.00 g/mol)
= 63.55 g/mol + 253.80 g/mol + 96.00 g/mol
= 413.35 g/mol
Now, let's calculate the solubility in g dm-3:
Solubility = (0.0159 mol dm-3) * (413.35 g/mol)
≈ 6.56 g dm-3
Therefore, the solubility of copper(II) iodate in g dm-3 is approximately 6.56 g dm-3. Hence the answer is option D: 6.56 × 10
To determine the solubility of copper(II) iodate in g dm-3, we need to use the solubility product constant (Ksp) and the molar mass of copper(II) iodate (Cu(IO3)2).
The equation for the solubility product of copper(II) iodate is:
Cu(IO3)2 ⇌ Cu2+ + 2IO3-
The Ksp expression for this reaction is:
Ksp = [Cu2+][IO3-]^2
Given that the solubility product (Ksp) of copper(II) iodate is 4 x 10-9 mol3 dm-9, we can set up the equation:
4 x 10-9 = [Cu2+][IO3-]^2
Since the molar mass of Cu(IO3)2 is not provided, we need to calculate it using the atomic masses of copper (Cu), iodine (I), and oxygen (O).
Cu(IO3)2:
Cu: atomic mass = 63.55 g/mol
I: atomic mass = 126.90 g/mol
O: atomic mass = 16.00 g/mol
Calculating the molar mass of Cu(IO3)2:
Cu(IO3)2 = Cu + 2(IO3)
Cu(IO3)2 = 63.55 + 2(126.90 + 3(16.00)) = 63.55 + 2(126.90 + 48.00) = 63.55 + 2(174.90) = 63.55 + 349.80 = 413.35 g/mol
Now, we can proceed to calculate the solubility of copper(II) iodate in g dm-3.
Let's assume the solubility of Cu(IO3)2 is "x" mol dm-3. Since the stoichiometry of Cu(IO3)2 is 1:2 (Cu2+ : IO3-), the concentration of Cu2+ is also "x" mol dm-3, and the concentration of IO3- is "2x" mol dm-3.
Substituting these concentrations into the Ksp expression:
4 x 10-9 = (x)(2x)^2 = 4x^3
Simplifying the equation:
4 x 10-9 = 4x^3
Dividing both sides of the equation by 4 gives:
x^3 = 1 x 10-9
Taking the cube root of both sides:
x = (1 x 10-9)^(1/3) = 0.1 x 10-3 mol dm-3
Finally, we can convert the solubility from mol dm-3 to g dm-3 using the molar mass of Cu(IO3)2:
Solubility = (0.1 x 10-3 mol dm-3) x (413.35 g/mol) = 41.34 x 10-6 g dm-3
Converting from scientific notation to standard notation:
Solubility = 4.134 x 10-5 g dm-3
Therefore, the solubility of copper(II) iodate in g dm-3 is approximately 4.134 x 10-5 g dm-3.
Among the given options:
A) 1.85 × 10-2
B) 4.14 × 10-1
C) 5.21 × 10-1
D) 6.56 × 10
The correct option would be A) 1.85 × 10-2, as it is the closest value to the calculated solubility.