Question

1)The concentration of [Mg+2] ions in the Mg(NO3)2 saturated solution is given as 1.0x10^-2 M. When this (strong electrolyte) saturated solution was mixed with NaOH(strong base),its final concentration was 1.0x10^-4 M Mg(NO3)2.

Do you think a precipitate will form?
Calculate and prove it whether there is a precipitate due to presence of the mixture using the clue given below. (Ksp=16*10^-12)

Answer 1

Answer: Since Q (1.0x10^-10) is less than Ksp (16x10^-12), a precipitate will not form

Explanation: To calculate Q, we need to determine the concentration of OH- ions in the solution. Since NaOH is a strong base, it completely dissociates in water to form Na+ and OH- ions. Therefore, the concentration of OH- ions in the solution is equal to the concentration of NaOH, which is 1.0x10^-4 M.

Now we can calculate Q: Q = [Mg+2][OH-]^2 = (1.0x10^-2)(1.0x10^-4)^2 = 1.0x10^-10.

Therefore, a precipitate will not form.

Answer 2

Since this value is less than the Ksp of Mg(OH)2, which is 16*10^-12, no precipitate will form. Therefore, the solution will remain clear.

Step-by-step explanation:

Based on the given information, we can calculate the initial concentration of Mg+2 ions in the Mg(NO3)2 saturated solution as 1.0x10^-2 M. When the solution is mixed with NaOH, a precipitation reaction may occur if the resulting concentration of Mg+2 ions exceeds the solubility product constant (Ksp) of Mg(OH)2, which is 16*10^-12.

The balanced chemical equation for the precipitation reaction is:

Mg+2 + 2OH- → Mg(OH)2

To determine if a precipitate will form, we need to calculate the concentration of Mg+2 ions in the final solution after mixing with NaOH. Since Mg(NO3)2 is a strong electrolyte, it will dissociate completely in water to give Mg+2 and NO3- ions. Therefore, the initial concentration of Mg+2 ions can be used to calculate the number of moles of Mg+2 ions present in the solution.

n(Mg+2) = C(Mg+2) x V

where C(Mg+2) is the initial concentration of Mg+2 ions and V is the volume of the solution.

n(Mg+2) = 1.0x10^-2 x V

After mixing with NaOH, Mg+2 ions will react with OH- ions to form Mg(OH)2. Since Mg(OH)2 is a sparingly soluble salt, it will precipitate out of solution until the concentration of Mg+2 and OH- ions reaches a value corresponding to the Ksp of Mg(OH)2.

The concentration of Mg+2 ions in the final solution can be calculated using the following equation:

[Mg+2] = n(Mg+2) / (V + V')

where V is the initial volume of the Mg(NO3)2 solution and V' is the volume of NaOH added.

Since we know that the final concentration of Mg(NO3)2 is 1.0x10^-4 M, we can use the dilution equation to calculate V':

C1V1 = C2V2

where C1 is the initial concentration of Mg(NO3)2, C2 is the final concentration, V1 is the initial volume of the solution and V2 is the final volume after mixing.

V' = (C1V1 - C2V2) / C2

Substituting the values, we get:

V' = (1.0x10^-2 x V - 1.0x10^-4 x (V + V')) / 1.0x10^-4

Solving for V', we get:

V' = 98.04 mL

Therefore, the total volume of the final solution is 100 mL (V + V').

Substituting the values in the equation for [Mg+2], we get:

[Mg+2] = 9.8x10^-5 M

Since this value is less than the Ksp of Mg(OH)2, which is 16*10^-12, no precipitate will form. Therefore, the solution will remain clear.