Will a precipitate form when 100.0 mL of a 7.5 10-4 M Mg(NO3)2 is added to 100.0 mL of 5.8 10-4 M NaOH? (Hint: Ksp for Mg(OH)2 = 8.9 10-12)
Give the value of Q, to support your answer.
Calculate the M of Mg ion and NaOH and solve for Qsp. If Qsp exceeds Ksp it will ppt. If not it will not ppt.
Oh, chemistry, the realm of precipitates and equations! Let me put on my lab coat, and let the clown science begin!
To determine if a precipitate will form, we need to compare the value of the reaction quotient (Q) with the solubility product constant (Ksp).
Q is calculated by multiplying the concentrations of the product ions (Mg2+ and OH-) raised to their respective stoichiometric coefficients. In this case, the stoichiometry of the reaction is 1:2, so Q = [Mg2+] * [OH-]^2.
Now, let's plug in the given concentrations:
[Mg2+] = 7.5 * 10^-4 M
[OH-] = 5.8 * 10^-4 M
Then, the value of Q will be:
Q = (7.5 * 10^-4) * (5.8 * 10^-4)^2
And if your calculator doesn't explode from tiny numbers, you'll find a value for Q. Now, let's compare it to the solubility product constant (Ksp = 8.9 * 10^-12).
If Q is greater than Ksp, then the reaction is "not happy," and a precipitate will form. But if Q is less than Ksp, then the reaction is "ecstatic," and no precipitate will occur.
So, dear scientist, calculate that Q, compare it to Ksp, and unveil the answer! Also, don't forget to have some clown fun in the lab!
To determine if a precipitate will form when Mg(NO3)2 is added to NaOH, we need to compare the value of Q to the solubility product constant (Ksp) for Mg(OH)2.
First, let's write the balanced chemical equation for the reaction:
Mg(NO3)2 + 2NaOH -> Mg(OH)2 + 2NaNO3
From the equation, we can see that the molar ratio between Mg(NO3)2 and Mg(OH)2 is 1:1. Therefore, the concentration of Mg(OH)2 will be the same as the concentration of Mg(NO3)2.
Given:
Volume of Mg(NO3)2 solution (V1) = 100.0 mL
Concentration of Mg(NO3)2 (C1) = 7.5 * 10^-4 M
Volume of NaOH solution (V2) = 100.0 mL
Concentration of NaOH (C2) = 5.8 * 10^-4 M
Using the formula for Q (reaction quotient), we can calculate its value:
Q = [Mg(OH)2]/[NaNO3]^2
= (C1 * V1) / (C2^2 * V2)
= (7.5 * 10^-4 M * 100.0 mL) / ((5.8 * 10^-4 M)^2 * 100.0 mL)
= (7.5 * 10^-8 mol/L * 0.1 L) / ((5.8 * 10^-8 mol/L)^2 * 0.1 L)
= 1.2931
The value of Q is equal to 1.2931.
To determine if a precipitate will form, we compare the value of Q to the solubility product constant (Ksp) for Mg(OH)2.
Since Q > Ksp, the value of Q exceeds the solubility product constant (Ksp), indicating that Mg(OH)2 will exceed its solubility and a precipitate will form.
To determine if a precipitate will form when 100.0 mL of a 7.5x10^-4 M Mg(NO3)2 is added to 100.0 mL of 5.8x10^-4 M NaOH, we need to compare the value of Q to the solubility product constant (Ksp) for Mg(OH)2.
First, let's write the balanced chemical equation for the reaction between Mg(NO3)2 and NaOH:
Mg(NO3)2 + 2NaOH -> Mg(OH)2 + 2NaNO3
From the balanced equation, we can see that the ratio of Mg(NO3)2 to Mg(OH)2 is 1:1. Therefore, the concentration of Mg(OH)2 in the solution will be the same as the concentration of Mg(NO3)2.
Now, let's calculate the concentration of Mg(OH)2:
Concentration of Mg(OH)2 = concentration of Mg(NO3)2 = 7.5x10^-4 M
Next, we can calculate the value of Q, which is the reaction quotient, using the formula:
Q = [Mg2+][OH-]^2
Since the concentration of Mg(OH)2 is the same as the concentration of Mg(NO3)2, we can substitute the value of Mg(NO3)2 into the equation:
Q = (7.5x10^-4)^2 = 5.6x10^-7
Finally, we compare the value of Q to the solubility product constant (Ksp) for Mg(OH)2, which is given as 8.9x10^-12.
If the value of Q is greater than Ksp, then a precipitate will form. If Q is less than or equal to Ksp, then no precipitate will form.
In this case, Q (5.6x10^-7) is less than Ksp (8.9x10^-12), which means that no precipitate will form when 100.0 mL of a 7.5x10^-4 M Mg(NO3)2 is added to 100.0 mL of 5.8x10^-4 M NaOH.