When 400ml of a 0.01M MgCl2 solution is mixed with a 300ml if a NaOH solution at pH 9.0, a precipitate will form. True or False?

To determine if a precipitate will form when 400 mL of a 0.01 M MgCl2 solution is mixed with a 300 mL NaOH solution at pH 9.0, we need to consider the solubility of magnesium hydroxide (Mg(OH)2).

First, let's calculate the number of moles of MgCl2 in 400 mL of a 0.01 M solution:
Moles of MgCl2 = concentration × volume
= 0.01 M × (400/1000) L
= 0.004 moles

Since MgCl2 dissociates into one Mg2+ ion and two Cl- ions when dissolved in water, the number of moles of Mg2+ ions is equal to the number of moles of MgCl2:
Moles of Mg2+ ions = 0.004 moles

Next, let's examine the solubility of Mg(OH)2. Magnesium hydroxide is sparingly soluble, with a solubility product constant (Ksp) of approximately 1.8 × 10^-11. When Mg(OH)2 dissolves in water, it dissociates into one Mg2+ ion and two OH- ions.

The pH of a solution is a measure of the concentration of hydrogen ions (H+). In an alkaline solution with a pH of 9.0, the concentration of OH- ions is higher than the concentration of H+ ions.

If the concentration of OH- ions exceeds the solubility product constant (Ksp), a precipitate will form. The concentration of OH- ions can be calculated using the following equation:

[OH-] = 10^(-pOH)

Given that the pH is 9.0, the pOH is 14 - pH = 14 - 9 = 5.

Therefore, [OH-] = 10^(-5) = 0.00001 M

Since the concentration of OH- ions is less than the concentration of Mg2+ ions (0.004 M), a precipitate will not form.

Hence, the statement "False" is correct.

To determine whether a precipitate will form when mixing the two solutions, we need to look at the chemical reaction that occurs between MgCl2 and NaOH. By balancing the equation, we can identify the products formed:

MgCl2 + 2NaOH --> Mg(OH)2 + 2NaCl

From the balanced equation, we can see that when MgCl2 reacts with NaOH, it forms Mg(OH)2 (magnesium hydroxide) and NaCl (sodium chloride).

Next, we need to consider the pH of the NaOH solution. A pH of 9.0 indicates that the NaOH solution is basic (alkaline). Magnesium hydroxide (Mg(OH)2) is sparingly soluble in water, and its solubility decreases with increasing pH. This means that as the pH gets higher, the solubility of Mg(OH)2 decreases, making it more likely for a precipitate to form.

Therefore, when you mix a 0.01M MgCl2 solution with a NaOH solution at pH 9.0, it is highly likely that a precipitate of Mg(OH)2 will form. Hence, the statement is most likely True.

(Mg^2+) = 0.01 x (400/700) = ?

pH of NaOH = 9 so pOH is 14-9 = 5 and (OH^-) = 1E-5M initially. After dilution it is
1E-5 x (300/700) = ?

Qsp = (Mg^2+)(OH^-)^2. Plug in the numbers to calculate Qsp then compare with Ksp for Mg(OH)2. If Qsp > Ksp a ppt will form.
Post your work if you get stuck.