Answer :
Final answer:
No, a precipitate will not form when 2.00 mL of 0.200 M KOH are added to 1.00 L of 0.100 M CaCl2.
Explanation:
To determine if a precipitate will form when 2.00 mL of 0.200 M KOH is added to 1.00 L of 0.100 M CaCl2, we need to compare the product of the concentrations of calcium ions and hydroxide ions with the Ksp value for Ca(OH)2.
First, we need to calculate the concentration of hydroxide ions ([OH-]) in the 2.00 mL of 0.200 M KOH solution. Since the volume is given in mL, we need to convert it to L:
2.00 mL = 2.00 x 10^-3 L
Next, we can use the formula:
[OH-] = Molarity x Volume
[OH-] = 0.200 M x 2.00 x 10^-3 L
[OH-] = 4.00 x 10^-4 M
Now, we can calculate the product of the concentrations of calcium ions and hydroxide ions:
[Ca2+][OH-]2 = (0.100 M) x (4.00 x 10^-4 M)^2
[Ca2+][OH-]2 = 1.60 x 10^-10
Since the product of the concentrations is less than the Ksp value (8.0 x 10^-6), a precipitate will not form.
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