Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ If 2.00 mL of 0.200 M of KOH are added to 1.00 L of

0.100 M CaCl2, will a precipitate

form? Given value Ksp =

[Ca2+][OH-]2 = 8.0 x

10-6

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.

Learn more about precipitation reactions here:

https://brainly.com/question/11081618

#SPJ14