[candiishop]

Calculate the solubility of Ca(OH)2 in a 1.950M NaOH solution at 53°C, given that the Ksp of Ca(OH)2 is 7.22 x 10-6 at that temperature.
Give your answer in micro Molar units.

a) 3.703 uM

b) 2.184 uM

c) 1.899 uM

d) 2.089 uM

e) 0.949 uM

Hi, sorry im so lost in chemistry. I tried this question above and turns out i got 1.899 micro molar units. I'm not sure this is right though.

Ca(OH)2 <-----> [Ca2+] [OH-]2
             <------

NaOH ------> [Na+] [OH-]

Ksp= [Ca2+] [OH-]^2
7.22x10-6 = [x ] [1.950]^2
[x ] = 1.866uM

[Borek]

That's OK. You have calculated it correctly, just your use of symbols is somewhat twisted

[Ca²⁺ ] usually means "concentration of Ca²⁺ cations", thus below equations look strange:

Ca(OH)2 <-----> [Ca2+] [OH-]2

NaOH ------> [Na+] [OH-]

These are neither formulas for K_so nor balanced reaction equations.

And you don't have to use some [x ] notation, as you are solving just for [Ca²⁺ ].