http://www.chemteam.info/Equilibrium/Calc-Ksp-From

Problem #2 is CaF2, so follow those steps, but using your PbBr2 data.

http://www.chemteam.info/Equilibrium/Calc-Ksp-From

[Pb2 ]= 1.0 x 10^-2 mol/L

[Br-]= 2 x 1.0 x 10^-2 mol/L = 2.0 x 10^-2 mol/L

Look here:

[Pb2 ]= 1.0 x 10^-2 mol/L

[Br-]= 2 x 1.0 x 10^-2 mol/L = 2.0 x 10^-2 mol/L

Look here:

Problem #2 is CaF2, so follow those steps, but using your PbBr2 data.

enable the molar solubilit be S Then [Ca2 ] = S [IO3]2- = 2S The solubility product = S x (2S)^2 = 4*S^3 = 7.a million x 10^-7 = 710 x 10^-9 which on fixing yields S = 8.9 x 10^-3 and the molar solubility For the subsequent party, you consider x because the molar solubility of Ca(IO3)2 in water and then the most objective of IO3 must be (2x 0.06) for the reason that NaIO3 is carefully soluble and then [Ca2 ] = x then writing the solubility product equation x * (2x 0.06) = 7.a million x 10^-7 and be certain for x, it quite is the molar solubility Ca(IO3)2 in the presence of 0.06 M NaIO3

This post is last updated on hrtanswers.com at Date : 1st of September – 2022