Question

Given that for `Ca(OH)_2`, `K_"sp"=4.8xx10^-6`, will precipitation occur if a `125*mL` volume of calcium chloride at `2.9xx10^-2*mol*L^-1` concentration is mixed with a `175*mL` volume of sodium hydroxide at `2.9xx10^-2*mol*L^-1` concentration?

Answer 1

We interrogate the equilibrium:

`Ca^(2+) + 2HO^(-) rightleftharpoons Ca(OH)_2(s)darr`

FOR which `K_"sp"=4.8xx10^-6`

Explanation:

Now we are given that `K_"sp"=4.8xx10^-6=[Ca^(2+)][HO^-]^2`.

And thus we must work out the concentrations with respect to `Ca^(2+)` and `HO^-`, and determine the initial ion product........`Q`

`Ca^(2+)=`

`"Moles of calcium ion"/"volume of solution"=(0.125*Lxx2.9xx10^-2*mol*L^-1)/(0.125*L+0.175*L)`

`=0.0121*mol*L^-1`

`HO^(-)=`

`"Moles of hydroxide ion"/"volume of solution"=(0.175*Lxx2.9xx10^-2*mol*L^-1)/(0.125*L+0.175*L)`

`=0.0169*mol*L^-1`

And the ion product `Q=[Ca^(2+)][HO^-]^2=3.46xx10^-6*mol*L^-1`

Since `K_"sp"=4.8xx10^-6` `>` `Q_"ion product"`, precipitation SHOULD NOT occur..........