Question

What is the `pH` of a solution that is prepared from a `4.08*g` mass of calcium hydroxide dissolved in a `1*L` volume?

Answer 1

See this old answer.

Explanation:

And so we know that in aqueous conditions,

`pH+pOH=14`

Here we assume a `1*L` volume of solution, for which `rho=1.02*g*mL^-1`....and thus `"solution mass"=1020*g`...and `0.4%` mass is due to the SOLUTE, `Ca(OH)_2`, i.e. a mass of `4.08*g`

And so we can assess `"solution concentration with respect to"` `"calcium hydroxide"`.

`[CaOH_2(aq)]=((4.08*g)/(74.09*g*mol^-1))/(1.0*L)=0.0551*mol*L^-1`.

And so `[HO^-]=0.110*mol*L^-1`, and `pOH=-log_10(0.110)=0.95`; and `pH=14-0.95=13.0`.

Whew, arithmetic......if there is a step that you do not follow, comment...