Question

What is `pH` of a solution prepared from a `313*mg` mass of barium hydroxide dissolved in a `1*L` volume of water...?

Answer 1

`pH=9.56`.......

Explanation:

By definition, `pH+pOH=14`, and thus.......

`pOH=14-pH`, and `pH=14-pOH`......

But `pOH=-log_10[HO^-]`

`=-log_10{((2xx3.13xx10^-3*g)/(171.34*g*mol^-1))/(1*L)}`

`=-log_10{3.65xx10^-5}=-(-4.44)=4.44`

And so `pH=14-pOH=14-4.44=9.56`

Why did I double the `[Ba(OH)_2]` to get `[HO^-]`?