Question

Calculate the POH of a solution made by mixing 50 ml of 0.01 M `Ba(OH)_2` solution with 50 ml water?

Answer 1

Well...`pOH=-log_10[HO^-]`

Explanation:

And we got barium hydroxide...

`Ba(OH)_2(s) stackrel(H_2O)rarrBa^(2+) + 2HO^-`

And to access `[HO^-]` we take....the quotient of the product...

`[HO^-]=(0.01*mol*L^-1xx50xx10^-3*Lxx2)/(100*mLxx10^-3*L*mL^-1)`

`=0.010*mol*L^-1`..

`pOH=-log_10(0.010)=-log_(10)10^(-2)=-(-2)=+2`

And what is `pH` of this solution....?