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

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Answer 1 · 2018-02-18T13:43:18

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

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....?