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....?

Calculate the POH of a solution made by mixing 50 ml of 0.01 M Ba(OH)_2Ba(OH)_2 solution with 50 ml water?