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?

1 Answer
anor277
Feb 18, 2018

Well...pOH=-log_10[HO^-]pOH=−log10[HO−]

Explanation:

And we got barium hydroxide...

Ba(OH)_2(s) stackrel(H_2O)rarrBa^(2+) + 2HO^-Ba(OH)2(s)H2O−−→Ba2++2HO−

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

[HO^-]=(0.01*mol*L^-1xx50xx10^-3*Lxx2)/(100*mLxx10^-3*L*mL^-1)[HO−]=0.01⋅mol⋅L−1×50×10−3⋅L×2100⋅mL×10−3⋅L⋅mL−1

=0.010*mol*L^-1=0.010⋅mol⋅L−1..

pOH=-log_10(0.010)=-log_(10)10^(-2)=-(-2)=+2pOH=−log10(0.010)=−log1010−2=−(−2)=+2

And what is pHpH of this solution....?

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