Question

calculate the ph of a solution made by mixing 50ml of 0.01 m of ba (oh)2 with 50 ml of water

Answer 1

Well...pOH=−log10[HO−]

Explanation:

And we got barium hydroxide...

Ba(OH)2(s)H2O−−→Ba2++2HO−

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

[HO−]=0.01⋅mol⋅L−1×50×10−3⋅L×2100⋅mL×10−3⋅L⋅mL−1

=0.010⋅mol⋅L−1..

pOH=−log10(0.010)=−log1010−2=−(−2)=+2

And what is pH of this solution....?