What is #pH# of a solution prepared from a #313mg# mass of barium hydroxide dissolved in a #1L# volume of water...?

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Answer 1 · 2017-07-01T04:34:07

#pH=9.56#.......

By definition, #pH+pOH=14#, and thus.......

#pOH=14-pH#, and #pH=14-pOH#......

But #pOH=-log_10[HO^-]#

#=-log_10{((2xx3.13xx10^-3*g)/(171.34*g*mol^-1))/(1*L)}#

#=-log_10{3.65xx10^-5}=-(-4.44)=4.44#

And so #pH=14-pOH=14-4.44=9.56#

Why did I double the #[Ba(OH)_2]# to get #[HO^-]#?

What is #pH# of a solution prepared from a #313*mg# mass of barium hydroxide dissolved in a #1*L# volume of water...?