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 pHpH of a solution prepared from a 313*mg313*mg mass of barium hydroxide dissolved in a 1*L1*L volume of water...?