Chemistry, 07.12.2020 08:30 gypsybrio1512
assuming complete dissociation, what is the pH of a 0.606 M Ba(OH)2 solution? round to 3 significant figures
Answers: 3
![[OH^-]=0.606\frac{molBa(OH)_2}{L}*\frac{2molOH^-}{1molBa(OH)_2} =1.21M](/tpl/images/0955/7598/217d3.png)
![pOH=-log([OH^-])=-log(1.12)=-0.0500](/tpl/images/0955/7598/bf2a1.png)
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assuming complete dissociation, what is the pH of a 0.606 M Ba(OH)2 solution? round to 3 significant...
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