Answer:
Molar
mass of \[Sr{{(OH)}_{2}}=88+34=122\]
Molarity, \[M=\frac{Strength}{Molar\,mass}=\frac{19.23}{122}=0.16\]
\[Sr{{(OH)}_{2}}\to S{{r}^{2+}}+2O{{H}^{-}}\]
\[[S{{r}^{2+}}]=0.16M,\,\,[O{{H}^{-}}]=2\times 0.16=0.32M\]
\[pOH=-\log [O{{H}^{-}}]\]
\[=-\log [0.32]=0.49\]
\[pH=14-pOH\]
\[=14-0.49=13.51\]
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