What is the pOH of a 1 mM solution of Ca(OH) | vedclass

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(A) The concentration of the solution is $1 \ mM = 10^{-3} \ M$. Since $Ca(OH)_2$ is a strong base,it dissociates completely as: $Ca(OH)_2 \rightarrow

Vedclass · Accepted Answer

$2.7$

Vedclass · Answer

(A) The concentration of the solution is $1 \ mM = 10^{-3} \ M$. Since $Ca(OH)_2$ is a strong base,it dissociates completely as: $Ca(OH)_2 ightarrow Ca^{2+} + 2OH^-$. For every $1 	ext{ mole}$ of $Ca(OH)_2$,$2 	ext{ moles}$ of $OH^-$ ions are produced. Therefore,$[OH^-] = 2 	imes 10^{-3} \ M$. $pOH = -\log_{10} [OH^-] = -\log_{10} (2 	imes 10^{-3})$. $pOH = -(\log_{10} 2 + \log_{10} 10^{-3}) = -(0.301 - 3) = 2.699 \approx 2.7$.

What is the $pOH$ of a $1 \ mM$ solution of $Ca(OH)_2$?

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