Question

Calculate $\left[{\mathrm{OH}}^{-}\right]$ in a$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{M}$ solution of$\mathbf{Ca}{\left(\mathbf{OH}\right)}_{\mathbf{2}}$ .

Short answer

The concentration of$O{H}^{-}$is calculated as

role="math" localid="1663754873733" $$\begin{gathered} \frac{1}{2}Ca{\left(OH\right)}_2=\frac{1}{2}\times 3.0\times {10}^{-7}M \\ =1.5\times {10}^{-7} M \end{gathered}$$

Thus, the concentration of $O{H}^{-}$in the solution is$1.5\times {10}^{-7} M$ .

Step-by-step solution

Define pH of the solution

The pH of the solution is prepared by taking the negative logarithm of hydrogen ion concentration, whereas the pOH of the solution is prepared by taking the negative logarithm of hydroxide ion concentration.

Calculate [OH-] concentration

The concentration of$Ca{\left(OH\right)}_2$is$3.0\times {10}^{-7}M$$Ca{\left(OH\right)}_2$.is dissociated as

$Ca{\left(OH\right)}_2\to C{a}^{2+}+2O{H}^{-}$

So, the concentration of$O{H}^{-}$is$\frac{1}{2}Ca{\left(OH\right)}_2$, whichis calculated as

role="math" localid="1663754910236" $$\begin{gathered} \frac{1}{2}Ca{\left(OH\right)}_2=\frac{1}{2}\times 3.0\times {10}^{-7}M \\ =1.5\times {10}^{-7} M \end{gathered}$$

Thus, the concentration of $O{H}^{-}$n the solution is$1.5\times {10}^{-7} M$ .