Answer:
[tex]0.00068M[/tex]
Explanation:
Hello there!
In this case, according to the ionization of calcium hydroxide, a strong base:
[tex]Ca(OH)_2\rightarrow Ca^{2+}+2OH^-[/tex]
Thus, since there is a 1:2 mole ratio of calcium hydroxide to hydroxide ions, we apply the following proportional factor to obtain:
[tex]0.00034\frac{molCa(OH)_2}{L}*\frac{2molOH^-}{1molCa(OH)_2} \\\\=0.00068\frac{OH^-}{L}\\\\=0.00068M[/tex]
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