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[tex]~~~~~\text{pH} + \text{pOH} = 14\\\\\implies \text{pOH} = 14- \text{pH}\\\\\implies \text{pOH} = 14 - 11.8\\\\\implies \text{pOH} = 2.2\\\\\text{Now,}\\\\~~~~~~~\text{pOH}= -\log\left[\text{OH}^{-1}\right]\\\\\implies \left[\text{OH}^{-1} \right] = 10^{-\text{pOH}} = 10^{-2.2}~ = 0.006309~M\\\\\text{Hence the concentration of OG}^{-1}~ \text{is 0.006309~M}[/tex]