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![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](/tpl/images/0398/9887/3eb8f.png)
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Calculate the oh− concentration after 53 ml of the 0.100 m koh has been added to 25.0 ml of 0.200 m...
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