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Which of the following is correct for 0.1 M BOH solution ${(K_{b} = 10^{- 5})}^{}$ )?

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Phenolphthalein indicator can be used during the titration of BOH with HCI.

pH of solution is 11

It's salt with HCl (i.e. BCI) form acidic solution in water

${OH}^{-}$ concentration is $10^{- 3} mol / L$

answer is A, B, C.

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Detailed Solution

It is known that,

$\begin{array}{l} pOH = \frac{1}{2} [{pK}_{b} - \log c] \\ pOH = = \frac{1}{2} [5 + 1] = 3 .0 \\ \Rightarrow pH = 11 \end{array}$

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