A weak base MOH of 0.1, N concentration | vedclass

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Question

$\mathrm{pOH}=5$

$\mathrm{OH}^{-}=10^{-5}$

$\mathrm{OH}=\mathrm{C} \alpha$

$\alpha=\frac{10^{-5}}{10^{-1}}=10^{-4} ; \alpha(	ext { in } \%)=

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

$\mathrm{pOH}=5$

$\mathrm{OH}^{-}=10^{-5}$

$\mathrm{OH}=\mathrm{C} \alpha$

$\alpha=\frac{10^{-5}}{10^{-1}}=10^{-4} ; \alpha(	ext { in } \%)=10^{-2}$

A weak base $MOH$ of $0.1\, N$ concentration shows a $pH$ value of $9$. What is the percentage degree of ionisation of the base ? ......$\%$

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