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0.1 mole of {C}{H}_{{3}}{N}{H}_{{2}} {({K}_{{b}}={5}{10}^{{-{4

Question

0.1 mole of $C H_{3} N H_{2}$ $(K_{b} = 5 \times 10^{- 4})$ is mixed with 0.08 mole of HCl and diluted to 10 litre, what will be the $O H^{-}$ concentration in the solution ?

4.25 \times 10^{- 4} M

1.25 \times 10^{- 4} M

7.9 \times 10^{- 4} M

2.5 \times 10^{- 4} M

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William Discussed

0.1 mole of

C H_{3} N H_{2}

(K_{b} = 5 \times 10^{- 4})

is mixed with 0.08 mole of HCl and diluted to 10 litre, what will be the

O H^{-}

concentration in the solution ?

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Question Text	0.1 mole of $C H_{3} N H_{2}$ $(K_{b} = 5 \times 10^{- 4})$ is mixed with 0.08 mole of HCl and diluted to 10 litre, what will be the $O H^{-}$ concentration in the solution ?
Answer Type	Video solution: 1
Upvotes	150