`BOH` is a weak base, molar concentration of `BOH` that provides a `[OH]^(-)` of `1.5xx10^(-3) M[K_(b)(BOH)=1.5xx10^(-5) M]` is

Question

`BOH` is a weak base, molar concentration of `BOH` that provides a `[OH]^(-)` of `1.5xx10^(-3) M[K_(b)(BOH)=1.5xx10^(-5) M]` is

`BOH` is a weak base, molar concentration of `BOH` that provides a `[OH]^(-)` of `1.5xx10^(-3) M[K_(b)(BOH)=1.5xx10^(-5) M]` is
A. `1.5xx10^(-5)M`
B. `0.015M`
C. `0.0015`
D. `0.15M`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`BOH hArr B^(+) + OH^(-) `
`K_(b)=([B^(+)][OH^(-)])/([BOH])`
`[B^(+)]=[OH^(-)]`
`:. K_(b)=([OH^(-)]^(2))/([BOH])`
`[BOH]=([OH^(-)]^(2))/(K_(b))`
`=((1.5 xx 10^(-3))^(2))/(1.5 xx 10^(-5))=1.5 xx 10^(-1)M`
`=0.15 M`

4 views Answered 2023-02-05 15:29