How do you determine the OH (aq) concentration in 1.0 M aniline ( $C_{6} H_{5} N H_{2}$ $C_6H_5NH_2$ ) solution. (The $K_{b}$ $K_b$ for aniline is $4.0 x 10^{- 10}$ $4.0 x 10^-10$ ?

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Answer 1 · 2017-01-08T23:14:26

You can do it like this:

For a weak base:

$sf(pOH=1/2[pK_b-logb])$

Where b is the concentration of the base.

$sf(pK_b=-logK_b=-log(4xx10^(-10))=9.34)$

$:.$ $sf(pOH=1/2[9.34-log(1)]$

$:.$ $sf(pOH=4.67)$

$:.$ $sf(-log[OH^-]=4.67)$

$:.$ $sf([OH^-]=2.14xx10^(-5)color(white)(x)"mol/l")$

How do you determine the OH (aq) concentration in 1.0 M aniline (C_6H_5NH_2C6H5NH2C_6H_5NH_2) solution. (The K_bKbK_b for aniline is 4.0 x 10^-104.0x10−104.0 x 10^-10?