Question

The ph of 0.010 m aqueous aniline is 8.32. What is the percentage protonated?

Answer 1

Answer : The percentage aniline protonated is, 0.0209 %

Explanation :

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-8.32\\\\pOH=5.68$

Now we have to calculate the hydroxide ion concentration.

$pOH=-\log [OH^-]$

$5.68=-\log [OH^-]$

$[OH^-]=2.09\times 10^{-6}M$

The equilibrium chemical reaction will be:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

From the reaction we conclude that,

Concentration of $OH^-$ ion = Concentration of $NH_4^+$ ion = $2.09\times 10^{-6}M$

Now we have to calculate the percentage aniline protonated.

$\text{percentage aniline protonated}=\frac{2.09\times 10^{-6}M}{0.010M}\times 100$

$\text{percentage aniline protonated}=0.0209\%$

Thus, the percentage aniline protonated is, 0.0209 %