A solution of 0.1 mol of CH_3NH_2 (K_b = 5 t | vedclass

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(B) The reaction is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^-$Initial moles: $0.1 \ mol$ of $CH_3NH_2$ and $0.08 \ mol$ of $HCl$.After reaction:

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$3.9$

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(B) The reaction is: $CH_3NH_2 + HCl ightarrow CH_3NH_3^+ + Cl^-$Initial moles: $0.1 \ mol$ of $CH_3NH_2$ and $0.08 \ mol$ of $HCl$.After reaction: $0.02 \ mol$ of $CH_3NH_2$ remains and $0.08 \ mol$ of $CH_3NH_3Cl$ is formed.This forms a basic buffer solution.The $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} ight)$.$pK_b = -\log(5 	imes 10^{-4}) = 4 - \log 5 = 4 - 0.7 = 3.3$.$pOH = 3.3 + \log \left( \frac{0.08}{0.02} ight) = 3.3 + \log(4) = 3.3 + 0.6 = 3.9$.

Buffer Solution	Volume (in $mL$) of $0.1 \ M$ Weak acid	Volume (in $mL$) of $0.1 \ M$ sodium salt of weak acid
$I$	$4.0$	$4.0$
$II$	$4.0$	$40.0$
$III$	$40.0$	$4.0$
$IV$	$0.1$	$10.0$

$A$ solution of $0.1 \ mol$ of $CH_3NH_2$ $(K_b = 5 \times 10^{-4})$ and $0.08 \ mol$ of $HCl$ is diluted to $1 \ L$. The $pOH$ of the resulting solution is $(\log 1.25 = 0.1)$.

Similar Questions

Which one of the following pairs of solutions is not an acidic buffer?

Study the following table:
Buffer Solution Volume (in $mL$) of $0.1 \ M$ Weak acid Volume (in $mL$) of $0.1 \ M$ sodium salt of weak acid
$I$ $4.0$ $4.0$
$II$ $4.0$ $40.0$
$III$ $40.0$ $4.0$
$IV$ $0.1$ $10.0$

Which of the two sets of buffer solutions have the least $pH$?

Which among the following pairs is not an acidic buffer?

Amongst the following solutions,the buffer solution is

When an acid or alkali is mixed with a buffer solution,the $pH$ of the buffer solution:

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