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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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When $0.1\; mol$ of $CH_3NH_2$ (ionisation constant, $K_b = 5 \times 10^{-4}$) is mixed with $0.08\;mol\; HCl$ and the volume is made up to $1\; L$ find the [$H^+$] of resulting solution,

(a) $8 \times 10^{-2}$

(b) $2 \times 10^{-11}$

(c) $1.23 \times 10^{-4}$

(d) $8 \times 10^{-11}$

Can you answer this question?
 
 

1 Answer

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Answer: $8 \times 10^{-11}$
 
$CH_3NH_2 + HCl \rightleftharpoons CH_3NH_3^+Cl^-$
 
When 0.1 mol of $CH_3NH_2$ is mixed with 0.08 mole HCl, the resulting solution contains 0.08 mol of salt ($CH_3NH_3^+Cl^-$) and 0.02 (0.1-0.08 = 0.02) mol of base ($CH_3NH_2$).
 
Using Henderson's equation,
$pOH = pK_b + log \frac{[salt]}{[base]}$
$= - log 5 \times 10^{-4} + log \frac{0.08}{0.02}$
$= -(\bar4.6990) + 0.6021$
$= 4 - 0.6990 + 0.6021 = 3.9031$
 
$pH = 14 - pOH =14 - 3.9031 = 10.0969$
 
$[H^+] = antilog(-10.0969) = 8 \times 10^{-11}$

 

answered Nov 27, 2013 by mosymeow_1
 

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