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Question: When 0.1 mole of \(C{{H}_{3}}N{{H}_{2}}\) (ionization constant, \({{K}_{b}}=5\times {{10}^{-4}}\)) i...

When 0.1 mole of CH3NH2C{{H}_{3}}N{{H}_{2}} (ionization constant, Kb=5×10−4{{K}_{b}}=5\times {{10}^{-4}}) is mixed with 0.08 mole HCl and the volume is made up to 1 litre, find the [H+]\left[ {{H}^{+}} \right] of the resulting solution.
(A) 8×10−28\times {{10}^{-2}}
(B) 2×10−112\times {{10}^{-11}}
(C) 1.23×10−41.23\times {{10}^{-4}}
(D) 8×10−118\times {{10}^{-11}}

Explanation

Solution

Start by writing the chemical equation for the reaction. Write the number of moles of both reactants and products before and after the reaction. Then, find the concentration of the acidic or basic species produced after the reaction. Proceed accordingly.

Complete step by step solution:
-Let’s start by writing the chemical equation for the reaction mentioned in the question.
CH3NH2+HCl→CH3NH3+Cl−C{{H}_{3}}N{{H}_{2}}+HCl\to C{{H}_{3}}NH_{3}^{+}C{{l}^{-}}
When methylamine reacts with HCl, methylammonium chloride salt is formed.
CH3NH2+HCl→CH3NH3+Cl−C{{H}_{3}}N{{H}_{2}}+HCl\to C{{H}_{3}}NH_{3}^{+}C{{l}^{-}}
Initially -0.1 - 0.08 -0
After completion of the reaction -0.02 -0 -0.08
So, after completion of the reaction, 0.08 mole of salt and 0.02 mole of unreacted methylamine is left behind.
This is an example of alkaline buffer solution.
Now, we need to calculate the [OH−]\left[ O{{H}^{-}} \right] using the formula,
[OH−]=Kb×[Base][Conjugate Acid] =5×10−4×0.020.08 [OH−]=1.25×10−4M\begin{aligned} & \left[ O{{H}^{-}} \right]={{K}_{b}}\times \dfrac{\left[ Base \right]}{\left[ Conjugate\,Acid \right]} \\\ & =5\times {{10}^{-4}}\times \dfrac{0.02}{0.08} \\\ & \left[ O{{H}^{-}} \right]=1.25\times {{10}^{-4}}M \end{aligned}
Now we know, [H+][OH−]=kw=10−14\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{k}_{w}}={{10}^{-14}}
∴[H+]=10−14[OH−]=10−141.25×10−4=8×10−11M\therefore \left[ {{H}^{+}} \right]=\dfrac{{{10}^{-14}}}{\left[ O{{H}^{-}} \right]}=\dfrac{{{10}^{-14}}}{1.25\times {{10}^{-4}}}=8\times {{10}^{-11}}M
Therefore, concentration of hydrogen ions is 8×10−11M8\times {{10}^{-11}}M.

Hence, the correct option is option (D) 8×10−11M8\times {{10}^{-11}}M.

Note: Remember the formula of Henderson-Hasselbalch equation. The problem can also be using that equation. Just once the concentration of hydroxyl ions is found, take the negative log of that and pOH is obtained. Then using the relation, pH + pOH=14, pH can be determined. Taking antilog of negative pH will give you the concentration of hydrogen ions present in the solution.