SATHEE: Chemical and Ionic Equilibrium 2 Question 15

Chemical and Ionic Equilibrium 2 Question 15

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15. $\mathrm{CH}{3} \mathrm{NH}{2}\left(0.1\right. $m o l e,$ \left.K_{b}=5 \times 10^{-4}\right) $i s a d d e d t o 0.08 m o l e o f$ \mathrm{HCl}$ and the solution is diluted to one litre, resulting hydrogen ion concentration is

======= ####15. $\mathrm{CH}{3} \mathrm{NH}{2}\left(0.1\right. $m o l e,$ \left.K_{b}=5 \times 10^{-4}\right) $i s a d d e d t o 0.08 m o l e o f$ \mathrm{HCl}$ and the solution is diluted to one litre, resulting hydrogen ion concentration is

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $1.6 \times 10^{- 11}$

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

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

$(2005, 1 M)$

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Solution:

$$ \mathrm{CH}{3} \mathrm{NH}{2}+\mathrm{HCl} \longrightarrow \mathrm{CH}{3} \mathrm{NH}{3}^{+}+\mathrm{Cl}^{-} $$

Initial : $0.10 0.08 0 0$

$\begin{array}{lllll} Final : & 0.02 & 0 & 0.08 & 0.08 \end{array}$

$\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{\left[\mathrm{CH}{3} \mathrm{NH}{3}^{+}\right]}{\left[\mathrm{CH}{3} \mathrm{NH}{2}\right]}$

$= - \log (5 \times 10^{- 4}) + \log \frac{0.08}{0.02} = 3.9$

$pH = 14 - pOH = 10.1$

$[H^{+}] = 8 \times 10^{- 11}$

Chemical and Ionic Equilibrium 2 Question 15

15. $\mathrm{CH}{3} \mathrm{NH}{2}\left(0.1\right. $m o l e,$ \left.K_{b}=5 \times 10^{-4}\right) $i s a d d e d t o 0.08 m o l e o f$ \mathrm{HCl}$ and the solution is diluted to one litre, resulting hydrogen ion concentration is

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Chemical and Ionic Equilibrium 2 Question 15

15. $\mathrm{CH}{3} \mathrm{NH}{2}\left(0.1\right.mole,\left.K_{b}=5 \times 10^{-4}\right)isaddedto0.08moleof\mathrm{HCl}$ and the solution is diluted to one litre, resulting hydrogen ion concentration is

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

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15. $\mathrm{CH}{3} \mathrm{NH}{2}\left(0.1\right. $m o l e,$ \left.K_{b}=5 \times 10^{-4}\right) $i s a d d e d t o 0.08 m o l e o f$ \mathrm{HCl}$ and the solution is diluted to one litre, resulting hydrogen ion concentration is