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Lewis Theory

0.1 mol of ...

Question

$0.1 m o l$ of $C H_{3} N H_{3} (K_{b} = 5 \times 10^{- 4})$ is mixed with $0.08 m o l$ of $H C l$ and diluted to $1 L$ . What will be the $H^{+}$ ion concentration in the solution?

8 \times 10^{- 2} M

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8 \times 10^{- 11} M

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1.6 \times 10^{- 11} M

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8 \times 10^{- 5} M

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Solution

The correct option is B $8 \times 10^{- 11} M$
$C H_{3} N H_{2} + H C l \to C H_{3} N H_{3}^{+} + C l^{-}$
At Initial 0.1 0.08
At equilibrium 0.02 0.08
$[O H^{-}] = \frac{K_{b} [C H_{3} N H_{2}]}{[C H_{3} N H_{3}^{+}]}$
$[O H^{-}] = \frac{5 \times 10^{- 4} \times 0.02}{0.08}$
$[O H^{-}] = 1.25 \times 10^{- 4}$
$[H^{+}] = \frac{k_{w}}{[O H^{-}]}$
$[H^{+}] = \frac{10^{- 14} \times 4}{5 \times 10^{- 4}}$
$[H^{+}] = 8 \times 10^{- 11}$

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Similar questions

Q. 0.1 mol of

C H_{3} N H_{2} (K_{b} = 6 \times 10^{- 4})

is added to 0.08 mol of

H C l

and the solution is diluted to

1 L

. The hydrogen ion concentration in the solution is:

$0.1 m o l$ of methyl amine ( $K_{b}$ = $5 \times 10^{- 4}$ ) is mixed with $0.08 m o l$ of $H C l$ and the solution is diluted to $1 l i t r e$ . The hydrogen ion concentration of the resulting solution is:

0.1 m o l

C H_{3} N H_{2}

(K_{b} = 5 \times 10^{- 4})

is mixed with

0.08 m o l

H C l

and diluted to

1 L

. Which statement(s) is/are correct?
(take

l o g 2 = 0.3)

0.1 mole of $C H_{3} N H_{2} (K_{b} = 5 \times 10^{- 4}$ is mixed with 0.08 mole of $H C l$ and

diluted to one litre. What will be the $H^{+}$ concentration in the solution?

Q. When

0.1

mol of

C H_{3} N H_{2} (K_{b} = 5 \times 10^{- 4})

is mixed with

0.08

mol of

H C l

and diluted to

1 L

, what will be the

H^{\oplus}

concentration in the solution?

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0.1 mol of CH3NH3(Kb=5×10−4) is mixed with 0.08 mol of HCl and diluted to 1 L. What will be the H+ ion concentration in the solution?

The correct option is B 8×10−11M CH3NH2+HCl→CH3NH+3+Cl−At Initial 0.1 0.08At equilibrium 0.02 0.08[OH−]=Kb[CH3NH2][CH3NH+3][OH−]=5×10−4×0.020.08[OH−]=1.25×10−4[H+]=kw[OH−][H+]=10−14×45×10−4[H+]=8×10−11

$0.1 m o l$ of $C H_{3} N H_{3} (K_{b} = 5 \times 10^{- 4})$ is mixed with $0.08 m o l$ of $H C l$ and diluted to $1 L$ . What will be the $H^{+}$ ion concentration in the solution?