$0.1 m o l$ 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:

$[H^{+}]$ = $3.68 \times 10^{- 11} M$

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$[H^{+}]$ = $1.79 \times 10^{- 11} M$

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$[H^{+}]$ = $6.53 \times 10^{- 11} M$

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

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Solution

The correct option is D
$[H^{+}]$ = $8 \times 10^{- 11} M$

$C H_{3} N H_{2}$ + $H C l$ $\to$ $C H_{3} N^{+} H_{3} C l^{-}$

At start
0.1 mol 0.08 mol 0 mol

On completion
0.02 mol 0 mol 0.08 mol

$p O H$ = $p K_{b}$ + $l o g \frac{[s a l t]}{[B a s e]}$ = $- l o g (5 \times 10^{- 4}) + l o g \frac{.08}{.02}$ = $3.9$

So $p H$ = $14$ - $3.9$ = $10.1$

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

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Similar questions

$0.1 m o l$ 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:

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 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?

0.1 m o l

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

is mixed with

0.08 m o l

H C l

and diluted to

1 L

. What will be the

H^{+}

ion concentration in the solution?

C H_{3} N H_{2}

(

0.1

mole,

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

is added to

0.08

moles of

H C l

and the solution is diluted to one litre. The resulting hydrogen ion concentration is :

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0.1 mol of methyl amine (Kb = 5 × 10−4) is mixed with 0.08 mol of HCl and the solution is diluted to 1 litre. The hydrogen ion concentration of the resulting solution is:

The correct option is D [H+] = 8 × 10−11 M CH3NH2 + HCl → CH3N+H3Cl− At start 0.1 mol 0.08 mol 0 mol On completion 0.02 mol 0 mol 0.08 mol pOH = pKb + log[salt][Base] = −log(5 × 10−4) + log.08.02 = 3.9 So pH = 14 - 3.9 = 10.1 [H+] = 8 × 10−11 M

$0.1 m o l$ 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: