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Henderson Hassleback Equation

HCl gas is pa...

Question

$H C l$ gas is passed through an aqueous solution of 0.1 M 1-aminopropane $(P r N H_{2})$ till the pH reaches 9.71. Calculate the ratio of $[P r N H_{2}] / [P r N H_{3}^{+}]$ in this solution. $[K_{a} (P r N H_{3}^{+}) = 1.96 \times 10^{- 11}]$
$l o g (1.96) = 0.292, l o g (0.1) = - 1$

10:1

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8:1

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4:1

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1:10

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Solution

The correct option is D 1:10
Given, $[P r N H_{2}] = 0.1, p H = 9.71, [K_{a} (P r N H_{3}^{+}) = 1.96 \times 10^{- 11}]$
Using the Henderson-Hasselbalch equation for weak bases:
$p H = p K_{a} + l o g [\frac{[c o n j u g a t e b a s e]}{[a c i d]}]$
$p H = p K_{a} + l o g [\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]}]$
$p H = - l o g K_{a} + l o g [\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]}]$
$p H + l o g K_{a} = l o g [\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]}]$
$9.71 + l o g (1.96 \times 10^{- 11}) = l o g [\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]}]$
$9.71 - 11 + l o g (1.96) = l o g [\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]}]$
$9.71 - 11 + 0.292 = l o g [\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]}]$
$l o g [\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]}] = - 1$ $\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]} = 0.1$
$\frac{[P r N H_{2}]}{[P r N H_{3}^{+}]} = 1 : 10$

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Similar questions

A buffer is a solution of weak acid and its conjugate base or a weak base and its conjugate acid. It resists the change in

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H C l

gas is passed through an aqueous solution of 0.1M 1-aminopropane

(P r N H_{2})

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[P r N H_{2}] / [P r N H_{3}^{+}]

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[K_{a} (P r N H_{3}^{+}) = 1.96 \times 10^{- 11}

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Henderson Hassleback Equation

Standard XIII Chemistry

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HCl gas is passed through an aqueous solution of 0.1 M 1-aminopropane (PrNH2) till the pH reaches 9.71. Calculate the ratio of [PrNH2]/[PrNH+3] in this solution. [Ka(PrNH+3)=1.96×10−11] log (1.96)=0.292, log (0.1)=−1

$H C l$ gas is passed through an aqueous solution of 0.1 M 1-aminopropane $(P r N H_{2})$ till the pH reaches 9.71. Calculate the ratio of $[P r N H_{2}] / [P r N H_{3}^{+}]$ in this solution. $[K_{a} (P r N H_{3}^{+}) = 1.96 \times 10^{- 11}]$
$l o g (1.96) = 0.292, l o g (0.1) = - 1$