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Salt of Weak Acid and Weak Base

What will be ...

Question

What will be the $p H$ of $0.1 M$ ${C H}_{3} C O O {N H}_{4}$ ? Dissociation constants of ${C H}_{3} C O O H$ and ${N H}_{4} O H$ are $K_{a} = 1.8 \times 10^{- 5}$ and $K_{b} = 1.8 \times 10^{- 5}$ respectively.

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Solution

The salt ${C H}_{3} {C O O N H}_{4}$ is formed from weak acid ${C H}_{3} C O O H$ and weak acid base ${N H}_{4} O H$ .
${C H}_{3} {C O O N H}_{4} + H_{2} O ⇌ {C H}_{3} C O O H + {N H}_{4} O H$
$p H = \frac{1}{2} {p K}_{w} - \frac{1}{2} {p K}_{a} - \frac{1}{2} {p K}_{b}$
$= \frac{1}{2} \times 14 + \frac{1}{2} \times 4 [- l o g 18 \times 10^{- 5}] - \frac{1}{2} [- l o g \times 1.8 \times 10^{- 5}]$
$= 7$

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Similar questions

K_{a}

C H_{3} C O O H

1.8 \times 10^{- 5}

K_{b}

N H_{4} O H

1.8 \times 10^{- 5}

. The pH of ammonium acetate will be:

Q. Calculate the pH of the following mixture given

K_{a} = 1.8 \times 10^{- 5}

K_{b} = 1.8 \times 10^{- 5} (p K_{a} = p K_{a} = 4.7447)

50 m L

0.05 M

N a O H + 50 m L

0.1 M

{C H}_{3} C O O H

Q. Dissociation constants of

C H_{3} C O O H

N H_{4} O H

in aqueous solution are

10^{- 5}

C H_{3} C O O H

3

. What will be the pH of

N H_{4} O H

?

%

0.1 M C H_{3} C O O N H_{4}

K_{a} = K_{b} = 1.8 \times 10^{- 5}

Calculate the e.m.f. of cell in Volts.

\begin{matrix} P t_{H_{2}} 1 a t m \end{matrix} ∣ ∣ ∣ \begin{matrix} C H_{3} C O O H 0.1 M \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} N H_{4} O H 0.01 M \end{matrix} ∣ ∣ ∣ \begin{matrix} P t_{H_{2}} 1 a t m \end{matrix}

K_{a}

C H_{3} C O O H = 1.8 \times 10^{- 5}

K_{b}

N H_{4} O H = 1.8 \times 10^{- 5}

x

, then write as nearest integer to

10 | x |

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Salt of Weak Acid and Weak Base

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