The pH of an aqueous solution of $1.0 M$ ammonium formate assuming complete dissociation is ______.

[ $p K_{a}$ of formic acid is 3.8 and $p K_{b}$ of ammonia is 4.8]

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Solution

$H C O O N H_{4} + H_{2} O ⟶ H C O O H + N H_{4} O H$

The expression for pH is as follows:
$p H = \frac{1}{2} [p K_{w} + p K_{a} - p K_{b}]$

Substituting the values, we get:

$p H = \frac{14 + 38.8 - 4.8}{2}$
$p H = \frac{13}{2}$

$p H = 6.5 \approx 7$

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1 M

(p K_{a} = 3.8, p K_{b} = 4.8)

({pK}_{a} {of formic acid = 3.8 and pK}_{b} of ammonium hydroxide = 4.8.)

p H

1.0 M

p K_{a} = 3.8

p K_{b} = 4.8

H C O O N H_{4}

p K_{a} (H C O O H) = 3.8 a n d p K_{b} = 4.8

2 \times 10^{- 4}

2 \times 10^{- 5}

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