What is the aq. ammonia concentration of a solution prepared by dissolving 0.15 mole of NH 4 + CH 3 COO − in 1 L H 2 O ⋅ [ K a ( CH 3 COOH ) = 1.8 × 1 0 − 5 ; K b ( NH 4 OH ) = 1.8 × 1 0

Question

What is the aq. ammonia concentration of a solution prepared by dissolving 0.15 mole of NH 4 + ​ CH 3 ​ COO − in 1 L H 2 ​ O ⋅ [ K a ( CH 3 ​ COOH ) ​ = 1.8 × 1 0 − 5 ; K b ( NH 4 ​ OH ) ​ = 1.8 × 1 0 − 5 ]

TestHub · Accepted Answer

The problem involves calculating the concentration of aqueous ammonia in a solution of ammonium acetate. Ammonium acetate hydrolyzes in water, and since $K_{a}$ of acetic acid equals $K_{b}$ of ammonia, the solution's pH is determined by the hydrolysis.

Hydrolysis reaction:

$C H_{3} CO O^{-} (a q) + N H_{4}^{+} (a q) + H_{2} O (l) ⇌ C H_{3} COO H (a q) + N H_{4} O H (a q)$

Hydrolysis constant:

$K_{h} = \frac{K _{w}}{K _{a} K _{b}} = \frac{[ C H _{3} COO H ] [ N H _{4} O H ]}{[ C H _{3} CO O ^{-} ] [ N H _{4}^{+} ]}$

Since $K_{a} = K_{b}$ , the degree of hydrolysis is:

$K_{h} = \frac{K _{w}}{K _{a}^{2}} = \frac{α ^{2}}{( 1 - α ) ^{2}}$

$α = \frac{K _{w} / K _{a}}{1 + K _{w} / K _{a}} \approx \frac{K _{w}}{K _{a}} = \frac{1 0 ^{- 7}}{1.8 \times 1 0 ^{- 5}} = 0.0055$

$[N H_{4} O H] = C α = 0.15 \times 0.0055 = 8.3 \times 1 0^{- 4} M$

Therefore, the concentration of aqueous ammonia is $8.3 \times 1 0^{- 4} M$ .

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