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sveticcg [70]
2 years ago
8

A 45.0 mL solution of 0.0450 M hydroxylamine is extracted with 125 mL of solvent. The distribution constant for the reaction is

5.00 and the pKa of the protonated form of hydroxylamine is 5.960. Calculate the concentration of hydroxylamine remaining in the aqueous phase at pH=4.50 and pH=6.50 .
Chemistry
1 answer:
julia-pushkina [17]2 years ago
8 0

Answer:

pH = 4.5, concentration = 0.045 M.

pH = 6.5, concentration = 0.175 M.

Explanation:

The ka for the can be calculated by using the formula below;

Ka = 10^-pka = 10^-5.960 = 1.1 × 10^-6

The concentration of hydrogen ion at pH = 4.50 can be calculated as given below;

{H^+ } = 10^-4.50 = 3.2 × 10^-5 M.

(NB=> 10 in this regards means the inverse of log).

The next step is to determine the distribution coefficient which can be calculated by using the formula below;

distribution coefficient = (partition coefficient) × ka / ka + ( concentration of Hydrogen ion,H^+).

distribution coefficient =( 5 × 1.1 × 10^-6 ) / 1.1 × 10^-6 + 3.2 × 10^-5 M. = 5.5 × 10^-6/ 3.2 = 0.00000171875

The fraction remaining from the compound = 45.0 mL / 45.0 mL + (0.00000171875 × 125).

= 0.999995.

Thus, the concentration at pH = 4.5 = 0.999995 × 0.0450 M = 0.045 M

(B). pH=6.50, thus the concentration of Hydrogen ion = 10^-6.5 = 3.2 × 10^-7 M.

distribution coefficient = (partition coefficient) × ka / ka + ( concentration of Hydrogen ion,H^+).

distribution coefficient = (5 × 1.1 × 10^-6)/ 1.1 × 10^-6 + 3.2 × 10^-7 M).

distribution coefficient = 5.5 × 10^-6/ 1.42 × 10^-6 = 3.9.

Therefore, the concentration = 3.9 × 0.0450 M = 0.175 M.

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