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Alik [6]
3 years ago
11

For the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression

is:_______a. K = [H+][NO2-]/[HNO2] b. K = [H+][N3+][O2-]^2/ [HNO2]c. K= [H+][NO2-] d. K = [H+]^2[NO-]/[HNO2]
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:

a. K = [H+][NO2-]/[HNO2]

Explanation:

The computation of the expression of the equilibrium constant is shown below:

Given that the weak acid is HNO_2 that exist in the solution that aqueous

The dissociation equation of HNO_2

HNO_2(aq) \rightarrow NO_{2}^{-} (aq) + H^{+}(aq)

Now

Acidionization constant i.e.

k_a = \frac{[NO_{2}^{-}][H^{+}]}{HNO_{2}}

Therefore the correct option is a.

Hence, the same is to be considered

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

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