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ki77a [65]
3 years ago
13

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

pe your answer using the format [NH4]+ for NH4+ and [Ni(CN)4]2- for Ni(CN)42-. Use the lowest possible coefficients.)
(A) HC2H3O2
HC2H3O2(aq) (twosidedarrow) [ ]H+(aq) + [ ] [ ](aq)
Ka = [ ][ ] / [ ]
(B) Co(H2O)63+
Co(H2O)63+(aq) (twosidedarrow) [ ]H+(aq) +[ ][ ](aq)
Ka = [ ][ ] / [ ]
(C) CH3NH3+
CH3NH3+(aq) (twosidedarrow) [ ]H+(aq) +[ ][ ](aq)
Ka = [ ][ ] / [ ]
Chemistry
1 answer:
anastassius [24]3 years ago
3 0

Answer :

(A) The dissociation reaction of HC_2H_3O_2 will be:

HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression :

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression :

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

(C) The dissociation reaction of CH_3NH_3^+ will be:

CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression :

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of HC_2H_3O_2 will be:

HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression of HC_2H_3O_2 will be:

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression of Co(H_2O)_6^{3+} will be:

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

(C) The dissociation reaction of CH_3NH_3^+ will be:

CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression of CH_3NH_3^+ will be:

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

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Answer:

case1.

The addition of acid and base leads to a change in pH of the water when adding to deionized water due to fact that acid and bases dissociated in dissolving in water. If the H+ ion increases in the water as acid addition hikes it, it will result in decreasing the pH value. The intensity of the acid also affects the dissociation of the ions.

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To answer this question, you will need to write the balanced equation and set up a BCA table. Using appropriate rounding rules,
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A  0             0.065

Explanation:

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number of moles can be known by putting the values in the formula,

n = \frac{2.5}{18}

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so, 0.13 moles of water will give x moles of oxygen on decompsition

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x = 0.065 moles of oxygen will be formed.

moles to gram will be calculated as

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