Question

Write the equation for the reaction associated with the ka2 of sulfuric acid h2so4

Answer 1

The reaction associated with ${{\text{K}}_{{{\text{a}}_{\text{2}}}}}$ of sulphuric acid $\left( {{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}} \right)$ is as follows:

$\boxed{{\text{HSO}}_4^ - \rightleftharpoons {{\text{H}}^ + } + {\text{SO}}_4^{2 - }}$

Further explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{P(g)}} + {\text{Q(g)}} \rightleftharpoons {\text{R(g)}} + {\text{S(g)}}$

Equilibrium constant is the constant that relates to the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:

${\text{K}} = \frac{{\left[ {\text{R}} \right]\left[ {\text{S}} \right]}}{{\left[ {\text{P}} \right]\left[ {\text{Q}} \right]}}$

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is a dibasic strong acid. Its first dissociation occurs as follows:

${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons {{\text{H}}^ + } + {\text{HSO}}_4^ -$

Here, ${{\text{K}}_{{{\text{a}}_{\text{2}}}}}$ is the equilibrium constant.

Now, ${\text{HSO}}_4^ -$ further dissociates and its dissociation occurs as follows:

${\text{HSO}}_4^ - \rightleftharpoons {{\text{H}}^ + } + {\text{SO}}_4^{2 - }$

The expression for ${{\text{K}}_{{{\text{a}}_{\text{2}}}}}$ is given as follows:

$\boxed{{{\text{K}}_{{{\text{a}}_{\text{2}}}}} = \frac{{\left[ {{{\text{H}}^ + }} \right]\left[ {{\text{SO}}_4^{2 - }} \right]}}{{\left[ {{\text{HSO}}_4^ - } \right]}}}$

Here,

${{\text{K}}_{{{\text{a}}_{\text{2}}}}}$ is the equilibrium constant.

$\left[ {{{\text{H}}^ + }} \right]$ is the concentration of hydrogen ion.

$\left[ {{\text{SO}}_4^{2 - }} \right]$ is the concentration of sulfate ions.

$\left[ {{\text{HSO}}_4^ - } \right]$ is the concentration of hydrogen sulfate ion.

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$(aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: H2SO4, HSO4-, SO42-, H+, equilibrium constant, P, Q, R, S, reactants, products, K, strong acid, dibasic, dissociation, chemical equilibrium, Ka1, Ka2.

Answer 2

Chemical reaction: HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq);
or HSO₄⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₄²⁻(aq)
Ka₂ = [H⁺]·[SO₄²⁻]/[HSO₄⁻]; Ka₂ = 1.0·10⁻².
[H⁺] is equilibrium concentration of hydrogen cations.
[SO₄²⁻] is equilibrium concentration of sulfate anions.
[HSO₄⁻] is equilibrium concentration of hydrogensulfate(bisulfate) anions.