Question

Assuming the first step of dissociation to be complete, find the concentration of all species in a 0.1 M of $\rm H_2SO_4$
If $\rm K_2 = 1.2 \times 10^{-2}$

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Answer 1

Answer :

★ First step of dissociation :

$\dag\:\boxed{\bf{H_2SO_4\:\longrightarrow\:H^+ + HSO_4^-}}$

At t = 0 :

conc. of H₂SO₄ = 0.1 M

conc. of $\sf{H^+}$ = 0

conc. of $\sf{HSO_4^-}$ = 0

At t = t :

Since H₂SO₄ is a strong acid, it will completely dissociate into $\sf{H^+}$ and $\sf{HSO_4^-}$.

conc. of H₂SO₄ = 0

conc. of $\sf{HSO_4^-}$ = 0.1M

conc. of $\sf{H^+}$ = 0.1M

★ Second step of dissociation :

$\dag\:\boxed{\bf{HSO_4^-\:\longleftrightarrow\:H^+ + SO_4^{-2}}}$

At t = 0 :

conc. of $\sf{HSO_4^-}$ = 0.1M

conc. of $\sf{H^+}$ = 0.1M

conc. of $\sf{SO_4^{-2}}$ = 0

At equilibrium :

conc. of $\sf{HSO_4^-}$ = 0.1 - x

conc. of $\sf{H^+}$ = 0.1 + x

conc. of $\sf{SO_4^{-2}}$ = x

Equilibrium constant of reaction :

$\leadsto\sf\:k=\dfrac{[H^+][SO_4^{-2}]}{[HSO_4^+]}$

$\leadsto\sf\:1.2\times 10^{-2}=\dfrac{(0.1+x)x}{(0.1-x)}$

$\leadsto\sf\:0.012=\dfrac{0.1x+x^2}{(0.1-x)}$

$\leadsto\sf\:0.0012-0.012x=0.1x+x^2$

$\leadsto\sf\:x^2+0.112x-0.0012=0$

On solving we get,

x = 0.00985 or x = -0.12185

We know that, x can't be negative.

∴ At equilibrium :

1) conc. of $\sf{HSO_4^-}$ :

• 0.1 - x
• 0.1 - 0.00985
• 0.09 M

2) conc. of $\sf{H^+}$ :

• 0.1 + x
• 0.1 + 0.00985
• 0.109 M

3) conc. of $\sf{SO_4^{-2}}$ :

• x
• 0.00985 M