What is the ph of a 0.40 m h2se solution that has the stepwise dissociation constants k a1 = 1.3 × 10-4 and k a2 = 1.0 × 10-11?

1 answer:

3 0

First, we can neglect the Ka2 value and use Ka1:

according to this equation and by using the ICE table:

H2Se ⇄ H+ +HSe-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X

so,

Ka1 = [H+][HSe-] / [H2Se]

so by substitution:

1.3 x 10^-4 = X*X / (0.4 -X) by solving for X

∴X = 0.00715

∴[H+] = 0.00715 m

∴PH = -㏒[H+]
= -㏒ 0.00715
= 2.15

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For the following reaction in aqueous solution, identify all those species that will be spectator ions. Select all that apply.

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The reaction is incorrect. The correct reaction is: Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃. The options are:

A. Hg₂SO₄

B. Na₂SO₄

C. Na⁺

D. NO₃⁻

E. SO₄⁻²

F. Hg₂(NO₃)₂

G. NaNO₃

H. Hg⁺²

Answer:

C, D, E, H

Explanation:

The ions are formed after dissociation of the compound in the solution. When they're negatively charged, they're called anions, when they're negatively charged, they're called cations. If the ion is presented on both sides of the reaction, it is called a spectator ion.

Thus, the reaction given:

Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃

So, let's do the dissociation.

Na₂SO₄ is formed by the ions Na⁺ and SO₄⁻²;

Hg(NO₃)₂ is formed by the ions Hg⁺² and NO₃⁻;

HgSO₄ is formed by the ions Hg⁺² and SO₄⁻²;

NaNO₃ is formed by the ions Na⁺ and NO₃⁻.

Thus, the spectator ions are Na⁺, SO₄⁻², Hg⁺², and NO₃⁻.

7 0

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Change for a liquid state to a solid state is known as what phase change?

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