NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
pOH=-lg[OH⁻]
pOH = 14-pH
14-pH = -lg[OH⁻]
[OH⁻]=10^(pH-14)
[OH⁻]=10^(11.8-14)=0.00631 = 6.31·10⁻³ mol/L
Answer:
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺
Explanation:
This is an easy redox reaction:
Br₂ + 2KI → I₂ + 2KBr
We determine the oxidation states.
0 for the elements at ground state.
K does not change the oxidation state during the reaction.
Bromine is reduced to bromide (oxidation state decreases)
and iodide is oxidized to Iodine (oxidation state increases)
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
In oxidation, electrons are released while in reduction, the electrons are gained. To make the ionic equation, we just add K⁺
So we sum both reactions
Br₂ + 2e⁻ + 2I⁻ + 2K⁺ ⇄ 2e⁻ + I₂ + 2Br⁻ + 2K⁺
We cancel the electrons on both sides of the equation:
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺
Just like how heat moves from a region of higher
temperature to a region of lower temperature, molecules also tend to move from
a region of higher concentration to a region of lower concentration. This is
called natural diffusion and is naturally happening to reach stability.
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The molar solubility is 7.4× M and the solubility is 7.4× g/L .
Calculation ,
The dissociation of silver bromide is given as ,
→ +
S
- S S
Ksp = [ ] [ ] = [S] [ S ] =
S = √ Ksp = √ 5. 5× = 7.4×
The solubility =7.4× g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of is dissociates and form one mole of each and ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4× mol/L = 7.4× M
To learn more about molar solubility ,
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