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Helga [31]
2 years ago
12

Calculate the pOH of an aqueous solution of .0.073 M LiOH

Chemistry
1 answer:
Novosadov [1.4K]2 years ago
7 0

Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.

LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:

[LiOH]= [OH-]= 0.073 M

Replacing in the definition of pOH:

pOH= -log (0.073 M)

<u><em>pOH= 1.14 </em></u>

In summary, the pOH of the aqueous solution is 1.14

Learn more:

  • <u>brainly.com/question/16032912?referrer=searchResults </u>
  • <u>brainly.com/question/13557815?referrer=searchResults</u>
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What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
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Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

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Now put all the given values in this formula, we get

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\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

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