Question

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

Answer 1

Answer:

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Explanation:

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Answer 2

The $pOH$ of the aqueous solution $0.073 M$ $LiOH$ is $1.14$

Explanation:

• The $pOH$ (or potential $OH$) is a measure of a solution's basicity or alkalinity, and it signifies the quantity of ion hydroxide present ($OH^-$).
• The logarithm of the concentration of $OH$ ions is given as $pOH$, with the sign reversed:
•   $pOH= - log [OH^-]$  
• A strong base, on the other hand, is one that entirely disintegrates seen between ion and indeed the $OH^-$in an aqueous medium.
• Because $LiOH$ is a strong base, the hydroxide proportion will be the same as the$OH^-$ concentration.
• This is:
• $[LiOH]=[OH^-]=0.073M$
• In the definition of $pOH$ , the following should be substituted:
• $pOH= -log (0.073)$
• $pOH= 1.14$