Question

What is the poh of 0.02 M Ca(OH)2

Answer 1

.for calcium hydroxide we got [HO−]=0.04⋅mol⋅L−1 ... ...and so pOH=−log10(0.04)=−(−1.40)=1.40.

please mark me brain mark list

Answer 2

Answer:

Explanation:

About $pOH$:

The concentration of $OH$ ions is used to determine the acidity or alkalinity of a solution. The negative logarithm of the$OH$ion concentration is $pOH$. $pOH = - log [OH-]$

Given:

The molecule is $pOH$ of a $0.02 M$ of $Ca(OH)_{2}$

Find:

Let us find the value of $pOH$of $0.02 M$ of $Ca(OH)_{2}$.

Solution:

Consider that

$p O H=-\log _{10}\left[H O^{-}\right]$

Here for the calcium hydroxide we got $\left[\mathrm{HO}^{-}\right]=0.04 \cdot \mathrm{mol} \cdot \mathrm{L}^{-1} \ldots$

and so $p O H=-\log _{10}(0.04)=-(-1.40)=1.40 \ldots$

Therefore the resultant value is $1.40$.

#SPJ3