# What is the resultant pOH when a 5*mL volume of NaOH of 0.165*mol*L^-1 concentration is diluted to a 450*mL volume?

May 13, 2017

$p O H = 2.74$

#### Explanation:

$p O H = - {\log}_{10} \left[H {O}^{-}\right]$; i.e. it is the basic counterpart to the $p H$ scale.

And so we find $\left[H {O}^{-}\right]$ of the given solution...........

$= \frac{5 \times {10}^{-} 3 \cdot L \times 0.165 \cdot m o l \cdot {L}^{-} 1}{0.450 \cdot L} = 1.83 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1.$

And so $p O H = - {\log}_{10} \left(1.83 \times {10}^{-} 3\right) = 2.74$.

And further we know that $p H + p O H = 14$ for aqueous solution......You should be able to tell $p H$ of this solution $\text{pdq}$. What is $p H$ here?