# 1)Calculate the pH of a 0.1M solution of NaOH ,knowing that the volume of the solution is 300mL. 2)Calculate the pH of a 0.3M solution of HCl , knowing that the volume of the solution is 250mL (?)

May 24, 2017

We use the relationship.........$p H + p O H = 14$............in aqueous solution under standard conditions............

#### Explanation:

$\left(i\right)$.........$p H + p O H = 14$............ and by definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$.

For a $0.1 \cdot m o l \cdot {L}^{-} 1$ solution of $N a O H$, we KNOW (do we?) that $H {O}^{-} \equiv 0.1 \cdot m o l \cdot {L}^{-} 1$, and thus $p O H = - {\log}_{10} \left(0.1\right) = - \left(- 1\right) = 1$.

And given $p H + p O H = 14$, $p H = 14 - p O H = 14 - 1 = 13$. The volume of the solution is irrelevant here.

$\left(i i\right)$.........$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(0.300\right) = 0.523$. Again the volume of the solution is a distractor............