# Question #27dc1

May 22, 2017

You add the acid to the water, and never the reverse..........

#### Explanation:

We want a $1 \cdot L$ volume of $H C l$ solution in water (of course I need to make these assumptions in order to answer the question).

By definition $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

And thus $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

And so I need to take a $1 \cdot m L$ volume of $H C l$, whose concentration is $1.0 \cdot m o l \cdot {L}^{-} 1$, and add this a $1 \cdot L$ volumetric flask full of $0.9 \cdot L$ water, and then make the volume to $1 \cdot L$.

$\text{Concentration"="Moles of solute"/"Volume of solution}$

$= \frac{1.0 \times {10}^{-} 3 \cdot L \times 1 \cdot m o l \cdot {L}^{-} 1}{1 \cdot L} = {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$.

$p H = - {\log}_{10} \left({10}^{-} 3\right) = - \left(- 3\right) = + 3$, as required..................